Elliptical orbits double velocity

[brijmohan3]

Homework Statement

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.

Homework Equations

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

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

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.

 

[TSny]

Hello, brijmohan3. Welcome to PF!

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

 

[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.

 

[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.

 

[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 :(

 

[TSny]

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

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

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

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.

 

[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?

 

[dauto]

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?

Tada

 

[TSny]

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?

Yes, that's right.

 

[brijmohan3]

Thank you!