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September 2, 2014

September 2, 2014

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

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 9:43pmThat is high. Did you work if from the Earth's surface, for from the center of Earth?

- PHYSICS -
**NIQUEFROMDALLAS**, Sunday, October 26, 2008 at 10:06pmI worked it from the center of the Earth so i have to go back and subtract the radius of the Earth from that answer.

- PHYSICS -
- PHYSICS -
**Damon**, Sunday, October 26, 2008 at 9:53pmwhere 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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