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

2. What is the speed of a geosynchronous satellite ?

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

Earth’s mass (m_E) = 5.97 x 10²⁴ kg

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

Solution

or use this equation

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