Question

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?

Answer 1

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