Newton's law of gravitation

CIE A-Level Physics· Unit 17: Gravitational fields· 15 min read

1. Newton's Law of Universal Gravitation★★☆☆☆⏱ 4 min

📘 Definition

Newton's law of universal gravitation

$F$ = force, $m_{1}, m_{2}$ = masses, $r$ = separation, $G$ = gravitational constant

Every particle of matter attracts every other particle with a force directly proportional to the product of their masses, and inversely proportional to the square of the distance between their centers.

Example:

Applies to all point masses, and uniform spherical masses when $r$ is center-to-center distance

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

📐 Worked Example

Calculate the gravitational force between two point masses of 10 kg and 20 kg separated by 0.5 m.

1
List known values:
2
$m_1 = 10 \text{ kg}, \quad m_2 = 20 \text{ kg}, \quad r = 0.5 \text{ m}, \quad G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$
3
Substitute into Newton's gravitation formula:
4
$F = \frac{(6.67 \times 10^{-11})(10)(20)}{(0.5)^2}$
5
Simplify and calculate:
6
$F = \frac{1.334 \times 10^{-8}}{0.25} = 5.34 \times 10^{-8} \text{ N} \approx 5.3 \times 10^{-8} \text{ N}$

2. The Inverse Square Relationship★★★☆☆⏱ 5 min

Gravitational force follows an inverse square law, meaning the magnitude of the force is proportional to $1/ r^{2}$ . This relationship is very common in field physics, and is frequently tested in CIE multiple choice questions.

📘 Definition

Inverse square law (gravitation)

If the distance between two masses increases by a factor $k$ , the gravitational force between them decreases by a factor $k^{2}$

Example:

Doubling distance ( $k = 2$ ) reduces force to $1/4$ of its original value

📐 Worked Example

The gravitational force between two masses at separation $r$ is $F$ . What is the new force when separation is increased to $2.5 r$ ?

1
Write the original and new force equations:
2
$F = \frac{G m_1 m_2}{r^2}, \quad F_2 = \frac{G m_1 m_2}{(2.5r)^2}$
3
Divide to eliminate all constant values ( $G, m_{1}, m_{2}$ ):
4
$\frac{F_2}{F} = \frac{r^2}{(2.5r)^2} = \frac{1}{2.5^2} = \frac{1}{6.25} = 0.16$
5
Final result:
6
$F_2 = 0.16F$

3. Applications to Extended Spherical Masses★★★☆☆⏱ 6 min

For uniform spherical objects like planets and stars, we can treat the entire mass as concentrated at the center of the sphere, so Newton's law of gravitation applies directly. The key rule is to always use the center-to-center distance for $r$ , not the distance between the surfaces of the objects.

📐 Worked Example

Calculate the gravitational force between Earth (mass $5.97 \times 1 0^{24}$ kg, radius 6370 km) and a 1000 kg satellite orbiting 200 km above Earth's surface.

1
Calculate the center-to-center distance:
2
$r = 6370 \text{ km} + 200 \text{ km} = 6570 \text{ km} = 6.57 \times 10^6 \text{ m}$
3
Substitute values into the formula:
4
$F = \frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})(1000)}{(6.57 \times 10^6)^2}$
5
Calculate numerator and denominator separately:
6
$\text{Numerator} = 3.98 \times 10^{17}, \quad \text{Denominator} = 4.32 \times 10^{13}$
7
Final result:
8
$F \approx 9.2 \times 10^3 \text{ N}$

✓ Quick check

Test your understanding of distance measurement:

What value of $r$ should you use for a 5 kg mass resting on Earth's surface?
- A: The radius of Earth (distance from Earth's center to the mass's center)
- B: The radius of the 5 kg mass only
- C: The distance between the surface of the mass and Earth's surface
Reveal answer
A —
Correct! Earth acts as a point mass at its center, so $r$ is the Earth's radius.

4. Common Pitfalls

Wrong move:

Using distance between surfaces instead of center-to-center distance

Why:

Most problems give orbit height above the surface, so students forget to add the planet's radius

Correct move:

Always add the radius of the planet/body to the surface height to get center-to-center separation

Wrong move:

Using a positive exponent for $G$ ( $6.67 \times 1 0^{11}$ instead of $6.67 \times 1 0^{- 11}$ )

Why:

The negative exponent is easy to miss when entering values into a calculator

Correct move:

Double-check the exponent of $G$ after entering it into your calculator

Wrong move:

Claiming the larger mass exerts a larger force on the smaller mass

Why:

Students confuse force with acceleration, forgetting Newton's third law

Correct move:

Gravitational force is an action-reaction pair: the force on each mass is equal in magnitude

Wrong move:

Squaring the product of masses in the formula

Why:

Students confuse the inverse square term with the product of masses term

Correct move:

Only the separation $r$ is squared in the formula $F = \frac{G m _{1} m _{2}}{r ^{2}}$

5. Quick Reference Cheatsheet

Concept	Formula/Value	Key Exam Note
Newton's gravitation law	$F = \frac{G m _{1} m _{2}}{r ^{2}}$	For point masses/uniform spheres
Gravitational constant $G$	$6.67 \times 1 0^{- 11} N m^{2} kg^{- 2}$	Always use negative exponent
Inverse square proportionality	$F_{2} = F_{1} (\frac{r _{1}}{r _{2}})^{2}$	Square the distance ratio
Center-to-center distance	$r = R_{planet} + h_{orbit}$	Add planet radius to surface height

6. Frequently Asked

Can I use Newton's law for non-spherical extended objects?

In CIE A-Level exams, you will only ever be asked to apply Newton's law to point masses or uniform spheres. For non-spherical objects, you will not need to perform calculations.

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · Paper 1
Inverse square proportionality question
2022 · Paper 2
Force calculation between Earth and satellite
2021 · Paper 1
Action-reaction force for gravitation

What's Next

Newton's law of gravitation is the foundational concept for all further topics in gravitational fields for CIE A-Level Physics. The force formula is used to derive gravitational field strength around planets, gravitational potential energy, and the rules for stable planetary and satellite orbits. Understanding the inverse square relationship is also critical for solving problems on variation of gravitational acceleration with altitude and escape velocity. Mastery of this sub-topic will let you confidently tackle all higher level gravitation problems in your exam.