Newton’s law of gravitation gives the attraction between two masses as F = GMm ÷ r². The detail that matters is that r is measured between the centres of the two masses, not from a surface.
This lesson is part of SPM Physics gravitation. It leads into explaining centripetal force in orbital motion. Use G = 6.67 × 10⁻¹¹ N m² kg⁻² and the values given in the question.
How do I apply the law step by step?
Use four steps.
- List M, m and r, converting all lengths to metres.
- Check that r runs from centre to centre.
- Substitute into F = GMm ÷ r², keeping scientific notation.
- Give the answer in newtons, in scientific notation if it is large or small.
The two steps that fail most often are the first and second, so write them out.
Worked example 1: two small masses
Two 1000 kg masses are 2.0 m apart, centre to centre. Find the force between them.
F = (6.67 × 10⁻¹¹ × 1000 × 1000) ÷ 2.0² = (6.67 × 10⁻⁵) ÷ 4.0 = 1.7 × 10⁻⁵ N. That is 0.000017 N, which is why you cannot feel the attraction between everyday objects.
Worked example 2: a satellite
A 500 kg satellite orbits 630 km above the Earth’s surface. The Earth’s mass is 5.97 × 10²⁴ kg and its radius is 6.37 × 10⁶ m. Find the gravitational force on the satellite.
First find r. The altitude is 630 km = 0.63 × 10⁶ m. So r = 6.37 × 10⁶ + 0.63 × 10⁶ = 7.00 × 10⁶ m.
Then F = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴ × 500) ÷ (7.00 × 10⁶)². The numerator is 1.99 × 10¹⁷ and the denominator is 4.90 × 10¹³, so F ≈ 4.06 × 10³ N.
Does the answer pass an inverse-square check?
A quick check uses the surface weight. The satellite’s weight on the surface would be 500 × 9.81 = 4905 N.
The satellite is at a distance of 7.00 ÷ 6.37 = 1.10 times the Earth’s radius, so the force should be 4905 ÷ 1.10² ≈ 4905 ÷ 1.21 ≈ 4050 N. This agrees with 4.06 × 10³ N, so the calculation is reasonable.
The mistake to avoid
The common mistake is to use the altitude as r, which treats the satellite as if it were 630 km from the Earth’s centre.
| Choice of r | Force | Comment |
|---|---|---|
| r = 6.30 × 10⁵ m (altitude only) | 5.0 × 10⁵ N | Wrong, more than 100 times too large |
| r = 7.00 × 10⁶ m (radius + altitude) | 4.06 × 10³ N | Correct |
The wrong force is larger than the surface weight, which is impossible for an object higher up. That is a quick signal that r is wrong. The units and significant figure checker helps you check the exponents.
Check yourself
A 60 kg person is at a distance of 2R from the Earth’s centre, where R is the Earth’s radius. Their weight on the surface is 589 N. What is the gravitational force at 2R?
Answer
The distance has doubled, so the force falls by a factor of 2² = 4. The force is 589 ÷ 4 = 147 N.
The mass of the person has not changed. Only the force of attraction has decreased.
What to study next
Go on to explaining centripetal force in orbital motion, where this force keeps the satellite in orbit. Practise unit handling in using orbital relationships with consistent units, and record your exponent slips in the mistake log and paper-error review tool.
If you would like a teacher to go through your substitutions with you, see online one-to-one Physics tuition.