# Elliptical orbits double velocity

1. Mar 26, 2014

### brijmohan3

1. The problem statement, all variables and given/known data

A rocket ship is in circular orbit of radius R around a planet. Its velocity is doubled by a sudden engine burst. Calculate the furthest distance from the planet on the new trajectory.

2. Relevant equations

Elliptical energy equation: E = Ek +Ep = -GMm/2a (a=semi major axis)

With Ek = (1/2)mv^2 and Ep = -GMm/R

3. The attempt at a solution

Sorry for not being very user friendly.

From applying the elliptical energy equation to the initial circular orbit, I deduced

(1) v^2 = GM/R where v = initial velocity, M = mass of planet.

After the impulse, I have:

(2) E = (1/2)*m*(2v)^2 -GMm/R = -GMm/2(r+R)
where r = length to planet from furthest distance

Combining (1) and (2) gives r = -2R

I have two questions in regards to this:

1) Is the method correct?

2) If the answer is right, how is it possible that I get a negative value? Does this just mean it is taken in the 'negative' direction to the original 'R'?

Thanks, and sorry for the lack of friendliness in the equations again.

2. Mar 26, 2014

### TSny

Hello, brijmohan3. Welcome to PF!

It might help to consider whether the total energy is positive, negative, or zero after the engine boost.

3. Mar 27, 2014

### brijmohan3

Thank you TSny.

I think the total energy after the impulse should still be negative due to the half potential at average distance: -GMm/2a?

Zero doesn't make sense to me as we are providing it with a positive impulse thus increasing the total energy in the process, but not equal to the original energy the rocket originally had.

I considered the total energy being positive [GMm/(r+R)] but that would tell me r=0, which is clearly not right since I believe the path of the orbit would increase in distance, not decrease. Unless my understanding is flawed and that is not the correct total energy.

4. Mar 27, 2014

### TSny

Here's one way to think about it.

In your first post you stated that you found v2 = GM/R for the circular orbit (before the engine burn).

So you can express the kinetic energy K in terms of G, M, m, and R before the burn.

If you double the speed, by what factor does K increase? Thus, you can easily express K immediately after the burn in terms of G, M, m, and R.

Then find the total energy just after the burn in terms of G, M, m, and R.

5. Mar 27, 2014

### brijmohan3

Hm.

I'm pretty sure that's what I did when I combined equations (1) and (2) -v2 = GM/R and E = (1/2)*m*(2v)2 -GMm/R = -GMm/(r+R) -------> AH sorry, I had "= -GMm/2(r+R)" in my original post as opposed to "= -GMm/(r+R)"

But just to share what I did:

So K (before impulse) = GMm/2R

Leads to K (after impulse) = 4* K (before impulse) = 2GMm/R

So total E (after impulse) = K (after impulse) + V (after impulse), but V (after impulse) = V (before impulse)

Therefore E = 2GMm/R + (-GMm/R) = GMm/R (which is what I had before from my equations)

So then by elliptical orbit equations, E = -GMm/(r+R) as before :/

I'm still not sure what I am doing wrong :(

6. Mar 27, 2014

### TSny

OK. So, after the boost the total energy is positive.

As you can see, any elliptical orbit must have a negative total energy.

So, the ship is not in an elliptical orbit after the boost. The trajectory will be some other conic section.

7. Mar 28, 2014

### brijmohan3

Ah! I didn't think about that. Thank you.

Does that mean that the furthest distance from the planet is therefore infinity since, with a positive total energy the ship is in an unbounded hyperbolic orbit?

8. Mar 28, 2014

### dauto

9. Mar 28, 2014

### TSny

Yes, that's right.

10. Mar 28, 2014

Thank you!