Looking up the night sky, a satellite is found to be in a circular orbit around the Earth.

Question

er the course of several nights, it is measured that the satellite has an orbital period of 0.7 days. The mass of the satellite is unknown.

The Earth has a mass of 5.97*10^24 kg and a radius of 6.37*10^6 meters.

(a) How far is the satellite from the center of the Earth?

 

(b) What is the average alttitude of the satellite?

Answer 1

Let

T = 0.7 days = 0.7×24×3600 s = 60480 s be the satellite's period,

M = 5.97×10²⁴ kg be the mass of the Earth,

R = 6.37×10⁶ m be the Earth's radius,

m be the mass of the satellite (unknown),

r be the distance of the satellite from the center of the Earth (to be computed),

G = 6.6743×10^-11 m³/kgs² be the gravitational constant.

The length of the satellites orbit is 2πr, and therefore its velocity is

v = 2πr/T

The only force acting on the satellite is gravity, and it is of magnitude

F = GMm/r²

The setellite is experiencing centripetal acceleration

a = v²/r = (2πr/T)²/r = 4π²r/T²

By Newton's Second Law

F = ma

Substitute the above quantities

GMm/r² = 4mπ²r/T²

Simplify into

r = (GMT²/4π²)^1/3

Substitute actual numbers

r = (6.6743×10^-11 × 5.97×10²⁴ × 60480²/(4π²))^1/3 ≈ 33.3 × 10⁶ m

(a) The satellite is about 33300 km from the center of the Earth.

(b) The average altitude is r - R = 33.3×10⁶ - 6.37×10⁶ ≈ 27×10⁶ m

The average altitude is about 27000 km.

That is a little lower that geostationary.