13.2 Gravitational Force Between Point Masses

2026 Syllabus Objectives

By the end of this topic, you should be able to:

  1. Understand that for a point outside a uniform sphere, the entire mass of the sphere can be treated as if it is concentrated at a single point at its centre.
  2. Recall and use Newton's Law of Gravitation: F = Gm₁m₂ / r²
  3. Analyse circular orbits in gravitational fields by linking gravitational force to the centripetal acceleration it produces.
  4. Understand the key features of a geostationary orbit.

1. The Point Mass Approximation

https://cambridgeassistant.com

What is a "point mass"?

A point mass is an object whose entire mass is assumed to be concentrated at a single point — like a tiny dot. In reality, planets and stars are huge, but we can still treat them as point masses under the right conditions.

When can we treat a sphere as a point mass?

Imagine a large, uniform sphere — like the Earth or the Sun. Uniform means the mass is spread evenly throughout the sphere; no part is denser than another.

Here is the key rule:

For any point located outside a uniform sphere, the gravitational effect of the entire sphere is exactly the same as if all its mass were squashed into a single point at its centre.

This means when we calculate the gravitational force between, say, the Earth and a satellite, we measure the distance r from the centre of the Earth to the satellite — not from the surface.

Why does this work?

Because the mass is distributed evenly, every part of the sphere pulls on an outside object. When you add all those tiny pulls together mathematically, they combine to act as one force — pulling exactly as if all the mass were at the centre.

When is it valid?

  • The sphere must be uniform (evenly distributed mass).
  • The point or object must be outside the sphere.
  • The distance between the objects must be much larger than the size of the objects themselves.

This is why planets, stars, and satellites can all be treated as point masses when calculating gravitational forces — the distances involved are enormous compared to the sizes of the objects.

2. Newton's Law of Gravitation

https://cambridgeassistant.com

The Law

Every mass in the universe attracts every other mass. Newton's Law of Gravitation tells us exactly how strong that attraction is.

The gravitational force between two point masses is:

  • Directly proportional to the product of the two masses (bigger masses → bigger force)
  • Inversely proportional to the square of the distance between their centres (further apart → much weaker force)

The Formula

F=Gm1m2r2F = \frac{G m_1 m_2}{r^2}

SymbolMeaningUnit
FGravitational force between the two massesNewtons (N)
GUniversal Gravitational Constant = 6.67 × 10⁻¹¹N m² kg⁻²
m₁Mass of the first objectkg
m₂Mass of the second objectkg
rDistance between the centres of the two massesmetres (m)

Important: r is always measured centre-to-centre, not surface-to-surface.

The Inverse Square Law

The "r²" in the denominator makes this an inverse square law. This means:

  • If the distance doubles (×2), the force becomes 4 times weaker (÷4), because 2² = 4.
  • If the distance triples (×3), the force becomes 9 times weaker (÷9), because 3² = 9.

Key features of the gravitational force

  • It is always attractive — gravity only pulls, never pushes.
  • It acts along the straight line joining the centres of the two masses.
  • It is a Newton's Third Law pair — if Earth pulls the Moon with force F, the Moon pulls Earth with the same force F in the opposite direction.
  • Gravity is an extremely weak force. Two 1 kg masses placed 1 m apart only attract each other with a force of about 6.67 × 10⁻¹¹ N — far too small to feel!

Worked Example 1

Calculate the gravitational force between a 60 kg person and the Earth.

Given: Mass of Earth M = 5.99 × 10²⁴ kg, radius of Earth R = 6.4 × 10⁶ m, G = 6.67 × 10⁻¹¹ N m² kg⁻²

F=Gm1m2r2=(6.67×10−11)(60)(5.99×1024)(6.4×106)2F = \frac{G m_1 m_2}{r^2} = \frac{(6.67 \times 10^{-11})(60)(5.99 \times 10^{24})}{(6.4 \times 10^6)^2}

F≈585 NF \approx 585 \text{ N}

This is simply the person's weight — the gravitational pull of the Earth on them.

Worked Example 2

A satellite of mass 6500 kg orbits at 2000 km above Earth's surface. The gravitational force is 37 kN. Calculate the mass of the Earth.

Step 1: Find r (centre-to-centre distance)

r = radius of Earth + height above surface = 6400 km + 2000 km = 8400 km = 8.4 × 10⁶ m

Step 2: Rearrange F = Gm₁m₂/r² for m₂ (mass of Earth)

m2=Fr2Gm1=(37×103)(8.4×106)2(6.67×10−11)(6500)m_2 = \frac{F r^2}{G m_1} = \frac{(37 \times 10^3)(8.4 \times 10^6)^2}{(6.67 \times 10^{-11})(6500)}

m2≈6.0×1024 kgm_2 \approx 6.0 \times 10^{24} \text{ kg}

Exam tip: A very common mistake is forgetting to add the planet's radius to the height above the surface. Always calculate r as: r = radius of planet + height above surface.

Sign in to view full notes