Question

A satellite moves in a circular orbit around the Earth at a speed of 5 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 x 10^24 kg. The value of the universal gravitational constant is 6.67259 x 10^−11 N* m^2 /kg^2 . Answer in units of km.

Answer 1

Answer:

$h=9590835m$

Explanation:

Writing Newton's 2nd Law and Newton's Gravitational Law on the satellite (of mass m, experimenting an acceleration a) orbiting Earth (of mass M) with r as the distance between their centers we have:

$F=ma=\frac{GMm}{r^2}$

Since this acceleration is centripetal, we can write:

$\frac{v^2}{r}=a=\frac{GM}{r^2}$

So we have:

$v^2=\frac{GM}{r}$

Or:

$r=\frac{GM}{v^2}$

This distance r is the sum of Earth's radius R and the satellite's altitude h (r=R+h), so for our values we have (in S.I.):

$h=\frac{GM}{v^2}-R=\frac{(6.67259 \times10^{-11}Nm^2/kg^2)(5.98\times10^{24}kg)}{(5000m/s)^2}-6370000m=9590835m$