[alright1234 (OP)]

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%

[alright1234 (OP)]

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/spacex-crew-dragon-explosion-nasa-memo.html 

[Kryptid]

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.

[Janus]

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 trans-lunar 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.

[alright1234 (OP)]

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.

[Petrochemicals]

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.

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

[alright1234 (OP)]

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 10-15 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 10-9 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 10-11 W2/m2. There is a four order of magnitude difference between the extrapolated sensitivity of the Parkes radio telescope and the 10-15 W2 /m2 s-band 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 10-22 W2/m2 at the earth that is 11 orders of magnitude less than the extrapolated sensitivity of the Parkes (10-11 W2/m2).

[Bored chemist]

Also, a radio signal cannot penetrate the Van Allen radiation belt that surrounds the earth.

Do you have satellite TV?

[alright1234 (OP)]

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?

[alancalverd]

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.

[Petrochemicals]

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.

[alancalverd]

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.

[Halc]

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 ping-pong 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.

[Janus]

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.

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.

[Kryptid]

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 10-15 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/m², not W²/m². 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 4πr²:

A = 4π(380,000,000)²
A = 4.54 x 10¹¹ square meters

Divide 50 watts by 4.54 x 10¹¹ 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 10-9 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 10-11 W2/m2. There is a four order of magnitude difference between the extrapolated sensitivity of the Parkes radio telescope and the 10-15 W2 /m2 s-band 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 10-22 W2/m2 at the earth that is 11 orders of magnitude less than the extrapolated sensitivity of the Parkes (10-11 W2/m2).

You are, again, ignoring antenna gain.

[alancalverd]

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!

[Petrochemicals]

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.

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 !

[alancalverd]

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 long-range aircraft, whose max landing mass can be less than half the takeoff limit.

This is O-level stuff in the UK, though clearly beyond the competence of Harvard emeriti

[Halc]

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 expressed-as-power 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.

[Bored chemist]

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?