# A rocket problem

1. Aug 9, 2015

### Abtinnn

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?

2. Aug 9, 2015

### A.T.

No, it goes into accelerating the fuel.

3. Aug 9, 2015

### Abtinnn

But the reaction force of that keeps the satellite in orbit, right?

4. Aug 9, 2015

### A.T.

Yes, force and energy are different things.

5. Aug 9, 2015

### Abtinnn

So you're saying that the rocket does no work on the satellite, but does work on the fuel by accelerating it?

6. Aug 9, 2015

Yes.

7. Aug 9, 2015

### Abtinnn

Oh alright.
Thanks a lot!

8. Aug 10, 2015

### tfr000

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.

9. Aug 11, 2015

### willem2

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. Aug 11, 2015

### A.T.

Of course it can, just like the work done by gravity is zero, for a constant speed circular orbit.

I think you should read the OP more carefully.

11. Aug 11, 2015

### tony873004

If the satellite is in circular orbit, then the force exerted on it by the planet is enough to keep it in orbit.
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. Aug 11, 2015

### Abtinnn

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. Aug 11, 2015

### Staff: Mentor

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.

14. Aug 11, 2015

### tfr000

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. Aug 11, 2015

### Staff: Mentor

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. Aug 11, 2015

### tfr000

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. Aug 11, 2015

### Abtinnn

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. Aug 11, 2015

### Staff: Mentor

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.
Sure, calculate the required force from thrust. Multiply by the exhaust velocity to get power.

Okay, doesn't change the result.

19. Aug 11, 2015

### MrAnchovy

The question in the OP was completely and correctly answered in post #2, why is the thread still running?