H2 Physics 8 - Gravitational Fields: Geostationary Orbit Radius

Step 1 - Set up the problem

A satellite is in geostationary orbit above the equator.

Step 1 - Set up the problem

Given: Earth's mass M = 5.97 times 10 to the 24 kg, G = 6.67 times 10 to the minus 11 N m squared per kg squared.

Step 1 - Set up the problem

A geostationary orbit has a period of exactly 24 hours.

Step 2 - Equate gravitational force to centripetal force

Gravity provides the centripetal force.

Step 2 - Equate gravitational force to centripetal force

GMm/r squared = m times 4 pi squared r / T squared.

Step 2 - Equate gravitational force to centripetal force

Mass m cancels - orbital radius is independent of satellite mass.

Step 3 - Rearrange for r

Rearranging: r cubed = G M T squared over four pi squared.

Step 3 - Rearrange for r

This is Kepler's third law for circular orbits.

Step 3 - Rearrange for r

The orbital radius depends only on the period and the central body's mass.

Step 4 - Substitute values

T = 86400 s, so T squared = 7.46 times 10 to the 9.

Step 4 - Substitute values

Numerator: G times M times T squared.

Step 4 - Substitute values

Powers of ten: -11 + 24 + 9 = 22.

Step 4 - Substitute values

Coefficient: 6.67 times 5.97 times 7.46 = 297. Numerator = 2.97 times 10 to the 24.

Step 5 - Find the orbital radius

Cube root of 7.52 times 10 to the 22.

Step 5 - Find the orbital radius

Cube root: 10 to the 7.33 times 1.96.

Step 5 - Find the orbital radius

r = 4.22 times 10 to the 7 m, or about 42,200 km.

Step 5 - Find the orbital radius

About 36,000 km above the surface - the well-known geostationary altitude.