Calculating Rocket Energy in Satellite Orbit Problem

AI Thread Summary
A satellite in a circular orbit around Earth requires sufficient gravitational force to maintain its trajectory, but in this scenario, it lacks enough gravitational pull and relies on a rocket engine to counteract this deficiency. The rocket, which points towards Earth, exerts a force that helps keep the satellite in orbit, but the work done by the rocket is considered zero since it acts perpendicular to the satellite's trajectory. As the rocket burns fuel, it accelerates the exhaust rather than the satellite directly, leading to energy being used in the system, but this does not contribute to the satellite's orbital energy. The discussion highlights the distinction between force and work, emphasizing that continuous thrust alters the orbit rather than maintaining it. Ultimately, the calculations for the energy involved can be derived from the thrust required and the exhaust velocity.
Abtinnn
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Assume that a satellite(mass m) is orbiting the Earth(mass M) at radius R and speed V (the orbit is circular). The satellite is in a position such that the force of gravity exerted on it by the planet is not enough to keep it in orbit. Therefore, the satellite has an engine, which always points at the centre of Earth. The rocket, when ignited, exerts an additional force on the satellite, enough to keep it in orbit with the help of Earth's gravity. Now the work done by the rocket is zero because it always points perpendicular to the trajectory and the total energy of the system stays constant. However, the rocket eventually runs out of fuel, meaning that some energy has been put in the rocket-satellite system. That energy goes into keeping the satellite in orbit, but is there a way to calculate it in terms of V,R,M and m?
 
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Abtinnn said:
That energy goes into keeping the satellite in orbit
No, it goes into accelerating the fuel.
 
A.T. said:
No, it goes into accelerating the fuel.
But the reaction force of that keeps the satellite in orbit, right?
 
Abtinnn said:
But the reaction force of that keeps the satellite in orbit, right?
Yes, force and energy are different things.
 
A.T. said:
Yes, force and energy are different things.
So you're saying that the rocket does no work on the satellite, but does work on the fuel by accelerating it?
 
Abtinnn said:
So you're saying that the rocket does no work on the satellite, but does work on the fuel by accelerating it?
Yes.
 
Oh alright.
Thanks a lot!
 
Abtinnn said:
Assume that a satellite(mass m) is orbiting the Earth(mass M) at radius R and speed V (the orbit is circular). The satellite is in a position such that the force of gravity exerted on it by the planet is not enough to keep it in orbit. Therefore, the satellite has an engine, which always points at the centre of Earth. The rocket, when ignited, exerts an additional force on the satellite, enough to keep it in orbit with the help of Earth's gravity. Now the work done by the rocket is zero because it always points perpendicular to the trajectory and the total energy of the system stays constant. However, the rocket eventually runs out of fuel, meaning that some energy has been put in the rocket-satellite system. That energy goes into keeping the satellite in orbit, but is there a way to calculate it in terms of V,R,M and m?

A couple of things... The work done by the rocket cannot be zero. And, a trajectory for which there is not enough attraction to stay in orbit cannot be circular - it would be parabolic or hyperbolic. Once the rocket has done enough work to circularize the orbit, you can shut it off and the satellite will stay in that orbit. In fact, you MUST shut it off at that time or the orbit will continue to change into something else - ellipse, parabola, hyperbola. The orbit continues to change as long as the rocket is firing.
 
tfr000 said:
A couple of things... The work done by the rocket cannot be zero. And, a trajectory for which there is not enough attraction to stay in orbit cannot be circular - it would be parabolic or hyperbolic. Once the rocket has done enough work to circularize the orbit, you can shut it off and the satellite will stay in that orbit. In fact, you MUST shut it off at that time or the orbit will continue to change into something else - ellipse, parabola, hyperbola. The orbit continues to change as long as the rocket is firing.
While accelerating the satellite until its speed is large enough for a circular orbit would be the sensible thing to do for a satellite operator, here the rocket is pointing towards earth, keeping it in a circular orbit by countering the gravity directly and reducing the acceleration towards Earth until it's just small enough for a circular orbit. Because the force of the rocket is perpendicular to the velocity, no work is done on the satellite.
 
  • #10
tfr000 said:
The work done by the rocket cannot be zero.
Of course it can, just like the work done by gravity is zero, for a constant speed circular orbit.

tfr000 said:
The orbit continues to change as long as the rocket is firing.
I think you should read the OP more carefully.
 
  • #11
Abtinnn said:
Assume that a satellite(mass m) is orbiting the Earth(mass M) at radius R and speed V (the orbit is circular). The satellite is in a position such that the force of gravity exerted on it by the planet is not enough to keep it in orbit.
If the satellite is in circular orbit, then the force exerted on it by the planet is enough to keep it in orbit.
Therefore, the satellite has an engine, which always points at the centre of Earth. The rocket, when ignited, exerts an additional force on the satellite, enough to keep it in orbit with the help of Earth's gravity. Now the work done by the rocket is zero because it always points perpendicular to the trajectory and the total energy of the system stays constant.
Initially, the work is 0, until the rocket gains some radial velocity as a result of the burn. Then you will have a component of the force in the direction of travel.

If you make a really fast burn you can have a large delta V in a negligible distance. In that case, the change in energy of the orbit is 0. This would still affect the orbit. It would introduce eccentricity. Your orbit is no longer round. Its semi-major axis may still be the same, but its perigee may now be subterranean, whether you fired the engine towards or away from the Earth.
 
  • #12
tony873004 said:
If the satellite is in circular orbit, then the force exerted on it by the planet is enough to keep it in orbit.

Sorry I may have used the wrong wording. What I meant is that the satellite cannot orbit the Earth in a circular trajectory with that V and R, and because of that a rocket it helping it stay in an orbit by exerting a force directed towards the centre of the Earth. So the satellite is not initially in a circular orbit, but with the help of the rocket it is.
 
  • #13
We can reduce it to an easier setup: the rocket is not moving at all (Earth-centered inertial reference frame), it simply hovers in place by using its engines.
 
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  • #14
mfb said:
We can reduce it to an easier setup: the rocket is not moving at all (Earth-centered inertial reference frame), it simply hovers in place by using its engines.
That would actually work something like the original post, unlike the rest of this... You can't continuously fire a rocket to maintain an orbit - you are changing the shape of the orbit by firing the rocket, and the shape continues to change until you stop. Probably it changes into something you don't want - like an escape orbit, or one that intersects the Earth.
 
  • #15
tfr000 said:
You can't continuously fire a rocket to maintain an orbit
Okay, skip the word "orbit" if that causes too much confusion.
You can continuously fire a rocket to maintain a circular motion at constant speed around an object. The thrust has to be radial, and the speed has to be slower or faster than the orbital speed at this radius.
 
  • #16
mfb said:
Okay, skip the word "orbit" if that causes too much confusion.
You can continuously fire a rocket to maintain a circular motion at constant speed around an object. The thrust has to be radial, and the speed has to be slower or faster than the orbital speed at this radius.
Now we're getting somewhere.
Rephrasing the original post:
Assume that a satellite(mass m) is orbiting the Earth(mass M) at radius R and speed V in a circular orbit. The satellite has an engine, which always points at the centre of Earth. The rocket, when ignited, exerts an additional force on the satellite, enough to counteract part of Earth's gravity. Now the work done by the rocket is zero because it always points perpendicular to the trajectory and the total energy of the system stays constant. However, the rocket eventually runs out of fuel, meaning that some energy has been put in the rocket-satellite system. That energy goes into keeping the satellite in an orbit which is different than that it would have without the rocket - as if the Earth were less massive. Is there a way to calculate it in terms of V,R,M and m?
 
  • #17
tfr000 said:
Now we're getting somewhere.
Rephrasing the original post:
The rocket, when ignited, exerts an additional force on the satellite, enough to counteract part of Earth's gravity.

That's not what I originally meant. What you're saying is that the speed of the satellite is too low to stay at orbit at R, and because of that a rocket pointing away from Earth exerts an outward radial force on the satellite to keep it in orbit.

What I meant is the exact opposite. The satellite is in an R such that its V is too high for it to stay in an orbit at that R.
 
  • #18
tfr000 said:
However, the rocket eventually runs out of fuel, meaning that some energy has been put in the rocket-satellite system.
No, it has been put in the rocket-satellite-exhaust system.
It has been used to accelerate the fuel downwards. You don't even need a rocket in space, you can run the engine in a lab, with the same result.
tfr000 said:
s there a way to calculate it in terms of V,R,M and m?
Sure, calculate the required force from thrust. Multiply by the exhaust velocity to get power.

Abtinnn said:
What I meant is the exact opposite. The satellite is in an R such that its V is too high for it to stay in an orbit at that R.
Okay, doesn't change the result.
 
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  • #19
The question in the OP was completely and correctly answered in post #2, why is the thread still running?
 

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