# Newton's law of gravitation

> CIE A-Level Physics · 9702
> Source: https://www.owlsprep.com/study/cie-9702-u17-newton-s-law-of-gravitation/

This module covers Newton's universal law of gravitation, the inverse square relationship between gravitational force and separation, and how to apply the law to calculate force between masses for CIE exam problems.

**Prerequisites:** [Newton's laws of motion](https://www.owlsprep.com/study/cie-9702-forces-newtons-laws/); [Introduction to gravitational fields](https://www.owlsprep.com/study/cie-9702-u17-introduction-to-gravitational-fields/)

## Learning objectives

- State Newton's universal law of gravitation
- Apply the inverse square relationship to gravitational force
- Calculate gravitational force between two masses
- Correctly find center-to-center separation for extended objects

## Newton's Law of Universal Gravitation

**Newton's law of universal gravitation** — 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.

*Notation:* $F$ = force, $m_1, m_2$ = masses, $r$ = separation, $G$ = gravitational constant

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

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

> **info**
>
> Gravitational force is always attractive, unlike electrostatic force which can be attractive or repulsive.

**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. $$F = \frac{1.334 \times 10^{-8}}{0.25} = 5.34 \times 10^{-8} \text{ N} \approx 5.3 \times 10^{-8} \text{ N}$$

## The Inverse Square Relationship

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.

**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 $2.5r$?

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. $$F_2 = 0.16F$$

> **Exam tip**
>
> Always square the distance ratio when using inverse proportionality. A common mistake is forgetting to square the ratio.

## Applications to Extended Spherical Masses

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.

> **warning**
>
> If a problem gives the height of a satellite above a planet's surface, you must add the planet's radius to get the correct center-to-center distance.

**Worked example:** Calculate the gravitational force between Earth (mass $5.97 \times 10^{24}$ 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. $$F \approx 9.2 \times 10^3 \text{ N}$$

**Check your understanding**

Test your understanding of distance measurement:

1. 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

   *Why:* Correct! Earth acts as a point mass at its center, so $r$ is the Earth's radius.

## Common pitfalls

- **Wrong:** Using distance between surfaces instead of center-to-center distance
  - Why it fails: Most problems give orbit height above the surface, so students forget to add the planet's radius
  - Correct: Always add the radius of the planet/body to the surface height to get center-to-center separation
- **Wrong:** Using a positive exponent for $G$ ($6.67 \times 10^{11}$ instead of $6.67 \times 10^{-11}$)
  - Why it fails: The negative exponent is easy to miss when entering values into a calculator
  - Correct: Double-check the exponent of $G$ after entering it into your calculator
- **Wrong:** Claiming the larger mass exerts a larger force on the smaller mass
  - Why it fails: Students confuse force with acceleration, forgetting Newton's third law
  - Correct: Gravitational force is an action-reaction pair: the force on each mass is equal in magnitude
- **Wrong:** Squaring the product of masses in the formula
  - Why it fails: Students confuse the inverse square term with the product of masses term
  - Correct: Only the separation $r$ is squared in the formula $F = \frac{G m_1 m_2}{r^2}$

## 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 $G$ | $6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ | Always use negative exponent |
| Inverse square proportionality | $F_2 = F_1 \left(\frac{r_1}{r_2}\right)^2$ | Square the distance ratio |
| Center-to-center distance | $r = R_{\text{planet}} + h_{\text{orbit}}$ | Add planet radius to surface height |

## 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.

- [Gravitational field strength](https://www.owlsprep.com/study/cie-9702-u17-gravitational-field-strength/)
- [Gravitational potential](https://www.owlsprep.com/study/cie-9702-u17-gravitational-potential/)
- [Orbital motion of satellites](https://www.owlsprep.com/study/cie-9702-u17-orbital-motion/)

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