Question

A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). At this distance, the free-fall acceleration is g/4.
(a) What is the satellite's orbital speed (m/s)?
(b) What is the period of revolution (min)?
(c) What is the gravitational force on the satellite (N) ?

Answer 1

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

$V_{orbital} = \sqrt{\frac{GM_E}{R}}$

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

$V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}$

$V_{orbital} = 5591.62m/s$

$V_{orbital} = 5.591*10^3m/s$

PART B) The period of satellite is given as,

$T = 2\pi \sqrt{\frac{r^3}{Gm_E}}$

$T = \frac{2\pi r}{V_{orbital}}$

$T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}$

$T = 238.61min$

PART C) The gravitational force on the satellite is given by,

$F = ma$

$F = \frac{1}{4} mg$

$F = \frac{270*9.8}{4}$

$F = 661.5N$