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November 20, 2014

November 20, 2014

Posted by **Adrianne** on Thursday, April 1, 2010 at 7:53pm.

an altitude h = [03] km above the earth's surface and has a mass

m = 3500 kg. You wish to calculate when the satellite will be back in the same position. From the second law of

motion and the gravitational force law, calculate the following:

(a) What is the satellite's velocity? (m/s)

(b) What is the period of the satellite's motion?

The only force acting on the satellite is the force of gravity (Fg) right? So by combining the two laws I have Fg = ma. I solved for Fg, but I need some pointers on where to go from there.

- Physics -
**Damon**, Thursday, April 1, 2010 at 8:28pmmass is m (it will not matter in the end)

if height above earth is h

then radius of orbit = r = Rearth + h

then

Fg = G m Mearth/r^2

a = v^2/r

so Fg = m v2/r

so

G m Mearth/r^2 = m v^2/r

note m cancels, feather drifts right beside the space station

G Mearth/r = v^2

there you go

get v then it goes 2 pi r at v

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