Naked Science Forum
On the Lighter Side => That CAN'T be true! => Topic started by: alright1234 on 03/05/2019 18:22:28

I was talk to a friend at Harvard and I came up with this small calculation. If you find fault or know the exact efficiency of a rocket engine that would be nice. I had to scientifically theorize (guess) based on the numerous different values.
It is not physically possible for the Apollo 11 mission to land on the moon and install the lunar reflector on the surface of the moon. The amount of fuel required to decelerate the Apollo 11 Command/Service Module (CSM) and Lander (L) after reaching the moon is calculated. The kinetic energy of Apollo 11 commandservice module and lander (CSML) that is propagating to the moon is calculated using the distance to the moon (363,104,000 m) and the time that the Apollo 11 space craft (CSML) propagated to the moon (4 days 6 hours and 45 minutes [364,900 seconds]),
v = (distance)/(time) = (363,104,000 m)/(364,900 s) = 983 m/s.......................................................85
The total mass of the command, service modules and lander (CSML) is,
(CM) + (SM) + (L) = 5,560 kg + 24,520 kg + 16,400 kg = 49,480 kg................................................86
Using equations 85 an 86, the kinetic energy of the Apollo 11 CSML is calculated,
1/2 mv2 = (.5)(49,480 kg)(983 m/s)2 = 2.39 x 1010 J.........................................................................87
Using the kinetic energy of the CSML (equ 87) and the energy of a kilogram of rocket fuel (4.2 x 107 J/kg), the minimum amount of fuel required to decelerate the CSML is calculated,
Fuel mass = (KE)/(fuel energy) = (2.39 x 1010 J)/(4.2 x 107 J/kg) = 569 kg...................................88
Using the rocket engine efficiency of 1% and the result of equation 88,
(569 kg)/X = .01 > X = 56,900 kg......................................................................................89
It would require approximately 50,000 kg of fuel (equ 89) to decelerate the 49,480 kg CSML after reaching the moon to allow the CSML to orbit the moon. On the return trip back to the earth using the velocity of 983 m/s and CSM weight of 30,000 kg, less the lander weight, it would require an additional 19,500 kg of fuel to accelerate the CSM for the return trip back to the earth and an additional 15,000 kg of fuel to decelerate the CSM at the earth which represents a total fuel load of approximately 100,000 kg yet according to NASA the CSM contains a total fuel load of 18,410 kg.
https://en.wikipedia.org/wiki/Apollo_Lunar_Module

"Does the Apollo 11 space craft contain the fuel required for the moon mission."
Yes.
Because they did.
I had to scientifically theorize (guess)
And the outcome of your guesswork is that you say that something which happened is impossible.
Either your guess is wrong, or your reasoning is wrong or you have forgotten some other factor..
Come back when you work out which.
In the meantime, here's a hint.
It is impossible for a ball to fall down stairs because the ball carries no fuel; true or false?

Yes, Apollo 11 did contain enough fuel because it actually made it there. That much is obvious.
Here are some things that you didn't take into consideration with your calculations:
(1) The spacecraft did not decelerate to zero velocity when it reached the Moon. Instead, it entered orbit around the Moon. That requires it to still be moving.
(2) The Moon itself is moving, which means you'd have to subtract its relative velocity from that of the spacecraft.
(3) A spacecraft becomes lighter as it burns fuel, thus requiring less and less fuel to slow it down.
(4) The spacecraft didn't travel in a straight line.
I had to scientifically theorize (guess)
Scientifically theorizing and guessing are not the same thing.

I by no means have any familiarity with the figures involved.
v = (distance)/(time) = (363,104,000 m)/(364,900 s) = 983 m/s
This assumes constant velocity the whole distance, which is unrealistic for an object in freefall in a gravitational field.
It was faster at first and slowed considerably along the way.
The total mass of the command, service modules and lander (CSML) is,
(CM) + (SM) + (L) = 5,560 kg + 24,520 kg + 16,400 kg = 49,480 kg
The SM figure is significantly variable as fuel is consumed. Nowhere is it stated when this figure is meaningful.
Using equations 85 an 86, the kinetic energy of the Apollo 11 CSML is calculated,
1/2 mv2 = (.5)(49,480 kg)(983 m/s)2 = 2.39 x 1010 J
Since it slows (and gets lighter) along the way, it does not need to discard most of this energy. It does not need to come to a stop, so not all the energy is discarded. It took a path to the inside of the moon's motion, which uses gravitational braking similar to swinging to the outside gets a gravitation slingshot acceleration for free. You seem to be taking none of this into account.
Using the rocket engine efficiency of 1%
From where is this taken? What exactly does it measure? A rocket efficiency has to do with the relative velocity of the reaction mass, and that figure isn't measured as a percentage of anything.

I was talk to a friend at Harvard and I came up with this small calculation. If you find fault or know the exact efficiency of a rocket engine that would be nice. I had to scientifically theorize (guess) based on the numerous different values.
It is not physically possible for the Apollo 11 mission to land on the moon and install the lunar reflector on the surface of the moon. The amount of fuel required to decelerate the Apollo 11 Command/Service Module (CSM) and Lander (L) after reaching the moon is calculated. The kinetic energy of Apollo 11 commandservice module and lander (CSML) that is propagating to the moon is calculated using the distance to the moon (363,104,000 m) and the time that the Apollo 11 space craft (CSML) propagated to the moon (4 days 6 hours and 45 minutes [364,900 seconds]),
v = (distance)/(time) = (363,104,000 m)/(364,900 s) = 983 m/s.......................................................85
The total mass of the command, service modules and lander (CSML) is,
(CM) + (SM) + (L) = 5,560 kg + 24,520 kg + 16,400 kg = 49,480 kg................................................86
Using equations 85 an 86, the kinetic energy of the Apollo 11 CSML is calculated,
1/2 mv2 = (.5)(49,480 kg)(983 m/s)2 = 2.39 x 1010 J.........................................................................87
Using the kinetic energy of the CSML (equ 87) and the energy of a kilogram of rocket fuel (4.2 x 107 J/kg), the minimum amount of fuel required to decelerate the CSML is calculated,
Fuel mass = (KE)/(fuel energy) = (2.39 x 1010 J)/(4.2 x 107 J/kg) = 569 kg...................................88
Using the rocket engine efficiency of 1% and the result of equation 88,
(569 kg)/X = .01 > X = 56,900 kg......................................................................................89
It would require approximately 50,000 kg of fuel (equ 89) to decelerate the 49,480 kg CSML after reaching the moon to allow the CSML to orbit the moon. On the return trip back to the earth using the velocity of 983 m/s and CSM weight of 30,000 kg, less the lander weight, it would require an additional 19,500 kg of fuel to accelerate the CSM for the return trip back to the earth and an additional 15,000 kg of fuel to decelerate the CSM at the earth which represents a total fuel load of approximately 100,000 kg yet according to NASA the CSM contains a total fuel load of 18,410 kg.
https://en.wikipedia.org/wiki/Apollo_Lunar_Module
No. You are basing your argument on a complete misunderstanding of the orbital mechanics involved. As point out by others, the Apollo craft did not maintain a constant velocity throughout it's trajectory. It lost velocity as it climbed away from the Earth ( just like a ball, tossed into the air loses velocity as it climbs into the air. Apollo 11 was put on a "free" return trajectory. This was a trajectory, which if it turned out to be needed would use the Moon's gravity to whip the craft around and return it to the Earth without the need to use its engines. All the craft had to do to enter Moon orbit when reaching the Moon was to kill the difference between this trajectory and Moon orbit trajectory. This worked out to be about 138 m/s of delta v.
Do you honestly believe that NASA would give a fuel amount for the craft that could so easily be shown to be insufficient? The USSR would have jumped on that discrepancy in a flash. They would have immediately been shouting it out from the mountain tops. If they thought that they could have even cast a doubt on whether the US actually landed on the Moon, they would have tried to. The fact that they didn't shows that they knew that they would looked foolish in the court of world opinion if they had tried.

I was talk to a friend at Harvard
Even if you had written that correctly, it wouldn't impress us.
That's one of the differences between a science discussion and many of the other fora on the web.
In some places "Harvard" would impress people.
But using it here just shows us that you can't spot a logical fallacy.
https://en.wikipedia.org/wiki/Argument_from_authority

My other Harvard story
Walmart, Boston MA. Bloke in the "10 items or less" checkout queue, with 15 items in his basket. Cashier says "Are you from Harvard and can't count, or from MIT and can't read?"
Now that's an argument from authority.
And rocket engines generally work in the 80  100% efficiency range, down to 20% when the vehicle is stationary. The science is about 4000 years old, though the liquid fuel technology is closer to 100.
Not that NASA is incapable of making mistakes. IIRC a Mars lander was programmed with altitude in meters (probably by scientists) but the sensor was calibrated (probably by aviators) in feet.

Engine efficiency 1 percent. If you raise this to ten percent, they only needed 10000kg, and had 8000 spare, which they didnt, spacecraft come back pretty empty.

A jet engine has an efficiency of 41%. An air bus 340 carries 300 passengers (20,000 kg) and contains 100,000 kg of fuel and has a range of 7,000 miles. The efficiency of a rocket is approximately 1% since the Saturn rocket has a payload of 50,000 kg and uses 3,000,000 kg of fuel and a range of 300 miles. All that smoke (lots and lots of rocket smoke) is unburnt fuel that reduces the efficiency. Do you have a link to the rocket efficiency greater than 1%.

A jet engine has an efficiency of 41%. An air bus 340 carries 300 passengers (20,000 kg) and contains 100,000 kg of fuel and has a range of 7,000 miles. The efficiency of a rocket is approximately 1% since the Saturn rocket has a payload of 50,000 kg and uses 3,000,000 kg of fuel and a range of 300 miles. All that smoke (lots and lots of rocket smoke) is unburnt fuel that reduces the efficiency. Do you have a link to the rocket efficiency greater than 1%.
Rocket engines do not have any one number that represents their efficiency. The efficiency changes depending on the speed and altitude of the rocket. Here is a page from NASA with a graph at the bottom. It shows that a rocket engine has an efficiency above 40% at a velocity near 9,000 kilometers per hour: https://history.nasa.gov/SP4404/appb4.htm
If you want to calculate the propulsive efficiency of a rocket engine, this may help: https://en.wikipedia.org/wiki/Propulsive_efficiency#Rocket_engines

A jet engine has an efficiency of 41%. An air bus 340 carries 300 passengers (20,000 kg) and contains 100,000 kg of fuel and has a range of 7,000 miles.
The efficiency of a jet engine is not a measure of its payloaddistance per fuel consumption. By that measure, an engine might achieve well over 100 efficiency.
I don't know where the 41% figure comes from, but I suspect it represents the chemical energy of the fuel converted to workable energy and not wasted as heat. That's a guess, but such a figure could in theory be applied to a rocket, even if better rockets had worse efficiency ratings because they're not chemical.
OK, a rocket used to put something into Earth orbit from a standstill on the ground has an efficiency rating of payload/mass. It has nothing to do with range. It is a percentage of total vehicle weight to the percentage of that weight that is payload, not vehicle or fuel. That is a percentage, and cannot be higher than 100%.
If the Saturn rocket numbers you gave are accurate, that efficiency is closer to 2%, but then you have to multiply that figure by its reliability, and that might reduce the number to closer to 1%.
The efficiency of a rocket is approximately 1% since the Saturn rocket has a payload of 50,000 kg and uses 3,000,000 kg of fuel and a range of 300 miles.
A child's toy water rocket could propel that same payload considerably further than 300 miles, so this is a misrepresentation. It's space. Any force at all will accelerate a payload to some speed and it will continue to move forever until a different force stops it again. So the toy rocket will not get the payload there fast, but it will get it there. It will not get any payload into orbit, and doing that seems to be what you were talking about.
A payload can be put into orbit without a vehicle. How would the efficiency of some sort of ballistic rail gun be measured? 100% I suppose if it's all payload without any container or something.
Rocket engines do not have any one number that represents their efficiency.
There is sort of one obvious one. I mentioned it in my prior post. It just isn't expressed as a percentage of anything since there is no theoretical ceiling to it. A chemical rocket achieves about 5 km/sec reaction mass velocity, and the far more efficient Hall thruster can achieve about 8x that much, but I don't think such an engine can be used to put something in orbit from the ground. For that you need power, not just efficiency. To go long distances, you need efficiency, not power.

All that smoke (lots and lots of rocket smoke) is unburnt fuel that reduces the efficiency.
Rockets don't "smoke" much in the conventional sense of unburned fuel.
The exhaust from a rocket depends on the type but it's likely to be mainly steam.
The SRBs on the shuttle burned a mixture of aluminium and ammonium perchlorate.
https://en.wikipedia.org/wiki/Space_Shuttle_Solid_Rocket_Booster
so some of teh white smoke was (fully burned) aluminium oxide.
It really would be better if you tried to learn a bit about a subject before getting all shouty about it.

A jet engine has an efficiency of 41%. An air bus 340 carries 300 passengers (20,000 kg) and contains 100,000 kg of fuel and has a range of 7,000 miles. The efficiency of a rocket is approximately 1% since the Saturn rocket has a payload of 50,000 kg and uses 3,000,000 kg of fuel and a range of 300 miles. All that smoke (lots and lots of rocket smoke) is unburnt fuel that reduces the efficiency. Do you have a link to the rocket efficiency greater than 1%.
The energy difference between one orbit level and another is classified as Useful Work Done , whilst the energy used to put something in another orbit is however inefficient it is. I do believe that to get the command module and lander into space is about 1 percent, as you are lifting the weight of the rocket, the weight of the lander and all of the fuel needed, so efficiency to put a small lander into orbit is terrible.
1 ton 1000kg 200km 2,000,000 metres (should be 200,000) thanks bored chemist:D
1000x200000x9.81
So for the energy difference between the surface and orbit is 1.962x10^9 joules per tonne
Kerosene has 45x10^9 per tonne.
Whilst not accurate as gravity becomes weaker as you get higher.

Would you like to say where some of the numbers came from?

Just another thought from an engineering point of iew, the saturn 5 rocket was the launch vehicle of the apollo 11 space craft. The mission of apollo 11 began after reaching orbit. So really it has noting to do with saturn 5 lift capabilities.
The extra potential energy is 3.4 MJ/kg
From https://en.m.wikipedia.org/wiki/Specific_orbital_energy
This also includesthe velovity of the iss at a lower orbit i believe
Strange to think that it only takes 23 litres (or 4.5 gallons) of diesil energy to put a tonne in orbit. Hence the horrendous efficiency figures. Once in orbit the actual spacecraft engines are far more efficient
Edit
make that 29 litres, 23 kg

The energy difference between one orbit level and another is classified as Useful Work Done , whilst the energy used to put something in another orbit is however inefficient it is. I do believe that to get the command module and lander into space is about 1 percent, as you are lifting the weight of the rocket, the weight of the lander and all of the fuel needed, so efficiency to put a small lander into orbit is terrible.
1 ton 1000kg 200km 2,000,000 metres
1000x200000x9.81
So for the energy difference between the surface and orbit is 1.962x10^9 joules per tonne
You're off by better than a factor of 10, the answer is closer to 3.15e10^10 joules/ton. It isn't enough to just get that craft to orbital height, it has to be still moving at some 7.79 km/sec when it arrives there in order to be in orbit. That is a lot of KE that also has to be accounted for. In fact, it is the vast majority of the energy that needs to be accounted for.
And even this value assumes that you are able to take the most energy efficient trajectory, ignore air resistance, etc. So in reality, it takes a bit more to actually put that craft into orbit.
The Saturn V had to get 140,000 kg to LEO. ( Service module + Lunar lander + third stage + fuel needed for translunar insertion). From LEO, they had to boost the craft by better than 3km/sec more in order to get to the Moon. This boost was provided by the third stage.

Would you like to say where some of the numbers came from?
In particular, would you like to tell us where 2,000,000 metres came from?

Would you like to say where some of the numbers came from?
In particular, would you like to tell us where 2,000,000 metres came from?
Thanks for that, its a typo in the figures but not the calculation.You're off by better than a factor of 10, the answer is closer to 3.15e10^10 joules/ton. It isn't enough to just get that craft to orbital height, it has to be still moving at some 7.79 km/sec when it arrives there in order to be in orbit. That is a lot of KE that also has to be accounted for. In fact, it is the vast majority of the energy that needs to be accounted for.
And even this value assumes that you are able to take the most energy efficient trajectory, ignore air resistance, etc. So in reality, it takes a bit more to actually put that craft into orbit.
The Saturn V had to get 140,000 kg to LEO. ( Service module + Lunar lander + third stage + fuel needed for translunar insertion). From LEO, they had to boost the craft by better than 3km/sec more in order to get to the Moon. This boost was provided by the third stage.
Well the iss has an energy difference as quoted as 3.4mj kg.
To geo syngc orbit and the average gravitational acceleration should settle it. Its not rocket science you know.

I am astonished at the naivety of some of the correspondents here. Just because you have played with fireworks, seen V2 rockets in museums, studied elementary physics and chemistry, and used rocketlaunched satellites to watch television, these facts cannot possibly outweigh the opinion of a man who someone met, because that man said he had actually walked on the grass at Harvard. Or maybe smoked it.

Would you like to say where some of the numbers came from?
In particular, would you like to tell us where 2,000,000 metres came from?
Thanks for that, its a typo in the figures but not the calculation.You're off by better than a factor of 10, the answer is closer to 3.15e10^10 joules/ton. It isn't enough to just get that craft to orbital height, it has to be still moving at some 7.79 km/sec when it arrives there in order to be in orbit. That is a lot of KE that also has to be accounted for. In fact, it is the vast majority of the energy that needs to be accounted for.
And even this value assumes that you are able to take the most energy efficient trajectory, ignore air resistance, etc. So in reality, it takes a bit more to actually put that craft into orbit.
The Saturn V had to get 140,000 kg to LEO. ( Service module + Lunar lander + third stage + fuel needed for translunar insertion). From LEO, they had to boost the craft by better than 3km/sec more in order to get to the Moon. This boost was provided by the third stage.
Well the iss has an energy difference as quoted as 3.4mj kg.
To geo syngc orbit and the average gravitational acceleration should settle it. Its not rocket science you know.
The math is quite simple:
An object in a circular orbit has an orbital energy of E= um/2r , where u is the gravitational parameter for the body it is orbiting, m is its mass, and r it orbital radius. This is the same as the Sum of its kinetic energy mv^2/2 and its gravitational potential energy um/r or E= mv^2/2u/r . It reduces to the first equation when you substitute sqrt(u/r), the circular orbital for v in the last equation.
For an object sitting on the surface of the Earth, its energy is just E= um/re, where re is the radius of the the Earth.
Thus the energy difference between an object sitting on the surface of the Earth and one in a circular orbit is:
E = um/2r(um/re) = um/reum/2r = um(1/re1/2r)
u = 3.987e14 for the Earth, re= 6378,000m and r = 6578000m
if m=1000kg, then energy difference is
E= 3.987e14(1000)(1/633780001/(2(6578000)) = 3.22e10J
This is a bit more than the value I gave earlier as that value was just for the transfer from surface to LEO, and neglected the boost needed to circularize the orbit.
It's both rocket science and orbital mechanics.

To measure the efficiency of a rocket requires the total payload weight the max velocity and the total fuel. The Saturn rocket efficiency I would guess is about .001 %. Also, the space efficiency of a rocket is estimated at 1%

While your at it tell me what is causing the halve circle path of the CSML and the CSML is propagating with a constant velocity since why would you use more fuel? Also, Space X thought that a rocket engine has a 50% efficiency.
https://www.space.com/spacexcrewdragonexplosionnasamemo.html

The Saturn rocket efficiency I would guess is about .001 %
Don't guess. Show the math.
Also, the space efficiency of a rocket is estimated at 1%
Give us a link to the source.
While your at it tell me what is causing the halve circle path of the CSML
Are you talking about an orbit? Surely you know what causes an orbit.
and the CSML is propagating with a constant velocity since why would you use more fuel?
I'm not sure what you mean by this.
Also, Space X thought that a rocket engine has a 50% efficiency.
Sounds like you are acknowledging that your 1% estimate is wrong. But that is beside the point, since I've already pointed out to you that you can't say "rockets have x% efficiency". You have to know the particulars.

While your at it tell me what is causing the halve circle path of the CSML and the CSML is propagating with a constant velocity since why would you use more fuel? Also, Space X thought that a rocket engine has a 50% efficiency.
If you are asking why why it follows a curved trajectory, then the answer is Earth's and Moon's gravity. Early on the Earth's gravity dominates, but as it moves further away this effect becomes weaker as the Moon's becomes stronger.
The craft would have to burn fuel in order to maintain a constant speed. As it is, it is on a ballistic trajectory and has to give up velocity in exchange for climbing higher in the Earth's gravity field. Upon the beginning of the translunar trajectory it is moving at something over 10.9 km/sec ( not that much shy of escape velocity) . By the time it gets to near Moon space, it will have given up almost all of that speed during the long climb against Earth's gravity.

While your at it tell me what is causing the halve circle path of the CSML and the CSML is propagating with a constant velocity since why would you use more fuel? Also, Space X thought that a rocket engine has a 50% efficiency.
If you are asking why why it follows a curved trajectory, then the answer is Earth's and Moon's gravity.
This statement is patently incorrect since the Earth and Moon gravitational effect is negligible.

The math is quite simple:
An object in a circular orbit has an orbital energy of E= um/2r , where u is the gravitational parameter for the body it is orbiting, m is its mass, and r it orbital radius. This is the same as the Sum of its kinetic energy mv^2/2 and its gravitational potential energy um/r or E= mv^2/2u/r . It reduces to the first equation when you substitute sqrt(u/r), the circular orbital for v in the last equation.
For an object sitting on the surface of the Earth, its energy is just E= um/re, where re is the radius of the the Earth.
Thus the energy difference between an object sitting on the surface of the Earth and one in a circular orbit is:
E = um/2r(um/re) = um/reum/2r = um(1/re1/2r)
u = 3.987e14 for the Earth, re= 6378,000m and r = 6578000m
if m=1000kg, then energy difference is
E= 3.987e14(1000)(1/633780001/(2(6578000)) = 3.22e10J
This is a bit more than the value I gave earlier as that value was just for the transfer from surface to LEO, and neglected the boost needed to circularize the orbit.
It's both rocket science and orbital mechanics.
Well if you do mgh to geosync orbit, at 35000km G=0.75 @~r=20000km you get something like 3x10 per tonne. This still seems a bit high for me, suppose it has to do with the fact that once out of the atmosphere the object already has significant velocity due to the fact it is orbiting at the speed of planet rotation and the planets solar orbit etc etc

The power of the radio signal produce by the Apollo 11 lander is estimated at 50 Watts. A radio signal's intensity is dependent on the inverse of the second order of the distance I = U2 = {[A cos(kr)]/r]}2 where A represents the power of the radio signal. After a radio signal propagates the distance of 238,000 miles (3.8 x 108 m) to the earth, a 50 W radio signal's intensity would diminish to I = A2 [cos2(kr)]/r2 = (50 W)2 (0.5) / (3.8 x 108 m)2 = 8.65 x 1015 W2/m2. The Parkes radio dish antenna located in Sydney Australia was used to communicate with the Apollo 11 mission. The sensitivity of the Parkes radio antenna is extrapolated using the radio signal intensity formed by a 300 km height communication satellite that emits a 20 W radio signal and produces an radio signal intensity at the surface of the earth of I = A2 [cos2(kr)]/r2 = (20 W)2 (.5) / (3 x 105m)2 = 2.22 x 109 W2/m2; adding two orders of magnitude to the satellite radio signal's intensity that forms at the surface of the earth, the sensitivity of the Parkes radio antenna is estimated at 1011 W2/m2. There is a four order of magnitude difference between the extrapolated sensitivity of the Parkes radio telescope and the 1015 W2 /m2 sband radio signal that originates from the Apollo 11 mission. Also, a radio signal cannot penetrate the Van Allen radiation belt that surrounds the earth. It is questionable how NASA communicated with the Apollo missions, Voyagers, and Mars probes using radio waves. It is argue that a satellite that is orbiting the earth at a height of 30,000 km above the earth that is passed the Van Allen belt justifies the functionality of the Apollo 11 communication system but the described satellite height is 10 % of the distance to the moon which is a doubtful magnitude for the height for orbiting communication satellite. NASA uses the Voyager to justify the communication system of the Apollo 11 mission but NASA states that the Voyager is sending back a 50 W radio signal from a distance of over 1 billion miles (1.61 x 1012m) from the earth which would produce a radio signal of I = (50 W)2 (0.5) / (1.61 x 1012 m)2 = 5 x 1022 W2/m2 at the earth that is 11 orders of magnitude less than the extrapolated sensitivity of the Parkes (1011 W2/m2).

Also, a radio signal cannot penetrate the Van Allen radiation belt that surrounds the earth.
Do you have satellite TV?

Also, a radio signal cannot penetrate the Van Allen radiation belt that surrounds the earth.
Do you have satellite TV?
What proof do you have that satellites exist outside the Van Allen belt? NASA?

Your Harvard philosopher (I can't imagine anyone else being quite so ignorant and arrogant  does he teach HSE inspectors?) really ought to study a bit of engineering before pronouncing on the physical impossibility of doing it.

To measure the efficiency of a rocket requires the total payload weight the max velocity and the total fuel. The Saturn rocket efficiency I would guess is about .001 %. Also, the space efficiency of a rocket is estimated at 1%
Efficiency of engine, efficiency of vehicle, efficiency of journey are 3 different things, like a completley efficient engine on a car with terrible aerodynamics trying to drive through a wall. Ie 100% efficient 50% efficient 0% efficient.

Payload to takeoff ratio is primarily a function of the surface gravity of the planet, which is why a much smaller rocket can lift the same payload from the moon than from the earth, and a return trip to Mars is a lot more difficult.
At MIT, and on every fuel truck at every airport, though apparently not at Harvard, they know the difference between engine efficiency and payload ratio.

Efficiency of engine, efficiency of vehicle, efficiency of journey are 3 different things, like a completley efficient engine on a car with terrible aerodynamics trying to drive through a wall. Ie 100% efficient 50% efficient 0% efficient.
Good summary. The list of different things is probably longer than that, and it just hasn't been made clear which ones we're talking about. I was probably talking more about the first one, and many of the others more about the wall. Given enough energy, one can measure the efficiency of driving through a wall by what percentage of the weight of a vehicle manages to make it to the other side of the wall, the rest being left as wreckage behind. That's a fair description of launch efficiency.
I had posed a similar problem once at the dinner table. Suppose a pingpong ball comes in as a meteor from space. How fast would it need to go to take out a submarine at 100m depth? Of course we don't expect the ball to survive intact.

At MIT, and on every fuel truck at every airport, though apparently not at Harvard, they know the difference between engine efficiency and payload ratio.
Payload ratio very much can be measured as a percentage, but what about engine efficiency? What exactly does that measure?
A 100% payload efficiency means the loaded vehicle is the same mass as the payload and can deliver that mass to its destination.
What could a 100% efficient engine do if such a thing were possible? Is it meaningfully expressed as a percentage?

The math is quite simple:
An object in a circular orbit has an orbital energy of E= um/2r , where u is the gravitational parameter for the body it is orbiting, m is its mass, and r it orbital radius. This is the same as the Sum of its kinetic energy mv^2/2 and its gravitational potential energy um/r or E= mv^2/2u/r . It reduces to the first equation when you substitute sqrt(u/r), the circular orbital for v in the last equation.
For an object sitting on the surface of the Earth, its energy is just E= um/re, where re is the radius of the the Earth.
Thus the energy difference between an object sitting on the surface of the Earth and one in a circular orbit is:
E = um/2r(um/re) = um/reum/2r = um(1/re1/2r)
u = 3.987e14 for the Earth, re= 6378,000m and r = 6578000m
if m=1000kg, then energy difference is
E= 3.987e14(1000)(1/633780001/(2(6578000)) = 3.22e10J
This is a bit more than the value I gave earlier as that value was just for the transfer from surface to LEO, and neglected the boost needed to circularize the orbit.
It's both rocket science and orbital mechanics.
Well if you do mgh to geosync orbit, at 35000km G=0.75 @~r=20000km you get something like 3x10 per tonne. This still seems a bit high for me, suppose it has to do with the fact that once out of the atmosphere the object already has significant velocity due to the fact it is orbiting at the speed of planet rotation and the planets solar orbit etc etc
Again, you are neglecting the energy needed for the craft to attain orbital velocity. If you were to assume no obstacles or atmosphere, you could put an object in orbit around the Earth just above its surface, but you would have to accelerate it up to a bit over 7.9 km/sec n order to do so. This, in of itself, would require 3.13e10 joules per 1000 kg of mass. mgh doesn't even come into play because you never raised the object any height in Earth's gravity field.
To attain GEO requires 5.78e10 joules per 1000 kg, 2.56e10 joules per 1000 kg more than the 200 km LEO orbit.

I'm still waiting for you to show the calculations that led you to believe that the Saturn V had an efficiency of 0.001%.
This statement is patently incorrect since the Earth and Moon gravitational effect is negligible.
Are you serious? What do you think holds the Moon in orbit around the Earth and the Earth in orbit around the Sun?
The power of the radio signal produce by the Apollo 11 lander is estimated at 50 Watts. A radio signal's intensity is dependent on the inverse of the second order of the distance I = U2 = {[A cos(kr)]/r]}2 where A represents the power of the radio signal. After a radio signal propagates the distance of 238,000 miles (3.8 x 108 m) to the earth, a 50 W radio signal's intensity would diminish to I = A2 [cos2(kr)]/r2 = (50 W)2 (0.5) / (3.8 x 108 m)2 = 8.65 x 1015 W2/m2.
Those calculations make no sense. If you're trying to calculate the radiation flux at the Earth's surface, that would be measured in W/m^{2}, not W^{2}/m^{2}. If that was merely a typo, then your calculated value is still much too low. In a worst case scenario, the antennae would be an isotropic radiator (which radiates its signal equally in all directions). Since all of the radio waves spread out evenly over a spherical volume, we can calculate the radiation flux at any given distance by dividing the total power by the surface area of the sphere at that distance.
So if we have a 50 watt source and want to know the radiation flux at a distance of 380,000,000 meters, we first calculate the surface area of a sphere with that radius, which is 4πr^{2}:
A = 4π(380,000,000)^{2}
A = 4.54 x 10^{11} square meters
Divide 50 watts by 4.54 x 10^{11} meters, and you get a radiation flux of 1.1013 x 10^{10} watts per square meter. Since actual antennae used in space communications are parabolic, not isotropic, the radiation flux is actually much, much higher. Parabolic antennae focus their signals into a tight beam instead of letting it spread out in all directions equally. That gives them a much, much larger radiation flux through the area they are beaming through. The value used to determine this is called the "antenna gain". When you multiply the antenna gain value by the radiation flux of an isotropic radiator of equal power, you get the actual radiation flux of the antenna.
The antennae used by the Lunar Module to communicate with the Manned Space Flight Network had a gain of 20.5 decibels: https://ntrs.nasa.gov/archive/nasa/casi.ntrs.nasa.gov/19720023255.pdf That is a gain factor of 112.202 (decibels are logarithmic). So if you multiply the radiation flux of 1.1013 x 10^{10} watts per square meter by 112.202, you end up with flux of about 1.2357 x 10^{8} watts per square meter. That is more than 1.4 million times stronger than what your calculations predict.
The Parkes radio dish antenna located in Sydney Australia was used to communicate with the Apollo 11 mission. The sensitivity of the Parkes radio antenna is extrapolated using the radio signal intensity formed by a 300 km height communication satellite that emits a 20 W radio signal and produces an radio signal intensity at the surface of the earth of I = A2 [cos2(kr)]/r2 = (20 W)2 (.5) / (3 x 105m)2 = 2.22 x 109 W2/m2; adding two orders of magnitude to the satellite radio signal's intensity that forms at the surface of the earth, the sensitivity of the Parkes radio antenna is estimated at 1011 W2/m2. There is a four order of magnitude difference between the extrapolated sensitivity of the Parkes radio telescope and the 1015 W2 /m2 sband radio signal that originates from the Apollo 11 mission.
You're making the same mistake as before. You have to take the antennae gain into consideration.
Also, a radio signal cannot penetrate the Van Allen radiation belt that surrounds the earth.
If that was true, then radio astronomy at Earth's surface would be impossible. Yet we have plenty of radio telescopes that work just fine sitting on the Earth's surface.
It is questionable how NASA communicated with the Apollo missions, Voyagers, and Mars probes using radio waves.
The only thing questionable about it is your understanding of the subject matter.
It is argue that a satellite that is orbiting the earth at a height of 30,000 km above the earth that is passed the Van Allen belt justifies the functionality of the Apollo 11 communication system but the described satellite height is 10 % of the distance to the moon which is a doubtful magnitude for the height for orbiting communication satellite. NASA uses the Voyager to justify the communication system of the Apollo 11 mission but NASA states that the Voyager is sending back a 50 W radio signal from a distance of over 1 billion miles (1.61 x 1012m) from the earth which would produce a radio signal of I = (50 W)2 (0.5) / (1.61 x 1012 m)2 = 5 x 1022 W2/m2 at the earth that is 11 orders of magnitude less than the extrapolated sensitivity of the Parkes (1011 W2/m2).
You are, again, ignoring antenna gain.

What could a 100% efficient engine do if such a thing were possible? Is it meaningfully expressed as a percentage?
It is the percentage of engine power input (rate of consumption of fuel x specific energy) available as thrust (expressed as power). Entirely meaningful and very important!

The math is quite simple:
An object in a circular orbit has an orbital energy of E= um/2r , where u is the gravitational parameter for the body it is orbiting, m is its mass, and r it orbital radius. This is the same as the Sum of its kinetic energy mv^2/2 and its gravitational potential energy um/r or E= mv^2/2u/r . It reduces to the first equation when you substitute sqrt(u/r), the circular orbital for v in the last equation.
For an object sitting on the surface of the Earth, its energy is just E= um/re, where re is the radius of the the Earth.
Thus the energy difference between an object sitting on the surface of the Earth and one in a circular orbit is:
E = um/2r(um/re) = um/reum/2r = um(1/re1/2r)
u = 3.987e14 for the Earth, re= 6378,000m and r = 6578000m
if m=1000kg, then energy difference is
E= 3.987e14(1000)(1/633780001/(2(6578000)) = 3.22e10J
This is a bit more than the value I gave earlier as that value was just for the transfer from surface to LEO, and neglected the boost needed to circularize the orbit.
It's both rocket science and orbital mechanics.
Well if you do mgh to geosync orbit, at 35000km G=0.75 @~r=20000km you get something like 3x10 per tonne. This still seems a bit high for me, suppose it has to do with the fact that once out of the atmosphere the object already has significant velocity due to the fact it is orbiting at the speed of planet rotation and the planets solar orbit etc etc
Again, you are neglecting the energy needed for the craft to attain orbital velocity. If you were to assume no obstacles or atmosphere, you could put an object in orbit around the Earth just above its surface, but you would have to accelerate it up to a bit over 7.9 km/sec n order to do so. This, in of itself, would require 3.13e10 joules per 1000 kg of mass. mgh doesn't even come into play because you never raised the object any height in Earth's gravity field.
To attain GEO requires 5.78e10 joules per 1000 kg, 2.56e10 joules per 1000 kg more than the 200 km LEO orbit.
But this, like the mgh discrepancy creates a problem. The energy content of diesil fuel being 4.9x10^10per tonne, halved ish for the oxidiser giving 2.5x10^10 per fuel tonne burned. You would never ever make orbit if it required any x10^10 number due to inefficiency and container weights. Same goes for the mgh to the moon !

I haven't attempted to follow the OP's ravings, but when we did rocket calculations in my schooldays (60 years ago) we remembered that the mass of a rocket decreases  significantly  as it flies. Interestingly, NASA and Roscosmos seem aware of the same basic equations. Also very important for longrange aircraft, whose max landing mass can be less than half the takeoff limit.
This is Olevel stuff in the UK, though clearly beyond the competence of Harvard emeriti

What could a 100% efficient engine do if such a thing were possible? Is it meaningfully expressed as a percentage?
It is the percentage of engine power input (rate of consumption of fuel x specific energy) available as thrust (expressed as power). Entirely meaningful and very important!
OK, that makes sense except for the expressedaspower part. It should be expressed as total work. Power is the rate that the work is performed, which is often quite low for the most efficient engines.
An engine with bit lower efficiency (defined this way) might be able to produce more overall work if it uses fuel with a better specific energy.
A 100% efficient engine then converts all its specific energy to thrust, wasting none to heat and unburned fuel and such. It cannot be 101% efficient, and thus the expression of this metric as a percentage makes sense.
Thanks.

Also, a radio signal cannot penetrate the Van Allen radiation belt that surrounds the earth.
Do you have satellite TV?
What proof do you have that satellites exist outside the Van Allen belt? NASA?
The fact that you can point a dish at them.
That proves that they are geosynchronous.
From that, and the mass and radius of the Earth etc, I can calculate how far up they are.
Of course, if you took the trouble to find out what you're talking about, rather than getting shouty, you would already know that.
So, now that you know you are wrong, are you going to apologise?

While your at it tell me what is causing the halve circle path of the CSML and the CSML is propagating with a constant velocity since why would you use more fuel? Also, Space X thought that a rocket engine has a 50% efficiency.
If you are asking why why it follows a curved trajectory, then the answer is Earth's and Moon's gravity. Early on the Earth's gravity dominates, but as it moves further away this effect becomes weaker as the Moon's becomes stronger.
The craft would have to burn fuel in order to maintain a constant speed. As it is, it is on a ballistic trajectory and has to give up velocity in exchange for climbing higher in the Earth's gravity field. Upon the beginning of the translunar trajectory it is moving at something over 10.9 km/sec ( not that much shy of escape velocity) . By the time it gets to near Moon space, it will have given up almost all of that speed during the long climb against Earth's gravity.
Checkmate king two.

While your at it tell me what is causing the halve circle path of the CSML and the CSML is propagating with a constant velocity since why would you use more fuel? Also, Space X thought that a rocket engine has a 50% efficiency.
If you are asking why why it follows a curved trajectory, then the answer is Earth's and Moon's gravity. Early on the Earth's gravity dominates, but as it moves further away this effect becomes weaker as the Moon's becomes stronger.
The craft would have to burn fuel in order to maintain a constant speed. As it is, it is on a ballistic trajectory and has to give up velocity in exchange for climbing higher in the Earth's gravity field. Upon the beginning of the translunar trajectory it is moving at something over 10.9 km/sec ( not that much shy of escape velocity) . By the time it gets to near Moon space, it will have given up almost all of that speed during the long climb against Earth's gravity.
Checkmate king two.
I think you just knocked your king over or are we playing pidgeon chess?

So, now that you know you are wrong, are you going to apologise?

The math is quite simple:
An object in a circular orbit has an orbital energy of E= um/2r , where u is the gravitational parameter for the body it is orbiting, m is its mass, and r it orbital radius. This is the same as the Sum of its kinetic energy mv^2/2 and its gravitational potential energy um/r or E= mv^2/2u/r . It reduces to the first equation when you substitute sqrt(u/r), the circular orbital for v in the last equation.
For an object sitting on the surface of the Earth, its energy is just E= um/re, where re is the radius of the the Earth.
Thus the energy difference between an object sitting on the surface of the Earth and one in a circular orbit is:
E = um/2r(um/re) = um/reum/2r = um(1/re1/2r)
u = 3.987e14 for the Earth, re= 6378,000m and r = 6578000m
if m=1000kg, then energy difference is
E= 3.987e14(1000)(1/633780001/(2(6578000)) = 3.22e10J
This is a bit more than the value I gave earlier as that value was just for the transfer from surface to LEO, and neglected the boost needed to circularize the orbit.
It's both rocket science and orbital mechanics.
Well if you do mgh to geosync orbit, at 35000km G=0.75 @~r=20000km you get something like 3x10 per tonne. This still seems a bit high for me, suppose it has to do with the fact that once out of the atmosphere the object already has significant velocity due to the fact it is orbiting at the speed of planet rotation and the planets solar orbit etc etc
Again, you are neglecting the energy needed for the craft to attain orbital velocity. If you were to assume no obstacles or atmosphere, you could put an object in orbit around the Earth just above its surface, but you would have to accelerate it up to a bit over 7.9 km/sec n order to do so. This, in of itself, would require 3.13e10 joules per 1000 kg of mass. mgh doesn't even come into play because you never raised the object any height in Earth's gravity field.
To attain GEO requires 5.78e10 joules per 1000 kg, 2.56e10 joules per 1000 kg more than the 200 km LEO orbit.
But this, like the mgh discrepancy creates a problem. The energy content of diesil fuel being 4.9x10^10per tonne, halved ish for the oxidiser giving 2.5x10^10 per fuel tonne burned. You would never ever make orbit if it required any x10^10 number due to inefficiency and container weights. Same goes for the mgh to the moon !
You are making the same mistake as some eminent scientist(I can't remember who)did when trying to claim that a rocket could never escape the Earth. He worked out how much energy would be needed to accelerate 1 kg to escape velocity, and compared to to amount of energy that could be released by 1 kg of Nitroglycerine, the most powerful explosive of the time. He showed that this was less than the energy needed to get the 1 kg of nitroglycerine up to escape velocity. He argued that if nitroglycerine couldn't even get its own mass up to escape velocity, how could you get a rocket up to it?
What he failed to realize is that with a rocket, you don't need to get the entire mass of your fuel up to escape velocity. The vast majority of the fuel mass gets left behind near the surface of the Earth.
For a rocket it is all about having enough thrust to lift your rocket and the exhaust velocity. Exhaust velocity determines what kind of massratio you need in terms of fuel to payload.
It comes down to the rocket equation
dV = Ve ln(MR)
dV is the velocity change of your rocket, Ve is its exhaust velocity, MR is the mass ratio ( Fueled rocket mass/ dry mass of rocket).
In addition, the RP1 rocket fuel (kerosene equivalent) was only used in the first stage of the Saturn V, the remaining stages used liquid Hydrogen, which allow for an ISP of 421 sec vs, the lower ISP of 263 sec for the first stage ( to get exhaust velocity from ISP, multiply it by g.)
Regardless of what you think, this is the actual science behind this.

I calculated the efficiency using the Saturn rocket and got .004%.

I calculated the efficiency using the Saturn rocket and got .004%.
I guess that explains why you don't want to do maths about the Cavendish experiment.

I'm saving it.

You seem to have misspelled "I'm saying sh**"

While your at it tell me what is causing the halve circle path of the CSML and the CSML is propagating with a constant velocity since why would you use more fuel? Also, Space X thought that a rocket engine has a 50% efficiency.
https://www.space.com/spacexcrewdragonexplosionnasamemo.html

I calculated the efficiency using the Saturn rocket and got .004%.
Show the math.
While your at it tell me what is causing the halve circle path of the CSML
The same thing that causes all orbits.

I calculated the efficiency using the Saturn rocket and got .004%.
Show the math.
While your at it tell me what is causing the halve circle path of the CSML
The same thing that causes all orbits.
The earth's gravity has a negligible effect on the 50,000 kg CSML. I'm saving it for a surprise. You like surprise don't you.

The earth's gravity has a negligible effect on the 50,000 kg CSML.
Given that the Earth's gravity can hold onto an object as distant as the Moon for billions of years, this statement is wrong.

Then why are astronaut in the ISS weightless?

Then why are astronaut in the ISS weightless?
The same reason that a ball dropped off of the Empire State Building is weightless: because they are in free fall. The astronauts are constantly falling towards the Earth, but since they are moving parallel to the Earth's surface, the Earth always curves away from them at the same rate that they fall. That is what an orbit is. Isaac Newton figured that out a long time ago:

Why does not a penny in the ISS stick to the walls?

Why does not a penny in the ISS stick to the walls?
There is no reason that it should.

OK, that makes sense except for the expressedaspower part. It should be expressed as total work. Power is the rate that the work is performed, which is often quite low for the most efficient engines.
Which is why I said "rate of consumption of fuel x specific energy" The product is kg/sec x joules/kg = joules/sec, i.e. watts.

Why does not a penny in the ISS stick to the walls?
Because Orbit chewing gum is a terrestrial breath freshener, not astronaut issue.

You are making the same mistake as some eminent scientist(I can't remember who)did when trying to claim that a rocket could never escape the Earth. He worked out how much energy would be needed to accelerate 1 kg to escape velocity, and compared to to amount of energy that could be released by 1 kg of Nitroglycerine, the most powerful explosive of the time. He showed that this was less than the energy needed to get the 1 kg of nitroglycerine up to escape velocity. He argued that if nitroglycerine couldn't even get its own mass up to escape velocity, how could you get a rocket up to it?
What he failed to realize is that with a rocket, you don't need to get the entire mass of your fuel up to escape velocity. The vast majority of the fuel mass gets left behind near the surface of the Earth.
For a rocket it is all about having enough thrust to lift your rocket and the exhaust velocity. Exhaust velocity determines what kind of massratio you need in terms of fuel to payload.
It comes down to the rocket equation
dV = Ve ln(MR)
dV is the velocity change of your rocket, Ve is its exhaust velocity, MR is the mass ratio ( Fueled rocket mass/ dry mass of rocket).
In addition, the RP1 rocket fuel (kerosene equivalent) was only used in the first stage of the Saturn V, the remaining stages used liquid Hydrogen, which allow for an ISP of 421 sec vs, the lower ISP of 263 sec for the first stage ( to get exhaust velocity from ISP, multiply it by g.)
Regardless of what you think, this is the actual science behind this.
Yes, all the fuel does not need to go all the way, but to put it in perspective diesel and oxygen combined have roughly a 2.35x10 to 10 energy content which is an energy dense for weight content. Minus inefficiencys and load, etc. To put a tonne in orbit at 400km requires you to put roughly HALF the fuel at 200km. The energy kinetic has to do with other factors such as orbit of the planet around the sun angular momentum etc. To put 1000kg in 400km range you have to lift enough energy to 200km range to put it the aditional 200km distance between it. Sort of a reverse 0f 9.81ms
The problem of fuel weights is thought of in interstellar travel, the bigger the fuel load the bigger the ammount of fuel needed to be carried.

Then why are astronaut in the ISS weightless?
The same reason that a ball dropped off of the Empire State Building is weightless: because they are in free fall.
What is the velocity of free fall of the ISS? This proves the Newton gravity equation is false since the mass should not be weightless. Right If it is falling (free fall) it must be falling downward which it is not. Pretty good there old chap. I was walking through the halls of Harvard and I thought of this.

What is the velocity of free fall of the ISS?
There is no one velocity of free fall in a vacuum. There is an acceleration, but not a velocity.
This proves the Newton gravity equation is false
No it doesn't. It proves that you have no idea how physics works.
since the mass should not be weightless.
Mass and weight are not the same thing. Try weighing a freefalling object on a freefalling scale in a vacuum. It won't measure anything because both will accelerate at the same rate.

What is the velocity of free fall of the ISS?
The question makes no sense. The acceleration of the ISS is nearly the same as if you jumped off a balcony, about 9 m/secē downward.
Right If it is falling (free fall) it must be falling downward which it is not.
An acceleration downward sounds like falling to me.

While your at it tell me what is causing the halve circle path of the CSML and the CSML is propagating with a constant velocity since why would you use more fuel? Also, Space X thought that a rocket engine has a 50% efficiency.
This entirely depends on how you measure "efficiency".
Let's take the Saturn V third stage for example. It has a "fuel" load of 109500 kg. This consists of both liquid Hydrogen and liquid Oxygen. Of this, it is the Hydrogen that provides the energy. They are burned together to form water( H2O) You use 2 kg of Hydrogen for 16 kg of Oxygen, so the Hydrogen makes up ~12167 kg of the load. At 141.86e6 joules per kg, this gives an available energy of 1.73e12 joules.
The rocket engine has an ISP 421 seconds, which equates to an exhaust velocity of 4130 m/sec. At 4130 m/sec relative to the rocket engine, the total KE of the exhaust gases will be 9.339e 11 joules. 9.33e11/1.73e12 =.54.
So if you rate your engine by what percentage of the energy released by burning the fuel goes into producing velocity for your exhaust and thus generating thrust, you get an efficiency of 54%.

. To put a tonne in orbit at 400km requires you to put roughly HALF the fuel at 200km.
No. The Saturn V first stage only got the rocket up to 67 km of the 170 km LEO altitude. Doing so, it used up ~80% of the total fuel load for the rocket. So it only had to lift 20% of its fuel to under 40% of its targeted orbit height.
Quit trying to guess at the answers.

Then why are astronaut in the ISS weightless?
The same reason that a ball dropped off of the Empire State Building is weightless: because they are in free fall.
What is the velocity of free fall of the ISS? This proves the Newton gravity equation is false since the mass should not be weightless. Right If it is falling (free fall) it must be falling downward which it is not. Pretty good there old chap. I was walking through the halls of Harvard and I thought of this.
It is "falling", in that its trajectory is always being curved to towards the Earth. Acceleration is a change in velocity. Velocity is a vector which includes both magnitude and direction. A body in a circular orbit is constantly changing its direction, but not its speed. This is still an acceleration At the ISS altitude, the acceleration due to gravity is ~8.66m/sec^2. The orbital velocity of the ISS is ~7.67 km/sec. Thus it "falls" ~ 4.33 m as it travels forward 7.67 km. This is a gentle curve*. So gentle that the Earth curves out from under it just as fast. Look up Newton's cannon.
* To a close approximation. The fact that gravity acts towards the center of the Earth, and not just perpendicular to the initial velocity vector of the ISS, alters this value by some.

. To put a tonne in orbit at 400km requires you to put roughly HALF the fuel at 200km.
No. The Saturn V first stage only got the rocket up to 67 km of the 170 km LEO altitude. Doing so, it used up ~80% of the total fuel load for the rocket. So it only had to lift 20% of its fuel to under 40% of its targeted orbit height.
Quit trying to guess at the answers.
Or another way of thinking of it.
It has to lift the craft. The fuel required to go half way requires no extra fuel, but the other half of the fuel has to be lifted half way, thus requiring fuel itself to provide energy to be lifted, this itself requires fuel. anonanonandariston. Parabolic usage.
The bigger the fuel load the larger the coontainment, as was realised by Isombard kingdom brunel, and the coal powered craft he made. The problem is that the Ss Great eastern didnt need to float up a pepetual hill, where as rockets are. The faster you can get them out of the atmosphere the more efficient they become.

Why does not a penny in the ISS stick to the walls?
The mass of the ISS is 420,000 kg. Even if all this mass were compacted into a point, an object held just 5 m away would take 59 min to reach it. That's with all the gravity of the ISS puling in one direction. Something inside the ISS would be surrounded by its mass, the gravity of which would be pulling it in all directions and for the most part canceling itself out. If the ISS were perfectly spherical with wall of uniform density, The pull of gravity would cancel out completely everywhere in the interior. The ISS is a bit more asymmetric, So you are going to likely end up with some net pull in different points of the interior, but it will be extremely small. Left alone for long enough, objects floating inside would eventually find resting spots, but not until a long , long time, But even after that, just turning on the ventilation system would be enough to dislodge them from their resting spots. The gravitational forces involved are just that small.
If you are asking why the penny doesn't stick to the "Earthside" wall, both the penny and ISS are "falling" along the same path due the same gravity. There is no reason for the penny to fall any differently than the ISS.
If you were in an elevator, and ti suddenly started to fall, between the time it started falling and hits the bottom of the shaft, you would float around inside the elevator just like the Astronauts do in the ISS. Both are examples of objects in free fall. You actually experience this partially in an elevator when it first starts downward (more so in fast elevators). As the elevator accelerates to reach full descent speed, there is a brief period were you will feel lighter. The downward acceleration of the elevator decreases weight you feel. If you normally weighed 160 lbs, and the elevator accelerated downward at 1 m/sec^2, then while the elevator was accelerating, you would only "weigh" 143.7 lbs ( this is what a spring or force reading scale would measure if you were standing on it.)

If you are asking why the penny doesn't stick to the "Earthside" wall, both the penny and ISS are "falling" along the same path due the same gravity. There is no reason for the penny to fall any differently than the ISS.
Tidal forces might make the penny tend to head for the near or far side of the ISS over the course of an hour or so, but again, the ventilation system would totally overpower that.

However, energies involved in such an endeavour would make it next to impossible. In a space ship the fuel needed for the later parts of the journey has to be carried aboard and thus also needs to be accelerated till it is utilized. Therefore the initial mass at the start of the journey is much more than the actual payload. The conventional chemical fuel for such an arduous journey will need a fuelmass of a whole galaxy. Even within the best possible scenario where almost 100% of mass is converted into energy (in a typical thermonuclear reaction only about 0.7 % of mass is converted into energy) one would require initial mass to be millions of times the mass of the final payload and the energy which required may be worth 200 years of total energy consumption of the whole world.
From https://arxiv.org/pdf/1308.4869

However, energies involved in such an endeavour would make it next to impossible. In a space ship the fuel needed for the later parts of the journey has to be carried aboard and thus also needs to be accelerated till it is utilized. Therefore the initial mass at the start of the journey is much more than the actual payload. The conventional chemical fuel for such an arduous journey will need a fuelmass of a whole galaxy. Even within the best possible scenario where almost 100% of mass is converted into energy (in a typical thermonuclear reaction only about 0.7 % of mass is converted into energy) one would require initial mass to be millions of times the mass of the final payload and the energy which required may be worth 200 years of total energy consumption of the whole world.
From https://arxiv.org/pdf/1308.4869
Since you did not include the specifics of what "arduous journey" was being contemplated here, its a bit hard to reply directly, but I am going to assume we are talking about an interstellar journey at some significant fraction of c
Fuel to payload ratios are determined by the rocket equation.
dV= Ve ln(MR)
where dV is the velocity change you get for your rocket
Ve is the exhaust velocity
ln(MR) is the natural log of the mass ratio ( fueled mass/dry mass) for the rocket.
If you need dV to be a significant fraction of the speed of light, there is a relativistic version.
To get the Mass ratio for any given dV, you rearrange to
MR= e^(dV/Ve)
where e is Eulers number = 2.718...
That being said, the quoted passage is neither here nor there as far as this thread is concerned, as we are not talking about sending craft to distant stars, but to the Moon. Applying the above equation to the Saturn V rocket shows that it was within its capabilities to do the job for which it was intended.

What is the velocity of free fall of the ISS?
There is no one velocity of free fall in a vacuum. There is an acceleration, but not a velocity.
This proves the Newton gravity equation is false
How can there be an acceleration without a velocity?

Also, I have began my calculation of the efficiency of a rocket using the finial velocity of 938 m/s and weight of 50,000 kg and the rocket fuel energy of 2.4 x 10^7 J/kg. Watson jr. is computing and will produce the final number.

Also, I have began my calculation of the efficiency of a rocket using the finial velocity of 938 m/s and weight of 50,000 kg and the rocket fuel energy of 2.4 x 10^7 J/kg. Watson jr. is computing and will produce the final number.
Why not do it correctly instead?

Show me the correct way.
The total fuel load is
109,500 + 456,100 + 2,160,000 = 2,725,600 kg

The kinetic energy of the CMSL propagating to the moon on a straight path is determined using the velocity of 938 m/s,
1/2 mv2 = (.5)(49,480 kg)(983 m/s)2 = 2.39 x 1010 J........................................................................................1
The energy of 2,725,600 kg of fuel using the energy of a kilogram of rocket fuel (4.2 x 107 J/kg),
Fuel Energy = (mass fuel) x (energy of fuel) = (2.72 x 106 kg)/(4.2 x 107 J/kg) = 1.142 x 1014 J....................2
Divided by kinetic of the CMSL and the energy of the fuel is,
Efficiency = (KE)/(enegy of fuel) = (2.39 x 1010 J)/(1.142 x 1014) = .0002 or .02 %........................................2
Multiply the efficiency by 100 times to normalize the effect of the earth's gravity forms an efficiency of free space of 2 %

Your calculations assume that all of the fuel energy went into accelerating the CMSL alone. It didn't. A significant portion went into accelerating the other stages of the vehicle as well. You need to take that into consideration.

However, energies involved in such an endeavour would make it next to impossible. In a space ship the fuel needed for the later parts of the journey has to be carried aboard and thus also needs to be accelerated till it is utilized. Therefore the initial mass at the start of the journey is much more than the actual payload. The conventional chemical fuel for such an arduous journey will need a fuelmass of a whole galaxy. Even within the best possible scenario where almost 100% of mass is converted into energy (in a typical thermonuclear reaction only about 0.7 % of mass is converted into energy) one would require initial mass to be millions of times the mass of the final payload and the energy which required may be worth 200 years of total energy consumption of the whole world.
From https://arxiv.org/pdf/1308.4869
Since you did not include the specifics of what "arduous journey" was being contemplated here, its a bit hard to reply directly, but I am going to assume we are talking about an interstellar journey at some significant fraction of c
Fuel to payload ratios are determined by the rocket equation.
dV= Ve ln(MR)
where dV is the velocity change you get for your rocket
Ve is the exhaust velocity
ln(MR) is the natural log of the mass ratio ( fueled mass/dry mass) for the rocket.
If you need dV to be a significant fraction of the speed of light, there is a relativistic version.
To get the Mass ratio for any given dV, you rearrange to
MR= e^(dV/Ve)
where e is Eulers number = 2.718...
That being said, the quoted passage is neither here nor there as far as this thread is concerned, as we are not talking about sending craft to distant stars, but to the Moon. Applying the above equation to the Saturn V rocket shows that it was within its capabilities to do the job for which it was intended.
I just thought you may want to go and tell them wahts what ?

The kinetic energy of the CMSL propagating to the moon on a straight path is determined using the velocity of 938 m/s,
1/2 mv2 = (.5)(49,480 kg)(983 m/s)2 = 2.39 x 1010 J........................................................................................1
The energy of 2,725,600 kg of fuel using the energy of a kilogram of rocket fuel (4.2 x 107 J/kg),
Fuel Energy = (mass fuel) x (energy of fuel) = (2.72 x 106 kg)/(4.2 x 107 J/kg) = 1.142 x 1014 J....................2
Divided by kinetic of the CMSL and the energy of the fuel is,
Efficiency = (KE)/(enegy of fuel) = (2.39 x 1010 J)/(1.142 x 1014) = .0002 or .02 %........................................2
Multiply the efficiency by 100 times to normalize the effect of the earth's gravity forms an efficiency of free space of 2 %
Where in the World did you pull the 938m/sec from? It is nowhere near the Delta v required to get the CMSL to the Moon and back. The CMSL left LEO moving at over 10,900 m/sec, and the energy required to do this also had to include the energy needed to climb to that altitude. If you are going to include the fuel in the CMSL, then you need to add any velocity changes performed by engine usage to that figure.
You can't just go by the difference between initial and final velocities. As also stated above, you also need to take into account the effects of staging.
So for example, upon reaching the proper point near the Moon, the CMSL fired its engines in order to brake into Lunar orbit. Upon leaving, it had to fire its engines to accelerate away from the Moon. Even if there were no net difference between the velocity just before braking and just after acceleration, the total delta v you would calculate for the craft would be equal to the sum of the two speed changes made. (For example, if it had to brake by 1000 m/sec, and then later accelerate by 1000 m/sec, the total dv budget for the engine would be 2000 m/sec.
Another problem with your calculation is that that you apply the joules/kg figure to the entire "wet load" of the rocket. So for example, when you say that the CMSL has "fuel" load of 109500 kg, this is not technically correct. It carries a combined mass of liquid Hydrogen and liquid Oxygen of 109500 kg. It is only the Hydrogen that the joules/kg figure applies to The Hydrogen only makes up ~ 11.11% of that 109500 kg or ~12155 kg. This is the mass you have to multiply by 4.2e7 joules/kg to get the energy content of the fuel, not the entire 109500 kg.

Why does not a penny in the ISS stick to the walls?
Thinking of spin?
Otherwise I would go with Kryptid

The fact that NASA is using a halve circle trajectory to propagate to the moon using the gravitational force of the earth is senseless logic that cannot be justified since an astronaut is weightless in the ISS.

The fact that NASA is using a halve circle trajectory to propagate to the moon using the gravitational force of the earth is senseless logic that cannot be justified since an astronaut is weightless in the ISS.
Your logic is flawed, please take the time to learn some orbital physics. You could start with Hoheman Transfer Orbit.

Really, please, please, educate me and explain how the 59,000 kg CSML changes direction 90 degree after propagating to the moon since I just cannot figure it out using your link.

Really, please, please, educate me and explain how the 59,000 kg CSML changes direction 90 degree after propagating to the moon since I just cannot figure it out using your link.
Imagine that there was a rope tied to the spacecraft, with a grappling hook on it.
When the ship got near the moon, they threw the hook out and it stuck in the rocks on the moon.
Do you see how that would change the direction of the ship (as long as the rope was strong enough?
Well, it's much the same, but, instead of a rope, they used gravity.

Really, please, please, educate me and explain how the 59,000 kg CSML changes direction 90 degree after propagating to the moon since I just cannot figure it out using your link.
Imagine that there was a rope tied to the spacecraft, with a grappling hook on it.
When the ship got near the moon, they threw the hook out and it stuck in the rocks on the moon.
Do you see how that would change the direction of the ship (as long as the rope was strong enough?
Well, it's much the same, but, instead of a rope, they used gravity.
What you are say is ridiculous since an astronaut the ISS is weightless which nullifies you theory. If the earth's gravitational force had an affect that you suggest the astronaut in the ISS would not be weightless.

Really, please, please, educate me and explain how the 59,000 kg CSML changes direction 90 degree after propagating to the moon since I just cannot figure it out using your link.
The craft is launched into a translunar trajectory. If it weren't for the gravitational effect of the Moon this trajectory would result in a long ellipse with a apogee that would come up a bit short of the Moon's distance. However, as the craft get fruther from the Earth, the Moon begins to have a significant effect on the trajectory deflecting towards the Moon. As it gets closer to the Moon, its gravity curves the path more and more, eventually swinging it around the far side of the Moon. If, at this point nothing is done, the craft would continue to swing around and eventually on a trajectory sending back to the vicinity of the Earth. Such a "free return" trajectory looks something like this:
(https://thingsmysonasks.files.wordpress.com/2017/07/apollo11flightpath.jpg?w=776)
This was used for the Apollo 11 mission as a safeguard, as once sent on its way to the Moon, it would return on its own if something forced them to abort the mission. On the other hand, if everything went to plan, the craft would do braking burn which would put it in lunar orbit once its reached the proper point in its swing around the Moon.
Your failure to grasp the intricacies of orbital mechanics is not proof or even evidence against them. "I don't understand it, so it must be wrong" is not a valid argument.

Really, please, please, educate me and explain how the 59,000 kg CSML changes direction 90 degree after propagating to the moon since I just cannot figure it out using your link.
Imagine that there was a rope tied to the spacecraft, with a grappling hook on it.
When the ship got near the moon, they threw the hook out and it stuck in the rocks on the moon.
Do you see how that would change the direction of the ship (as long as the rope was strong enough?
Well, it's much the same, but, instead of a rope, they used gravity.
What you are say is ridiculous since an astronaut the ISS is weightless which nullifies you theory. If the earth's gravitational force had an affect that you suggest the astronaut in the ISS would not be weightless.
We have already gone over this. Your assertion shows a profound ignorance of the actual physics involved.

What you are say is ridiculous since an astronaut the ISS is weightless which nullifies you theory. If the earth's gravitational force had an affect that you suggest the astronaut in the ISS would not be weightless.
If you were to jump off of a building, you would be weightless until you hit the ground. Being in free fall is the same as being weightless. Please learn basic physics before continuing.

What you are say is ridiculous since an astronaut the ISS is weightless which nullifies you theory. If the earth's gravitational force had an affect that you suggest the astronaut in the ISS would not be weightless.
If you were to jump off of a building, you would be weightless until you hit the ground. Being in free fall is the same as being weightless. Please learn basic physics before continuing.
I disagree with this point. In space, ie deep space or if the solar system where moving at a high enough speed, like a comet , you could forever not encounter other mass to give you said weight, where as if you are in orbit in the iss, you will quickly degrade and impact the earth, if you do not have any external forces to stop you. One is literally freefall as you are falling toward, the other is self sustained !

I disagree with this point.
Please learn basic physics before continuing.

I disagree with this point. In space, ie deep space or if the solar system where moving at a high enough speed, like a comet , you could forever not encounter other mass to give you said weight
Merely being near a mass won't give you weight.
where as if you are in orbit in the iss, you will quickly degrade and impact the earth, if you do not have any external forces to stop you.
Given that the ISS does not "quickly degrade and impact the Earth", this is wrong. Centrifugal force/momentum keeps objects in orbit from doing that (ignoring atmospheric drag, of course).
One is literally freefall as you are falling toward, the other is self sustained !
You would be weightless in both scenarios:

What you are say is ridiculous since an astronaut the ISS is weightless which nullifies you theory. If the earth's gravitational force had an affect that you suggest the astronaut in the ISS would not be weightless.
If you were to jump off of a building, you would be weightless until you hit the ground. Being in free fall is the same as being weightless. Please learn basic physics before continuing.
I disagree with this point. In space, ie deep space or if the solar system where moving at a high enough speed, like a comet , you could forever not encounter other mass to give you said weight, where as if you are in orbit in the iss, you will quickly degrade and impact the earth, if you do not have any external forces to stop you. One is literally freefall as you are falling toward, the other is self sustained !
Any trajectory that is free to respond to gravity is a freefall one. They come in four basic shapes. Circular, elliptical, parabolic, and hyperbolic.*
Everything, From a tossed ball to an object falling in from deep space and leaving again never to return, is on one of these trajectories. The tossed ball follows an elliptical path, while the deepspace object follows a hyperbolic one.
The only real difference between the elliptical path of the ball and that of the ISS is that the ellipse the ball follows is very narrow and intersects with the surface of the Earth, with the vast majority of it being underground and thus we only see a small section of it , while the ISS trajectory is much closer to circular and doesn't even pass deep into the atmosphere, which means the whole path remains above ground.** The ISS is, in a sense. "falling towards the Earth", It's just moving so fast horizontally that the Earth curves out from under it as fast as its path curves towards the Earth, and thus it keeps "missing".
* technically one could say that the circle is just a special type of ellipse and the parabola a special type of hyperbola, so their are only two types of trajectory.
** if you were able to compress the Earth to a small object, say less than 1 m in radius, then a ball tossed at just 40 mph horizontally from the same distance from its center as the surface is now, it would fall in, whip around the center mass and return to its starting point repeatedly following that very narrow elliptical orbit.