Gravitational potential (ϕ) is the energy per unit mass that a small test mass would have at a point in a gravitational field. Think of it as the “height” of a hill in a landscape of gravity. The higher the hill, the more energy a ball would have if placed there.
For a point mass \$M\$ the potential at a distance \$r\$ is given by
\$\phi = -\frac{GM}{r}\$
Here \$G\$ is the gravitational constant (\$6.674\times10^{-11}\,\text{N}\,\text{m}^2\text{/kg}^2\$). The negative sign shows that the potential is lower (more negative) closer to the mass.
Imagine a ball on a hill. The higher the hill, the more potential energy the ball has. In gravity, the “hill” is the distance from a massive object. The closer you are, the steeper the hill and the more negative the potential.
A satellite orbits Earth at a distance of \$7.0\times10^6\,\text{m}\$ from its centre. Calculate the gravitational potential energy per kilogram at that orbit. (Use \$G = 6.674\times10^{-11}\,\text{N}\,\text{m}^2\text{/kg}^2\$, \$M_{\text{Earth}} = 5.97\times10^{24}\,\text{kg}\$.)
\$\phi = -\frac{(6.674\times10^{-11})(5.97\times10^{24})}{7.0\times10^6} \approx -5.7\times10^7\,\text{J/kg}\$
| Tip | Why It Helps |
|---|---|
| Remember the negative sign! | It indicates that potential energy decreases as you approach the mass. |
| Check units (J/kg) | Ensures you used the correct formula and constants. |
| Use the hill analogy in explanations | Makes your answer clear and memorable. |