A spy satellite is in circular orbit around Earth. It makes one revolution in 4.00 h. Mass of Earth is 5.974 × 1024 kg, radius of Earth is 6371 km and Gravitational constant G is = 6.674 × 10−11 N·m2/kg2.1) How high above Earth’s surface is the satellite?2) What is the satellite’s acceleration?

A. 6419 km

B. 9.822 m/s²

Explanation:

Parameters given:

Mass of earth, M = 5.974 * 10^24 kg

Radius of earth, R = 6371 km = 6.371 * 10^6 m

Gravitational constant, G = 6.674 * 10^(-11) Nm²/kg²

Orbital period, T = 4 hours = 14400 secs

A. Height of the satellite above the earth's surface can be found by subtracting the earth's radius, R, from the radius of the orbit of the satellite, Ro.

The radius of the orbit, Ro, is given as:

Ro = ³√(GMT²/4π²)

Ro = ³√[(6.674 * 10^(-11) * 5.974 * 10^24 * 14400²)/4π²]

Ro = ³√[(8.268 * 10²¹)/3.948]

Ro = ³√(2.084 * 10²¹)

Ro = 1.279 * 10^7 m = 12790 km

Height = Ro - R

Height = 12790 - 6371

Height = 6419 km

B. Orbital acceleration:

The orbital acceleration is equal to the acceleration due to gravity. This can be proven with the formula

g = GM/R²

g = (6.674 * 10^(-11) * 5.974 * 10^24)/(6.371 * 10^6)²

g = 9.822 m/s²

d is the correct answer