[alright1234 (OP)]

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 command-service 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

[Bored chemist]

"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?

[Kryptid]

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.

[Halc]

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.

[Janus]

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 command-service 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.

[Bored chemist]

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

[alancalverd]

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.

[Petrochemicals]

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.

[alright1234 (OP)]

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 un-burnt fuel that reduces the efficiency. Do you have a link to the rocket efficiency greater than 1%.

[Kryptid]

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 un-burnt 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/SP-4404/app-b4.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

[Halc]

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 payload-distance 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.

[Bored chemist]

  All that smoke (lots and lots of rocket smoke) is un-burnt 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.

[Petrochemicals]

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 un-burnt 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.

[Bored chemist]

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

[Petrochemicals]

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

[Janus]

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 trans-lunar 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.

[Bored chemist]

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?

[Petrochemicals]

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 trans-lunar 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.

[alancalverd]

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 rocket-launched 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.

[Janus]

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 trans-lunar 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^2/2-u/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/re-um/2r = um(1/re-1/2r)
u = 3.987e14 for the Earth, re= 6378,000m and r = 6578000m   
if m=1000kg,  then energy difference is
E= 3.987e14(1000)(1/63378000-1/(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.