Question

What is the period of a satellite in a circular orbit just above the surface of the Moon? The Moon’s mass is 7.36 × 1022 kilograms and its radius is 1.738 × 106 meters.

Answer 1

The period of a satellite in a circular orbit just above the surface of the Moon can be calculated using Kepler's third law.

Step-by-step explanation:

The period of a satellite in a circular orbit just above the surface of the Moon can be calculated using Kepler's third law. The period of an orbit is related to the radius of the orbit. For the Moon, the orbital radius is 3.84 × 10^8 m and the period is 27.3 days. To calculate the period of the satellite, we need to add the height of the satellite above the surface of the Earth to the radius of the Moon. Given that the height is 1,500 km, the total radius becomes 7,880 km. With these values, we can calculate the period of the satellite.

Answer 2

F / m = a = G M / R^2 formula for gravitational attraction

a = ω^2 R formula for centripetal force for an object with radius R

G M / R^2 = ω^2 R equating accelerations

ω = (G M / R^3)^1/2

ω = 2 pi f = 2 pi / P for circular motion

P = 2 pi (R^3 / (G M))^1/2

R^3 / (G M) = (1.738E6)^3 / (6.67E-11 * 7.36E22) = 1.07E6

P = 6.28 * 1034 sec = 6500 sec = 1.80 hrs