Newton's Law of Gravitation Revision Notes | A Level AQA Physics

1. Why Does Gravity Reach Across Space?

The Moon is roughly 384,000 km away from Earth. Nothing physically connects them — no rope, no rod, no contact at all. Yet the Moon stays in orbit, pulled continuously towards Earth. The question Newton asked in the 17th century was: what is the rule that governs this pull, and does the same rule apply to an apple falling from a tree?

The answer he found is one of the most powerful ideas in physics: gravity is a universal attractive force. It acts between every pair of objects that have mass, anywhere in the universe, regardless of what those objects are made of or how far apart they are.

This universality is worth pausing on. The same force that pulls you towards the ground also holds the International Space Station in orbit, keeps the planets circling the Sun, and binds galaxies together. The law does not change depending on where you are — that is what makes it universal.

2. The Equation: Quantifying the Gravitational Pull

Knowing that gravity is universal is useful, but physics demands precision. How large is the force? Newton showed that the magnitude depends on exactly two things: the masses involved and the distance between them.

Notice that $r$ is the separation between the centres of the two masses, not between their surfaces. This matters enormously when you are working with planets or stars, where the radius of the object itself is significant.

The constant $G = 6.67 \times 10^{-11}$ N m² kg⁻² is extraordinarily small. That is why you do not feel a gravitational pull towards the person sitting next to you — the masses involved in everyday life are far too small to produce a noticeable force. Only when at least one mass is planetary in scale does gravity become significant.

3. Unpacking the Inverse-Square Law

The $r^2$ in the denominator is the most important feature of the equation. It means the force does not halve when you double the distance — it drops to a quarter. This is the inverse-square law, and it appears in several areas of A-level physics, including electric fields and radiation intensity.

Picture the gravitational influence of a mass spreading outward in all directions, like light from a lamp. As you move further away, that influence is spread over a larger and larger spherical surface. The surface area of a sphere is $4\pi r^2$, so the intensity of the effect falls off with $r^2$. That geometric argument is why the inverse-square form appears naturally.

The table above is worth memorising as a pattern. Halving the separation quadruples the force; tripling the separation reduces it to one ninth. Examiners frequently test this proportional reasoning without requiring a full numerical calculation.

4. The Point Mass Assumption

The equation $F = \frac{Gm_1m_2}{r^2}$ is derived for point masses — objects whose physical size is negligible compared to their separation. In practice, real objects have finite size, so when is it valid to use this formula?

For a uniform sphere, Newton proved that the gravitational effect outside the sphere is identical to that of a point mass of the same total mass located at the sphere's centre. This means the formula applies to planets and stars as long as $r$ is measured from their centres and you are outside their surfaces.

The assumption breaks down when the separation is comparable to the size of the objects themselves — for example, calculating the force between two touching spheres of similar size. At A-level, you will almost always be working with separations much larger than object radii, so the point mass model is appropriate.

5. Worked Calculations

Theory becomes exam-ready through practice. The calculation below follows the structure AQA mark schemes reward: state the formula, substitute with units, then evaluate. Work through each step before revealing the next.

Worked example: Calculating the Gravitational Force Between Earth and the Moon

Calculate the gravitational force between the Earth and the Moon. Use: mass of Earth $m_1 = 5.97 \times 10^{24}$ kg, mass of Moon $m_2 = 7.34 \times 10^{22}$ kg, centre-to-centre separation $r = 3.84 \times 10^{8}$ m, $G = 6.67 \times 10^{-11}$ N m² kg⁻².

1. $F = \frac{Gm_1m_2}{r^2}$ — State Newton's Law of Gravitation. This earns the method mark even if arithmetic errors follow.
2. $r^2 = (3.84 \times 10^{8})^2 = 3.84 \times 3.84 \times 10^{16} = 14.7456 \times 10^{16} = 1.47456 \times 10^{17}$ m² — Square the separation carefully. $3.84 \times 3.84 = 14.7456$, and $(10^8)^2 = 10^{16}$. Never skip this step — squaring errors are the most common source of lost marks.
3. $Gm_1m_2 = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 7.34 \times 10^{22}$\n$= (6.67 \times 5.97 \times 7.34) \times 10^{-11+24+22}$\n$= (6.67 \times 5.97 \times 7.34) \times 10^{35}$\n$6.67 \times 5.97 = 39.8199$\n$39.8199 \times 7.34 = 292.278$\n$= 292.278 \times 10^{35} = 2.92278 \times 10^{37}$ N m² — Collect powers of ten separately: $10^{-11} \times 10^{24} \times 10^{22} = 10^{35}$. Multiply the decimal parts step by step to get 292.278. Since $292.278 = 2.92278 \times 10^2$, the full result is $2.92278 \times 10^{35+2} = 2.92278 \times 10^{37}$ N m².
4. $F = \frac{2.92278 \times 10^{37}}{1.47456 \times 10^{17}} = \frac{2.92278}{1.47456} \times 10^{37-17} = 1.9823 \times 10^{20}$ N — $2.92278 \div 1.47456 \approx 1.98$. Subtract the powers: $10^{37} \div 10^{17} = 10^{20}$.
5. $F \approx 1.98 \times 10^{20}$ N — Round to 3 significant figures to match the data given. Always include the unit — a bare number scores zero for the accuracy mark.

Answer: $F \approx 1.98 \times 10^{20}$ N

Now try a proportional reasoning question — the type that does not require a full numerical calculation but tests whether you understand the structure of the law.

Worked example: Proportional Reasoning with Newton's Law

Two satellites, X and Y, orbit the same planet. Satellite X has mass $m$ and orbits at distance $r$. Satellite Y has mass $3m$ and orbits at distance $2r$. Calculate the ratio $\frac{F_Y}{F_X}$.

1. $F_X = \frac{GmM}{r^2}$ \quad $F_Y = \frac{G(3m)M}{(2r)^2}$ — $M$ is the planet's mass, which is the same for both. Keep it as a symbol — it will cancel.
2. $(2r)^2 = 2 \times 2 \times r^2 = 4r^2$ $F_Y = \frac{3GmM}{4r^2}$ — Squaring $2r$ gives $4r^2$, not $2r^2$. This is the most common error in proportional questions.
3. $\frac{F_Y}{F_X} = \frac{\frac{3GmM}{4r^2}}{\frac{GmM}{r^2}} = \frac{3GmM}{4r^2} \times \frac{r^2}{GmM} = \frac{3}{4}$ — $GmM$ and $r^2$ cancel completely. The ratio depends only on the numerical factors introduced by the changes in mass and distance.
4. $\frac{F_Y}{F_X} = \frac{3}{4} = 0.75$ — Satellite Y experiences three quarters of the force on Satellite X. The tripling of mass increased the force, but the doubling of distance (which quadruples $r^2$) reduced it more, giving a net decrease.

Answer: $\frac{F_Y}{F_X} = \frac{3}{4}$

6. Newton's Third Law and the Symmetry of Gravity

Here is something that surprises many students: the gravitational force the Earth exerts on you is exactly equal in magnitude to the force you exert on the Earth. The formula $F = \frac{Gm_1m_2}{r^2}$ is symmetric — swap $m_1$ and $m_2$ and you get the same value of $F$. This is Newton's Third Law applied to gravity.

The forces are equal in magnitude but opposite in direction. Earth pulls you downward; you pull Earth upward with the same force. The reason Earth does not visibly accelerate towards you is that its mass is so enormous that the resulting acceleration ($a = F/m$) is negligibly small — far below any measurable threshold.

7. Exam Technique and Mark-Winning Strategies

AQA questions on Newton's Law of Gravitation fall into three predictable categories: direct calculation, proportional reasoning, and conceptual explanation. Each rewards a slightly different approach, and knowing which type you are facing saves time under exam conditions.

For calculate questions, always write the formula first — this earns the method mark even if your arithmetic goes wrong. Substitute values with units in the same line, then evaluate. Never substitute a value for $r$ without first checking whether you need $r^2$; forgetting to square the separation is the single most penalised error in this topic.

For explain or describe questions about how force changes with distance or mass, use the proportionality statements explicitly. Write $F \propto \frac{1}{r^2}$ and then state the consequence in words. Examiners award marks for the proportionality statement and the correct direction of change separately.

Newton's Law of Gravitation — $F = \frac{Gm_1m_2}{r^2}$ — encapsulates two spec requirements in a single equation: gravity as a universal attractive force between all matter, and the precise inverse-square dependence on separation. Master the proportional reasoning and the centre-to-centre distance convention, and this topic becomes one of the most reliably mark-scoring areas of the Fields unit.