Question

For a satellite to be in a circular orbit 950 km above the surface of the earth, what orbital speed must it be given?

Answer 1

1.) 7382 m/s

The gravitational attraction between the satellite and the Earth provides the centripetal force that keeps the satellite in circular motion, so we can write

$\frac{GMm}{(R+h)^2}=m\frac{v^2}{R+h}$

where

G is the gravitational constant

$M=5.98\cdot 10^{24}kg$ is the Earth's mass

m is the satellite's mass

$R=6370 km = 6.37\cdot 10^6 m$ is the Earth's radius

$h=950 km = 0.95\cdot 10^6 m$ is the altitude of the satellite above the Earth's surface

v is the orbital speed

Solving the formula for v, we find

$v=\sqrt{\frac{GM}{R+h}}=\sqrt{\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24} kg)}{(6.37\cdot 10^6 m +0.95\cdot 10^6 m)}}=7382 m/s$

2) 1.73 hours

The period of the orbit is the time taken to complete one revolution around the Earth, therefore:

$T=\frac{2\pi (R+h)}{v}$

where the numerator is the circumference of the orbit and v the orbital speed, therefore we find

$T=\frac{2\pi (6.37\cdot 10^6 m+0.95e6 m)}{7382 m/s}=6227 s$

Converting into hours,

$T=\frac{6227 s}{3600 s/h}=1.73 h$