The Hubble Space Telescope (HST) orbits 569,000m above Earth's surface. Given that Earth's mass is 5.97 * 10^24 kg and its radius is 6.38 * 10^6 m, what is HST's tangential speed?

Refer to the diagram shown below.

M = 5.97 x 10²⁴ kg, mass of the earth

h = 5.69 x 10⁵ m, height of HST above the earth's surface

R = 6.38 x 10⁶ m, radius of the earth

Note that

G = 6.67 x 10⁻¹¹ (N-m²) / kg², gravitational acceleration constant.

R + h = 6.38 x 10⁶ + 5.69 x 10⁵ = 6.949 x 10⁶ m

The force between the earth and HST is

F = (GMm) / (R+h) ²

Let v = tangential velocity of the HST.

The centripetal force acting on HST is equal to F.

Therefore

m*[v² / (R+h) ] = (GMm) (R+h) ²

v² = (GM) / (R+h)

= [ (6.67 x 10⁻¹¹ (N-m²) / kg²) * (5.97 x 10²⁴ kg) ] / (6.949 x 10⁶ m)

= 5.7303 x 10⁷ (m/s) ²

v = 7.5699 x 103 m/s

Answer

The tangential speed of HST is about 7,570 m/s