Homework Help: College Physics

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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