Basic Physics

# 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 m2/kg2

M = Earth’s mass = 5.97 x 1024 kg

T = 24 hours = 24 (3600 seconds) = 86,400 seconds = 8.64 x 104 seconds

Solution Satellite’s height above Earth’s surface = 4.22 x 107 m = 4.22 x 104 km = 42,200 km

Satellite’s height above Earth’s surface = 6.38 x 106 m = 6.38 x 103 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

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2. What is the speed of a geosynchronous satellite ?

Universal constant (G) = 6.67 x 10-11 N m2/kg2

Earth’s mass (mE) = 5.97 x 1024 kg

The distance between satellite and Earth’s center (r) = 4.22 x 107 m

Solution or use this equation [wpdm_package id=’949′]

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