Question

The Hubble telescope’s orbit is 5. 6 × 10^5 meters

above Earth’s surface. The telescope has a mass

of 1. 1 × 10^4 kilograms. Earth exerts a gravitational

force of 9. 1 × 10^4 newtons on the telescope. The

magnitude of Earth’s gravitational field strength

at this location is

(1) 1. 5 × 10^−20 N/kg (3) 8. 3 N/kg

(2) 0. 12 N/kg (4) 9. 8 N/kg

Answer 1

F = G M m / R^2 where R is distance from center of earth

F2 / F1 = (R1 / R2)^2 equating force at R2 to force at R1 (the problem is not too clear - will assume that force at R2 is 9.1E4 N/m^2

R2 = (6.37E6 + .091E6) m = 6.46E6

R1 = 6.37E6 m

(R2 / R1)^2 = 1.03

F2 = 1 / 1.03 * 9.1E4 = 8.85E4 N gravitational force on telescope

E = F / m = 8.85E4 / 1.1E4 = 8.04 N/kg

(3) appears to be closest