Satellite motion

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Greetings,

I have a question asking which one of the following two takes more fuel:

1) Sending a satellite from its orbit around the earth to the moon
2) Sending the same satellite from moon to its orbit around the earth

I am really confused how to proceed in this situation. I think both cases should require the same amount of work. How can I have a sound reasoning here and prove it? This is not a homework question and I just ask for insight here.

Many thanks
 

Answers and Replies

  • #2
DaveC426913
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Are you looking for a rigorous answer, or an intuitive answer?
Intuitively, the gravitational influence of a body is often compared to a well or funnel i,e, a gravity well - out of which one must climb.
Which one would you rather climb out of - the Moon's gravity well, or the Earth's gravity well?
 
  • #3
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Are you looking for a rigorous answer, or an intuitive answer?
Intuitively, the gravitational influence of a body is often compared to a well or funnel i,e, a gravity well - out of which one must climb.
Which one would you rather climb out of - the Moon's gravity well, or the Earth's gravity well?
Actually I want a more rigorous answer, yet I cannot think how to approach the problem. Obviously escaping from the Earth's gravity is harder; however we are not at the surface of the Earth. The period of the satellite around the world is 8 hours, so the radius of the orbit can also be found from there. I tried integrating the work done against the forces of gravity by the Moon and the Earth, yet I cannot link this to the fuel consumption. This is an old exam question and the original is:
http://imgur.com/TQUZbe4
Also note that the answer should be interpreted and a fully rigorous calculation is not possible as the numerical values for the mass of the earth etc. are not provided and calculators were not allowed on this exam.
 
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  • #4
A.T.
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Actually I want a more rigorous answer, yet I cannot think how to approach the problem.
Try calculating the combined graviational potential of Earth and Moon (plus centrifugal, if you to it in the rotating rest frame of Earth and Moon centers). From this you get the work done by integration between two distances (plus change in kinetic energy in the appropriate frame).

The question doesn't specify the orbit around the Moon. Assume the lowest possible?
 
  • #5
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Try calculating the combined graviational potential of Earth and Moon (plus centrifugal, if you to it in the rotating rest frame of Earth and Moon centers). From this you get the work done by integration between two distances (plus change in kinetic energy in the appropriate frame).

The question doesn't specify the orbit around the Moon. Assume the lowest possible?
I think braking to get out of and accelerating to get into the orbit around the Earth requires the same amount of energy, so I treat the satellite as standing still between the centers of the Moon and the Earth(The Moon, the Earth and the satellite is on the same line to put it in other words). Yes you are right, but actually the question statement is a bit vague and I do not know how you can compare the two values without knowing the relationship between the given parameters.

I think what the question means is that the satellite lands on the Moon and stays there.
 
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  • #6
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When going from the moon to low earth orbit, you can use the atmosphere to slow down. That won't work when going from low earth orbit to the moon.
 
  • #7
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When going from the moon to low earth orbit, you can use the atmosphere to slow down. That won't work when going from low earth orbit to the moon.
I do not think the question takes into account subtleties such as air friction. The reasoning should be based entirely on mechanical energy considerations as far as I understand because this is an introductory physics class. However, the question is still a bit vague to me as you just interchange the initial point and the final point since the gravitational potential energy is conservative I just expect a change in the sign of the work done. Furthermore, we can also assume that it goes without acceleration and at a constant speed right? I am certainly missing something here but I cannot pinpoint it.
 
  • #8
jbriggs444
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However, the question is still a bit vague to me as you just interchange the initial point and the final point since the gravitational potential energy is conservative I just expect a change in the sign of the work done.
Are you interested in "work done" or "fuel used". The two are very different notions.
 
  • #9
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Are you interested in "work done" or "fuel used". The two are very different notions.
I am interested in the fuel used and I should be able to provide a valid proof by demonstrating my reasoning with equations.
 
  • #10
jbriggs444
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I am interested in the fuel used and I should be able to provide a valid proof by demonstrating my reasoning with equations.
I prefer to do it without equations. Two words: "time reversal".
 
  • #11
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I prefer to do it without equations. Two words: "time reversal".
So what do you think the answer is?
 
  • #12
jbriggs444
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Same fuel requirement either way.
 
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  • #13
DaveC426913
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I prefer to do it without equations. Two words: "time reversal".
I see where you're going with that but it is valid?
If one did a time reversal analysis of a ball falling off a ledge onto the floor, one would not get the same energy output.
 
  • #14
jbriggs444
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I see where you're going with that but it is valid?
If one did a time reversal analysis of a ball falling off a ledge onto the floor, one would not get the same energy output.
Indeed, but we are not talking energy. We are talking delta-v.
 
  • #15
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Okay I am sorry for posting it on the wrong forum first of all. I solved the question and I attach the solution here. The required fuel should be the same in both cases taking into account that it lands on the lunar surface instead of crashing into it.

I made the approximation that the satellite moves with constant velocity except the initial and final change in velocity(acceleration, deceleration) along the line connecting the centers of the Moon and the Earth. In order for the satellite to move with constant velocity its engines should provide a thrust of exactly the same amount of the gravitational force exerted on it, but in the opposite direction. This is valid for the satellite traveling in both directions. Please let me know if you see a loophole in my reasoning.
imgur.com/fotiFBC
 
  • #16
jbriggs444
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The trajectories that you chose are simple ones that ignore some of the realities of orbital mechanics. But the two trajectories have the important feature, that they are time-reversed images of each other. You have made the key observation, that both acceleration and deceleration take the same amount of fuel. You have not carefully written down an argument that the fuel used to fight gravity and maintain a constant velocity is the same in both directions, but that is an easy argument to make.

A more realistic minimum energy maneuver would use a "Hohmann Transfer orbit". That involves a rapid burn to leave the original circular Earth orbit and accelerate into an elliptical orbit that will intersect the desired orbit around the Moon tangentially. Upon reaching the intersection point, a second rapid burn is used to decelerate into the final circular orbit.

Edit: I missed an important loophole in your argument. It assumes that the fuel used is a negligible fraction of the rocket's total mass.

In a real rocket, you have to use more fuel to get a given change of velocity when the tanks are full than when the tanks are nearly empty. The force of gravity is also greater when the tanks are full than when the tanks are empty. So you have an asymmetry between the outbound and the returnbound trip.

There is a way to rescue the argument. I've hinted obliquely at it earlier in this thread.
 
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  • #17
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The trajectories that you chose are simple ones that ignore some of the realities of orbital mechanics. But the two trajectories have the important feature, that they are time-reversed images of each other. You have made the key observation, that both acceleration and deceleration take the same amount of fuel. You have not carefully written down an argument that the fuel used to fight gravity and maintain a constant velocity is the same in both directions, but that is an easy argument to make.

A more realistic minimum energy maneuver would use a "Hohmann Transfer orbit". That involves a rapid burn to leave the original circular Earth orbit and accelerate into an elliptical orbit that will intersect the desired orbit around the Moon tangentially. Upon reaching the intersection point, a second rapid burn is used to decelerate into the final circular orbit.

Edit: I missed an important loophole in your argument. It assumes that the fuel used is a negligible fraction of the rocket's total mass.

In a real rocket, you have to use more fuel to get a given change of velocity when the tanks are full than when the tanks are nearly empty. The force of gravity is also greater when the tanks are full than when the tanks are empty. So you have an asymmetry between the outbound and the returnbound trip.

There is a way to rescue the argument. I've hinted obliquely at it earlier in this thread.
Yes I agree with the fact that the fuel used should be a negligible fraction of the total mass of the rocket for my argument to hold. However, I am still oblivious to the way of saving the argument.
 
  • #18
jbriggs444
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Yes I agree with the fact that the fuel used should be a negligible fraction of the total mass of the rocket for my argument to hold. However, I am still oblivious to the way of saving the argument.
The argument holds if you shift from counting fuel to counting delta-v. The delta-v budget is the same for either trajectory. But if you know how much delta-v you need, you know how much fuel you will use. The Tsiolkovsky rocket equation can tell you that.
 
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