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.
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.
\$F = G \dfrac{M m_2}{r^2}\$
where \$M\$ is the source mass.
\$g = \dfrac{F}{m2} = \dfrac{G \dfrac{M m2}{r^2}}{m_2}\$
\$g = G \dfrac{M}{r^2}\$
\$\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.
Imagine a pond with a stone dropped in the centre. The ripples spread outwards.
🌍 This visual helps remember that the field is a property of space around the mass, not the mass itself.
Calculate the gravitational field at a point 10 000 km from the centre of Earth.
| Parameter | Value |
|---|---|
| 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.
📚 Good luck, and keep practising!