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

Gravitational Field Strength of a Point Mass

Newton’s Law of Gravitation

The force between two point masses is given by

\$F = G \dfrac{m1 m2}{r^2}\$

where \$G\$ is the gravitational constant, \$m1\$ and \$m2\$ are the masses, and \$r\$ is the separation between them.

💡 Think of this as the “pull” that every mass exerts on every other mass.

Definition of Gravitational Field

The gravitational field \$g\$ at a point is the force that a unit mass would feel there:

\$g = \dfrac{F}{m_2}\$

This removes the dependence on the test mass \$m_2\$ and tells us how strong the field is at that location.

Deriving \$g = \dfrac{GM}{r^2}\$

  1. Start with Newton’s law:

    \$F = G \dfrac{M m_2}{r^2}\$

    where \$M\$ is the source mass.

  2. Insert this into the definition of the field:

    \$g = \dfrac{F}{m2} = \dfrac{G \dfrac{M m2}{r^2}}{m_2}\$

  3. Cancel the test mass \$m_2\$:

    \$g = G \dfrac{M}{r^2}\$

  4. Result: the gravitational field produced by a point mass is

    \$\boxed{g = \dfrac{GM}{r^2}}\$

📚 The key step is the cancellation of \$m_2\$, showing that the field depends only on the source mass and distance.

Analogy: The Gravitational Field as a Map

Imagine a pond with a stone dropped in the centre. The ripples spread outwards.

  • Each ripple represents a field line of gravity.
  • The closer you are to the stone, the faster the ripples move—just like \$g\$ increases as \$r\$ decreases.
  • Just as the ripples affect a floating leaf, the gravitational field affects any mass placed in its vicinity.

🌍 This visual helps remember that the field is a property of space around the mass, not the mass itself.

Example Problem

Calculate the gravitational field at a point 10 000 km from the centre of Earth.

ParameterValue
Mass of Earth \$M\$\$5.97\times10^{24}\,\text{kg}\$
Distance \$r\$\$1.0\times10^{7}\,\text{m}\$
Gravitational constant \$G\$\$6.674\times10^{-11}\,\text{N·m}^2\text{/kg}^2\$

Plug into \$g = \dfrac{GM}{r^2}\$:


\$g = \dfrac{(6.674\times10^{-11})(5.97\times10^{24})}{(1.0\times10^{7})^2} \approx 9.8\,\text{m/s}^2.\$

🎓 This matches the familiar surface gravity of Earth.

Exam Tips

  • Always write the full expression for \$F\$ before cancelling \$m_2\$ to avoid mistakes.
  • Remember that \$g\$ is a vector pointing towards the mass; the magnitude follows the derived formula.
  • When given a distance in km, convert to meters before using the formula.
  • Use the symbol \$M\$ for the source mass and \$m\$ for the test mass to keep variables clear.
  • Practice deriving the formula from scratch; examiners may ask you to show the steps.

📚 Good luck, and keep practising!