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.
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.
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