15.1 Newton's Law of Universal Gravitation

Newton's Law of Universal Gravitation states:

Mathematical Form

$F=\frac{G{m}_1{m}_2}{r^2}$

where:

• $F$ = gravitational force between the two masses (N)
• $m_1$, $m_2$ = masses of the two objects (kg)
• $r$ = distance between their centres (m)
• $G$ = Universal Gravitational Constant $=6.67\times 10^{-11}\mathrm{N}{\mathrm{m}}^2\mathrm{k}{\mathrm{g}}^{-2}$

Vector Form

${\vec{F}}_{21}=-\frac{G{m}_1{m}_2}{r^2}{\hat{r}}_{12}$

Here ${\hat{r}}_{12}$ is the unit vector pointing from $m_1$ to $m_2$. The negative sign indicates the force is attractive — it acts opposite to ${\hat{r}}_{12}$, i.e., directed towards the source mass.

By Newton's Third Law, the force exerted by $m_2$ on $m_1$ is equal in magnitude and opposite in direction: ${\vec{F}}_{12}=-{\vec{F}}_{21}$

Universal Gravitational Constant $G$

Cavendish measured $G$ approximately 100 years after Newton proposed the law.

Why is it Called a 'Universal' Law?

The law is called universal because:

• It applies to every pair of objects in the universe.
• It is independent of the size, composition, or location of the objects.
• It governs motion from everyday objects on Earth to the motion of planets, stars, and galaxies.

Inverse-Square Relationship

Since $F\propto \frac{1}{r^2}$:

Worked Example

Problem: Calculate the gravitational force between the Earth ($m_E=5.97\times 10^{24}\mathrm{kg}$) and the Moon ($m_M=7.35\times 10^{22}\mathrm{kg}$), given that the average Earth–Moon distance is $r=3.84\times 10^8\mathrm{m}$.

Solution:

$F=\frac{G{m}_E{m}_M}{r^2}$

$F=\frac{(6.67\times 10^{-11})(5.97\times 10^{24})(7.35\times 10^{22})}{(3.84\times 10^8{)}^2}$

$F=\frac{6.67\times 10^{-11}\times 4.39\times 10^{47}}{1.475\times 10^{17}}$

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

This large force keeps the Moon in its orbit around Earth.