# Orbits question

1. Apr 28, 2010

### bon

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

Ok so I'm told that there is a satellite mass m in a circular orbit of radius ro around the earth (of mass M). I'm told the speed of the satellite is vo.

So i've shown that Eo = total energy of satellite = -1/2mvo^2

Now im told that whilst the satellite is in its circular orbit, there is an impulse that leaves its angular momentum unchanged but increases its kinetic energy by 1%

I'm told to find the difference between the maximum and minimum distances from the centre of the earth..

2. Relevant equations

3. The attempt at a solution

So I know that after the impulse, it will follow an elliptical path.

Energy eqn for non-circular orbit is:

1/2 m (dr/dt)^2 + J^2/2mr^2 - GMm/r = E

So I know J = J initial = mvoro

but all i know is that its kinetic energy has increased by 1%, not its total, so how can i get an expression for E to solve the problem?

thanks!

2. Apr 28, 2010

### bon

anyone?

3. Apr 28, 2010

### Filip Larsen

From the constant angular momentum, h, for the initial circular orbit you can derive h expressed as a function of v0. Since the maneuver is considered impulsive, radius does not change while speed does, so establishing an expression for the total orbital energy after the maneuver and replacing the kinetic energy term with the "1% scaled" kinetic energy for v0 you can get an expression that relates the semi-major axis, a, of the new elliptic orbit with v0. Knowing h and a and some geometric relationship for an elliptic orbit, you can derive the eccentricity and finally the apogee and perigee radius as a function of v0 and the "1% factor" (and mu).

4. Apr 28, 2010

### bon

Ok..thanks..but is there another way to do this - just using energy considerations - i.e. the equations I have above, as we haven't really done anything involving semi major axis etc..

thanks

5. Apr 28, 2010

### ehild

At the minimum and maximum distances dr/dt =0. That means that the radial component of the velocity is zero, the velocity is normal to the radius, so the angular momentum is J = mrv at the maximum and minimum distance from Earth. The angular momentum is the same as it was for the circular orbit, so you can write v in terms of r using the conservation of angular momentum.
You can get the new energy after the impulse. The radius did not change so the potential energy stayed -mv0^2 = -GmM/r0, but the KE became 1.01/2 mv0^2. You have the equation for the energy at arbitrary r along the orbit, and you know that dr/dt = 0 and the angular momentum is J=mrv. Make E(r) equal to the energy you got after the impulse, this is an equation for r, solve.

ehild

6. Apr 28, 2010

### D H

Staff Emeritus
This problem is essentially asking you to find the (new) semi-major axis and eccentricity. You cannot determine both of those from energy alone. You need energy and angular momentum.

7. Apr 28, 2010

### bon

Thanks ehild..so I get that before Jo = mrovo

so by conservation of angular momentum i get J = mr^2 theta dot = m ro vo

I understand what you say about energy also..

im just not sure what to equate to what..

for the orbit after the collision, the orbit equation 1/2m(dr/dt)^2 + J^2/2mr^2 - GMm/r = E applies..

i know J = Jo...but I don't see how to get E..

please could you just explain the steps more clearly? thank you very much!

8. Apr 28, 2010

### ehild

You know KE and PE just before the impulse: KE =1/2 mv0^2, PE = -GmM/r0. After that, KE increased by 1 %, PE stayed the same, so E=1.01 * 1/2 mv0^2-GmM/r0 = -0.99 * 1/2 mv0^2.

ehild

9. Apr 28, 2010

### bon

Thank yo so much. very clear now!

Ehild = the best

10. Apr 28, 2010

### bon

cool okay so i solved the equation and got rmin = 0.42ro, rmax = 2.44ro.

so distance between them is 2.86ro

Is this right?

The answer says it should be 0.2ro, but this can't be right - the distance between maximum and minimum distances from the centre of the earth cant be less than ro right?

thanks

11. Apr 28, 2010

### bon

sorry found my mistake... :S

12. Apr 28, 2010

### ehild

Have you got the correct solution? Congratulation!

ehild

13. Apr 29, 2010

### Filip Larsen

Now I'm curious to hear the result. I get around 0.90 and 1.11 when going via semi-major axis and eccentricity.

14. Apr 29, 2010

### ehild

They are r0/1.1 and r0/0.9

ehild