A GPS satellite orbits at an altitude of 20,200 km above the surface of Earth. What is the speed of the satellite? (Recall that RE = 6.37 ✕ 106 m.)

Answer 1

Because the only force that acts on the satellite is force of gravity, centripetal acceleration of the body equal to the acceleration due to gravity on that height, and we can write

a_c = g

We know that centripetal acceleration

a_c = v² / r,

here r = R + h - radius of orbital motion with respect to the center of Earth,

h = height of the satellite above the surface of the Earth.

Hence,

v = √ g·r

Acceleration due to gravity on this height

g = GM / r²

On the surface of the Earth

g₀ = GM / R² = 9.8 m / s²

So, now I will express

GM = g₀ R².

Then for g we have:

g = g₀ R² / r ²

Now we get for velocity

v = √ g₀ R² / (R + h)

Finally, v = R √ g₀ / (R + h)

If you put in this formula R = 6.37 x 10⁶ m , g₀ = 9.8 m / c² and h = 2.02 x 10⁷ m, then you will find numerical value for the velocity.

Pay attention that your number is less than 7.9 km /s - the so - called I cosmic speed, when the body rotates close to the Earth's surface.