Newton's Law of Gravitation — Apple and Earth (Worked Example)

Given

• m₁ = 0.18 kg (apple)

• m₂ = 5.97 × 10²⁴ kg (Earth)

• r = 6.371 × 10⁶ m

• G = 6.674 × 10⁻¹¹ N·m²/kg²

Find

The gravitational force F.

Principle

Newton's gravitation law applied at Earth's surface — it must reproduce the familiar weight mg, a good self-check.

Formula

Solution

1. Substitute: $F = \frac{(6.674 \times 10^{-11})(0.18)(5.97 \times 10^{24})}{(6.371 \times 10^6)^2}$

2. Numerator: $7.17 \times 10^{13}$; denominator: $4.059 \times 10^{13}$

3. Divide: $F = 1.77\ \mathrm{N}$

Unit check

N·m²/kg² × kg² / m² = N — a force, correct.

What this means physically

1.77 N — exactly the apple's weight mg = 0.18 × 9.8 = 1.76 N (rounding). Newton's universal law and the local rule F = mg are the same physics; g is just Gm_earth/r² in disguise.