Question

A satellite is in a circular orbit 8200 km above the Earth’s surface; i.e., it moves on a circular path under the influence of nothing but the Earth’s gravity. Find the speed of the satellite. The radius of the Earth is 6.37 × 106 m, and the acceleration of gravity at the satellite’s altitude is 1.87965 m/s 2 . Answer in units of km/s

Answer 1

5.23km/s

Explanation:

Given

Radius of Earth = 6.37 * 10^6 m

Altitude of Satellite = 8200km = 8200 * 10³m = 8.2 * 10^6 m

Gravity Acceleration on Satellite Altitude = 1.87965m/s²

For a satellite to remain in circular orbit, then it means the acceleration of gravity must be exact as the centripetal acceleration.

Centripetal Acceleration = V²/R

So, Acceleration of Gravity (A)= Centripetal Acceleration = V²/R

Make V the subject of formula

A = V²/R

V² = AR

V = √AR

Where R = (radius of earth) + (altitude of satellite)

R = 6.37 * 10^6 + 8.2 * 10^6

R = 14.57 * 10^6m

A = 1.87965m/s²

V = √(1.87965 * 14.57x10^6)

V = √27386500.5

V = 5233.211299001789

V = 5233.2113 m/s ------- Approximated

V = 5.23km/s