recall and use Newton’s law of gravitation F = Gm1m2 / r2 for the force between two point masses

Gravitational Force Between Point Masses

In this topic you will recall and apply Newton’s law of universal gravitation to calculate the force between two point masses.

Newton’s Law of Gravitation

The magnitude of the attractive force between two point masses \$m{1}\$ and \$m{2}\$ separated by a distance \$r\$ is given by

\$F = \frac{G\,m{1}m{2}}{r^{2}}\$

where \$G\$ is the universal gravitational constant.

Key \cdot ariables and Units

Direction of the Force

The force acts along the line joining the two masses and is always attractive. Each mass experiences a force directed towards the other mass.

Derivation Sketch (Conceptual)

1. Assume the force depends only on the masses and the separation.
2. Experimental evidence shows the force is proportional to each mass: \$F \propto m{1}m{2}\$.
3. Measurements of planetary motion reveal an inverse‑square dependence on distance: \$F \propto 1/r^{2}\$.
4. Combine the proportionalities and introduce a constant of proportionality \$G\$ to obtain the law.

Worked Example

Calculate the gravitational force between the Earth (\$m_{E}=5.97\times10^{24}\,\text{kg}\$) and a 1 kg satellite orbiting at a height of \$400\,\text{km}\$ above the Earth’s surface.

Solution:

1. Find the centre‑to‑centre distance:

   \$r = R_{E} + h = 6.371\times10^{6}\,\text{m} + 4.0\times10^{5}\,\text{m}=6.771\times10^{6}\,\text{m}\$

2. Insert values into Newton’s law:

   \$F = \frac{(6.674\times10^{-11})\,(5.97\times10^{24})\,(1)}{(6.771\times10^{6})^{2}}\$

3. Calculate:

   \$F \approx 8.7\,\text{N}\$

Common Misconceptions

• Force depends on distance linearly. The correct dependence is inverse‑square (\$1/r^{2}\$).
• Only massive bodies attract. Any two masses, no matter how small, exert a gravitational force.
• Gravitational force is a contact force. It acts at a distance, without physical contact.

Practice Questions

1. Two 2 kg masses are placed 0.5 m apart. Calculate the magnitude of the gravitational force between them.
2. A planet of mass \$8.0\times10^{24}\,\text{kg}\$ exerts a gravitational force of \$3.2\times10^{22}\,\text{N}\$ on a spacecraft of mass \$2.0\times10^{3}\,\text{kg}\$. Find the distance between the planet’s centre and the spacecraft.
3. Explain why the gravitational force between two electrons is negligible compared with the electrostatic force, using the values \$G=6.674\times10^{-11}\,\text{N·m}^{2}\text{/kg}^{2}\$ and \$k_{e}=8.988\times10^{9}\,\text{N·m}^{2}\text{/C}^{2}\$.

Summary

Newton’s law of universal gravitation provides a simple, quantitative description of the attractive force between any two point masses. Remember the inverse‑square dependence, the role of the constant \$G\$, and that the force always acts along the line joining the masses. Mastery of this law is essential for solving a wide range of A‑Level physics problems, from orbital mechanics to simple laboratory calculations.