Newton's Law of Gravitation

In 1687, in his Principia Mathematica, Isaac Newton proposed what is still one of the most famous equations in science: the inverse-square law of gravitation. It states that every pair of masses in the Universe attracts every other pair with a force proportional to the product of the masses and inversely proportional to the square of the distance between their centres. With this one equation, Newton explained both the fall of an apple on Earth and the orbit of the Moon around the Earth — unifying terrestrial and celestial mechanics for the first time in history.

This lesson is OCR A-Level Physics A Module 5.4 (Gravitational fields). We state Newton's law, apply it to point masses and spheres, derive the gravitational field strength of a planet, and work through several exam-style calculations.

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Statement of the Law

The gravitational force between two point masses is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

In symbols:

F = −GMm / r²

Where:

• F is the gravitational force on each mass due to the other (in newtons),
• M and m are the two masses (in kilograms),
• r is the distance between their centres of mass (in metres),
• G = 6.67 × 10⁻¹¹ N m² kg⁻² is the universal gravitational constant,
• the minus sign indicates that the force is attractive — it acts to pull the two masses together.

Gravity is always attractive: there is no "negative mass" and no repulsive gravity in classical physics.

The Gravitational Constant G

G is one of the fundamental constants of nature, along with the speed of light c and the Planck constant h. It was first measured by Henry Cavendish in 1798 using a delicate torsion-balance experiment — an extraordinary feat given the weakness of gravity.

The modern value is

G = 6.674 × 10⁻¹¹ N m² kg⁻²

To three significant figures (as used in A-Level exams) you can take G = 6.67 × 10⁻¹¹ N m² kg⁻². OCR quotes this value in the data booklet, so you do not need to memorise it — but you will certainly need to use it.

Exam Tip: Students often forget how tiny G is. Gravity is by far the weakest of the four fundamental forces. It only dominates on astronomical scales because all masses are positive and the force is always attractive, so the contributions all add up. For single small objects, electromagnetic forces are overwhelmingly stronger.

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Inverse-Square Geometry

The 1/r² dependence comes from geometry. The total "gravitational flux" emanating from a mass M is constant; it spreads out over a sphere of area 4πr² at distance r. The flux per unit area — the field strength — therefore falls as 1/r².

graph TD
    A[Mass M] --> B[Spherical shell, radius r<br/>area = 4πr²]
    A --> C[Spherical shell, radius 2r<br/>area = 16πr²]
    A --> D[Spherical shell, radius 3r<br/>area = 36πr²]

Doubling the distance reduces the force to 1/4. Tripling the distance reduces it to 1/9. Halving the distance quadruples the force.

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Force Between Extended Bodies

Newton's law is stated for point masses. What about real planets and stars? Newton himself proved — using his newly invented calculus — a theorem known as the shell theorem:

A spherically symmetric body acts, from outside, as if all its mass were concentrated at its centre.

This is an extraordinary simplification. It means that the gravitational pull of the Earth on an apple at the surface is exactly the same as the pull of a point mass of 5.97 × 10²⁴ kg located 6.37 × 10⁶ m below (at the Earth's centre). You can therefore use the simple point-mass formula for any spherical planet, star, or moon — provided the point is outside the body.

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Gravitational Field Strength of a Spherical Mass

Combining Newton's law F = GMm/r² (ignoring the minus sign and taking magnitudes) with the definition g = F/m:

g = GM / r²

This is the field strength at a distance r from the centre of a spherical body of mass M. Memorise it — it is one of the most important equations in OCR A-Level gravitational fields.

Because g ∝ 1/r², the field strength falls off very fast. Double your distance from the Earth's centre and g drops to a quarter.

Calculating g at the Earth's Surface

Let us verify the value 9.81 N kg⁻¹. The Earth's mass is M = 5.97 × 10²⁴ kg and its radius is R = 6.37 × 10⁶ m.

g = GM / R² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.37 × 10⁶)²
           = 3.98 × 10¹⁴ / 4.06 × 10¹³
           = 9.81 N kg⁻¹ ✓

The agreement is exact (to 3 sig. fig.), which is impressive confirmation that Newton's law and the Earth's measured size are mutually consistent.

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Worked Example 1 — Field Strength at Altitude

Calculate the gravitational field strength at the orbit of the International Space Station, 400 km above the Earth's surface. Take R_E = 6.37 × 10⁶ m, M_E = 5.97 × 10²⁴ kg.

The distance from the Earth's centre is

r = 6.37 × 10⁶ + 0.40 × 10⁶ = 6.77 × 10⁶ m

Then

g = GM / r² = (6.67 × 10⁻¹¹ × 5.97 × 10²⁴) / (6.77 × 10⁶)²
           = 3.98 × 10¹⁴ / 4.58 × 10¹³
           = 8.69 N kg⁻¹