Posted by **NIQUEFROMDALLAS** on Sunday, October 26, 2008 at 9:36pm.

This is the exact problem from the paper and I got an answer of 4.2 * 10^7 and I was wondering if this was the right answer.

What is the height above the earth's surface at which all synchronous satellites (regardless of mass) must be placed in orbit? The period of synchronous satellites is one day. The radius of the earth 6.36(10)6m

- PHYSICS -
**bobpursley**, Sunday, October 26, 2008 at 9:43pm
That is high. Did you work if from the Earth's surface, for from the center of Earth?

- PHYSICS -
**Damon**, Sunday, October 26, 2008 at 9:53pm
where is m v^2/r = G M m/r^2

as you said, m cancels

v^2 = G M/r

T = 24 h *3600 s/h = 86,400 s

2 pi r = v T = 86,400 v

so

r = 13,751 v

v^3 = G M/13,751 = 6.67*10^-11 * 6*10^24/1.38*10^4

v^3 = 29*10^9

v = 3.07 * 10^3

then r = 4.24 * 10^7

eight = (10^7 )(4.24 - .636) = 3.6 *10&

remember to subtract the radius of earth to get the altitude.

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