# Geosynchronous satellite – problems and solutions

1. What is the height of a geosynchronous satellite above Earth’s surface?

G = universal constant = 6.67 x 10^{-11} N m^{2}/kg^{2}

M = Earth’s mass = 5.97 x 10^{24} kg

T = 24 hours = 24 (3600 seconds) = 86,400 seconds = 8.64 x 10^{4} seconds

Solution

Satellite’s height above Earth’s surface = 4.22 x 10^{7} m = 4.22 x 10^{4 }km = 42,200 km

Satellite’s height above Earth’s surface = 6.38 x 10^{6} m = 6.38 x 10^{3 }km = 6380 km

Satellite’s height above Earth’s surface = satellite’s distance from Earth’s center – Earth’s radius

Satellite’s height above Earth’s surface = 42,200 km – 6380 km

Satellite’s height above Earth’s surface = 35,820 km

2. What is the speed of a geosynchronous satellite ?

Universal constant (G) = 6.67 x 10^{-11} N m^{2}/kg^{2}

Earth’s mass (m_{E}) = 5.97 x 10^{24 }kg

The distance between satellite and Earth’s center (r) = 4.22 x 10^{7} m

Solution

or use this equation

[wpdm_package id=’949′]

- Newton’s law of universal gravitation problems and solutions
- Gravitational force, weight problems, and solutions
- Acceleration due to gravity problems and solutions
- Geosynchronous satellite problems and solutions
- Kepler’s law problems and solutions problems and solutions