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August 20, 2014

August 20, 2014

Posted by **Anonymous** on Monday, October 31, 2011 at 11:10pm.

(a) Find the altitude of the satellite.

- physics -
**drwls**, Monday, October 31, 2011 at 11:50pmEquate the gravitational force,

GmM/R^2, to the centripetal force, m V^2/R

M is the Earth's mass and G is the universal constant, which you need to look up. Satellite mass m cancels out.

Substitute 2 pi R/(139*60) for the velocity, V (in m/s).

Solve for the remaining variable R.

The satellite altitude is H = R - Rearth

- physics -
**tchrwill**, Tuesday, November 1, 2011 at 10:46amThe orbital period of a satellite derives from

T = 2(Pi)sqrt(r^3/µ) where T = the period in seconds

Pi = 3.14

r = the orbital radius in feet and

µ = the earth's gravitational constant = 1.407974x10^16 ft^3/sec^2.

139(60)= 2(3.14)sqrt(r^3/1.407974x10^16)

Solving for r = 5523.6 miles making the altitude 1560.6 miles = 2511km.

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