Question

The Hubble telescope’s orbit is 5.6 × 105 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth exerts a gravitational force of 9.1 × 104 Newtons on the telescope. The magnitude of Earth’s gravitational field strength at this location is

Answer 1

The gravitational field is the Force divided by the mass

Call g the gravitational fiel, F the force exerted by the earth and m the mass of the telescope.

g = F / m


g=9.1x10^4 N / 1.1 x 10^4 kg = 8.27 N/kg

Note that the unit N/kg is equivalent to m/s^2