use ϕ = –GM / r for the gravitational potential in the field due to a point mass

Gravitational Potential

What is Gravitational Potential?

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.

Formula for a Point Mass

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.

Analogy: The Hill and the Ball ⚽️

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.

Key Points to Remember

• Potential is a scalar quantity.
• For a point mass, $\phi = -GM/r$.
• Closer to the mass → more negative potential.
• Potential difference gives work done by gravity.

Examples

• Earth’s surface: $M_{\text{Earth}} = 5.97\times10^{24}\,\text{kg}$, $r \approx 6.37\times10^6\,\text{m}$ → $\phi \approx -6.3\times10^7\,\text{J/kg}$.
• Moon’s surface: $M_{\text{Moon}} = 7.35\times10^{22}\,\text{kg}$, $r \approx 1.74\times10^6\,\text{m}$ → $\phi \approx -1.9\times10^6\,\text{J/kg}$.
• Near a black hole: $M = 10\,M_{\odot}$, $r = 10\,R_s$ → $\phi$ becomes hugely negative, showing how strong the gravity is.

Practice Question

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 $G = 6.674\times10^{-11}\,\text{N}\,\text{m}^2\text{/kg}^2$, $M_{\text{Earth}} = 5.97\times10^{24}\,\text{kg}$.)

Answer

$$\phi = -\frac{(6.674\times10^{-11})(5.97\times10^{24})}{7.0\times10^6} \approx -5.7\times10^7\,\text{J/kg}$$

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