Cambridge Notes, Past Papers, Revision Questions

Cambridge A-Level Physics 9702 – Gravitational Force Between Point Masses

Gravitational Force Between Point Masses

Learning Objective

Derive, from Newton’s law of gravitation and the definition of gravitational field, the equation

\$g = \frac{GM}{r^{2}}\$

for the gravitational field strength due to a point mass.

Key Concepts

Newton’s law of universal gravitation

Gravitational field (field strength) definition

Vector nature of the force and field

Relationship between force, mass and field strength

Symbols and Units

Symbol	Quantity	SI Unit	Typical \cdot alue (if applicable)
\$G\$	Universal gravitational constant	m³·kg⁻¹·s⁻²	\$6.674 \times 10^{-11}\$
\$M\$	Mass of the source (point mass)	kg	–
\$m\$	Test mass placed in the field	kg	–
\$r\$	Distance between the centres of the two masses	m	–
\$\mathbf{F}\$	Gravitational force on the test mass	N (kg·m·s⁻²)	–
\$\mathbf{g}\$	Gravitational field strength at distance \$r\$	m·s⁻²	–

Derivation Steps

Start with Newton’s law of universal gravitation for two point masses:
\$\mathbf{F} = -\,\frac{G M m}{r^{2}}\,\hat{\mathbf{r}}\$
where \$\hat{\mathbf{r}}\$ is a unit vector directed from the source mass \$M\$ to the test mass \$m\$, and the negative sign indicates that the force is attractive.

Define the gravitational field (or field strength) \$\mathbf{g}\$ at a point in space as the force per unit test mass:
\$\mathbf{g} = \frac{\mathbf{F}}{m}\$

Substitute the expression for \$\mathbf{F}\$ from step 1 into the definition of \$\mathbf{g}\$:
\$\$\mathbf{g} = \frac{-\,G M m}{r^{2}}\,\frac{\hat{\mathbf{r}}}{m}
= -\,\frac{G M}{r^{2}}\,\hat{\mathbf{r}}\$\$

Since the magnitude of the field is usually denoted by \$g\$ (without the vector sign), take the absolute value of the vector expression:
\$g = \frac{G M}{r^{2}}\$
The direction of \$\mathbf{g}\$ is radially inward toward the mass \$M\$.

Physical Interpretation

The equation \$g = GM/r^{2}\$ tells us that the gravitational field strength:

is directly proportional to the source mass \$M\$;

decreases with the square of the distance from the source;

has the same units as acceleration, reflecting that a free‑falling test mass experiences an acceleration \$a = g\$.

Example Calculation

Find the gravitational field strength at the surface of the Earth (radius \$R{\oplus}=6.37\times10^{6}\,\text{m}\$, mass \$M{\oplus}=5.97\times10^{24}\,\text{kg}\$).

Using \$g = GM/r^{2}\$:

\$\$g_{\oplus} = \frac{(6.674\times10^{-11})\,(5.97\times10^{24})}{(6.37\times10^{6})^{2}}

\approx 9.81\ \text{m·s}^{-2}\$\$

This matches the standard value of the acceleration due to gravity at sea level.

Common Misconceptions

“Gravity is a force only near the Earth.” – The equation applies to any point mass, not just the Earth.

“\$g\$ is constant everywhere.” – \$g\$ varies with \$r\$; it is only approximately constant near the Earth's surface because \$r\$ changes very little over typical heights.

“The field points outward.” – The field points toward the source mass (inward), indicated by the negative sign in the vector form.

Suggested Diagram

Suggested diagram: A point mass \$M\$ at the origin, a test mass \$m\$ at distance \$r\$ on the positive \$x\$‑axis, showing the radial vector \$\hat{\mathbf{r}}\$, the attractive force \$\mathbf{F}\$ directed toward \$M\$, and the gravitational field \$\mathbf{g}\$ at that point.

Summary

Starting from Newton’s law of gravitation and the definition of gravitational field, we derived the simple yet powerful relation \$g = GM/r^{2}\$. This expression underpins many calculations in astrophysics, orbital mechanics, and everyday problems involving gravity.

derive, from Newton’s law of gravitation and the definition of gravitational field, the equation g = GM / r 2 for the gravitational field strength due to a point mass