Newton's Law of Gravitation — Satellite in Orbit (Worked Example)

Medium

Problem

A 1200 kg communications satellite orbits 400 km above Earth's surface. What gravitational force does Earth exert on it? (Earth mass 5.97 × 10²⁴ kg, Earth radius 6371 km, G = 6.674 × 10⁻¹¹ N·m²/kg²)

Given

m = 1200 kg
M = 5.97 × 10²⁴ kg
r = 6371 + 400 = 6771 km = 6.771 × 10⁶ m
G = 6.674 × 10⁻¹¹ N·m²/kg²

Find

The gravitational force F.

Principle

Orbital motion is free fall: Earth's gravity provides the entire centripetal force, computed with r measured from Earth's centre.

Formula

F = G\frac{Mm}{r^2}

F = GmM/r²

Solution

Orbital radius: $r = 6.371 \times 10^6 + 4.0 \times 10^5 = 6.771 \times 10^6\ \mathrm{m}$
Substitute: $F = \frac{(6.674 \times 10^{-11})(5.97 \times 10^{24})(1200)}{(6.771 \times 10^6)^2}$
Numerator: $4.781 \times 10^{17}$ ; denominator: $4.585 \times 10^{13}$
Divide: $F = 1.04 \times 10^4\ \mathrm{N}$

1.04 × 10⁴ N

Unit check

N·m²/kg² × kg² / m² = N — correct.

What this means physically

About 10,400 N — nearly 90% of the satellite's surface weight. Astronauts feel weightless not because gravity vanished, but because they are in continuous free fall around Earth.

Common mistake

Using 400 km as r. The 400 km is the altitude; r must include Earth's 6371 km radius — the single most common error in orbit problems.