Posted by **Lelia** on Tuesday, November 27, 2012 at 10:51pm.

A satellite in an elliptical orbit has a speed of 9.00km/s when it is at its closes approach to the Earth(perigee). The satellite is 7.00x10^6 m from the center of the Earth at this time. When the satellite is at its greatest distance from the center of the Earth (apogee), its speed is 3.66km/s. Find the distance from the satellite to the center of the Earth at apogee. (assume any energy losses are negligible.)

- College Physics -
**PhysicsPro**, Tuesday, November 27, 2012 at 10:53pm
Post your work thus far and I'll continue/correct/guide

- College Physics -
**Lelia**, Tuesday, November 27, 2012 at 10:56pm
Kfinal + Ugravfinal = Kinitial + Ugravinitial

which i ended up with

Vfinal^2 + [2GM/r] = Vinitial^2 + [2GM/r]

I believe I need to solve for r but this is where I got stuck.

- College Physics -
**PhysicsPro**, Tuesday, November 27, 2012 at 11:06pm
Well since it's an elliptical orbit you can set them the distances as functions of Kvff^3

<Kf^3>(2GM^4)=vf%

vf%<6.76[FL(3.33/Ki)]=d%

d%=<4.1882[Ug]

d=(1.333<pi^4>)(d%)

Tell me what you get

- College Physics -
**PhysicsPro**, Tuesday, November 27, 2012 at 11:17pm
Quick explanation because I have to go;

Kvff^3

Rearrange values so that kf^3=vf%

vf% is of 6.76 because of orbit while 3.33 is the continual. Solve for d%

d% the continual converted to elliptical motion is going to be roughly 4.1882 (4.188? I believe it's two, but shouldn't matter)

By Ugravinitial

and you have d when you apply the standard orbital motion functions.

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