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?
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
