A geosynchronous satellite is one that stays above the same point on the equator

Question

A geosynchronous satellite is one that stays above the same point on the equator of the earth. Determine the height above the Earth's surface such a satellite must orbit and find it's speed. (Note that the radius in equation is measured from the center of the earth, not the surface.) You may use the following constants:

The universal gravitational constant G is 6.67 x 10-11 N m2 / kg2 .

The mass of the earth is 5.98 x 1024 kg.

The radius of the earth is 6.38 x 106 m.

Hint: In order for the satellite to remain in orbit, the gravitational force must equal the centripetal force:

Explanation / Answer

universal gravitational constant G is 6.67 x 10-11 N m2 / kg2

The mass of the earth is 5.98 x 1024 kg

The radius of the earth is 6.38 x 106 m.

We assume that the satellite is in Uniform Circular Motion (UCM). If this is true, the acceleration will be given by

a = v^2/r. (1)

This is the "a" in the force equation,

F = ma. (2)

However, the force on the satellite is also given by the gravitational force it due to the Earth:

F = GMm/(r^2) (3)

Combining (2) and (3) above, we get

Gm/r = v^2 (4)

Since the satellite goes around the Earth once each day (T = 1 day = 86400 s), its speed is

v = 2*pi*r/T (5).

Plugging (4) into (5) gives us

GM/r = (2*pi*r)^2/(T^2) (6).

We want to solve this for r, so we have

r^3 = GMT^2/[4*(pi)^2]. (7)

It's plug-and-chug time. Substituting values for G, M = 5.98 x 10^24 kg, and T, we get

= 7.54*10^22

r = 4.23 x 10^7 m. (8).

The Earth's radius is 0.64 x 10^7 meters, so that the answer is 3.59 x 10^7 m for the height above the surface.

Using the value from (8) in equation (4) and solving for v, we find that

v = 3070 m/s (9)

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