Posted by **tanya** on Saturday, December 10, 2011 at 6:00pm.

A satellite moves in a circular orbit around the Earth at a speed of 6.1 km/s.

Determine the satellite’s altitude above

the surface of the Earth. Assume the

Earth is a homogeneous sphere of radius

6370 km and mass 5.98 × 10

24 kg. The value of the universal gravitational constant is 6.67259 × 10^−11N · m2/kg2.

- physics -
**tchrwill**, Saturday, December 10, 2011 at 7:26pm
When orbiting at 6.3km/s:

From Vc = sqrt(µ/r) where Vc = the velocity of an orbiting body, µ = the gravitational constant of the earth and r the radius of the circular orbit,with µ = GM, G = the universal gravitational constant and M = the mass of the central body, the earth in this instant,

r = µ/Vc^2

= 6.67259x10^-11(5.98x10^24)/6300^2

The altitude is therefore

(r - 6370)/1000 km.

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