understand how the concept of gravitational potential leads to the gravitational potential energy of two point masses and use EP = –GMm / r

Gravitational Potential

Gravitational potential (\$\phi\$) at a point in space is defined as the gravitational potential energy per unit mass of a test particle placed at that point, when the reference point for zero potential is taken at infinity.

Mathematically,

\$\phi(r) = \frac{U{\text{grav}}(r)}{m{\text{test}}}\$

where \$U{\text{grav}}(r)\$ is the gravitational potential energy of the test mass \$m{\text{test}}\$ at a distance \$r\$ from the source mass \$M\$.

Derivation of the Potential of a Point Mass

1. Consider a point mass \$M\$ creating a spherically symmetric field.
2. The gravitational force on a test mass \$m\$ at distance \$r\$ is \$F = -\dfrac{GMm}{r^{2}}\$ (directed towards \$M\$).
3. Work done in moving \$m\$ from infinity to \$r\$ is the negative of the integral of the force:

\$\$U{\text{grav}}(r) = -\int{\infty}^{r}\frac{GMm}{r^{\prime 2}}\,dr^{\prime}

= -\frac{GMm}{r}\$\$

Dividing by \$m\$ gives the gravitational potential of the source mass \$M\$:

\$\phi(r) = -\frac{GM}{r}\$

Gravitational Potential Energy of Two Point Masses

If two point masses \$M\$ and \$m\$ are separated by a distance \$r\$, the gravitational potential energy of the system is obtained by multiplying the potential of one mass by the other mass:

\$U_{\text{grav}} = m\,\phi(r) = -\frac{GMm}{r}\$

This expression is the basis of many A‑Level calculations involving orbital motion, escape speed, and energy changes in gravitational fields.

Key Symbols

Worked Example

Calculate the gravitational potential energy of the Earth–Moon system. Use \$M{\text{Earth}} = 5.97\times10^{24}\,\text{kg}\$, \$M{\text{Moon}} = 7.35\times10^{22}\,\text{kg}\$, and average separation \$r = 3.84\times10^{8}\,\text{m}\$.

1. Write the formula: \$U = -\dfrac{GM{\text{E}}M{\text{M}}}{r}\$.
2. Substitute the values:

\$U = -\frac{(6.674\times10^{-11})(5.97\times10^{24})(7.35\times10^{22})}{3.84\times10^{8}}\$

Evaluating gives \$U \approx -7.6\times10^{28}\,\text{J}\$.

Conceptual Points to Remember

• The gravitational potential is always negative because work must be done against the attractive force to separate the masses to infinity.
• Potential energy depends only on the relative positions of the masses, not on the path taken.
• When \$r\$ becomes very large, \$\phi \to 0\$ and \$U_{\text{grav}} \to 0\$, which defines the reference point.
• For a uniform spherical body of mass \$M\$ and radius \$R\$, the potential outside the sphere is the same as that of a point mass at its centre.

Summary

Gravitational potential \$\phi = -GM/r\$ provides a convenient way to calculate the gravitational potential energy of a pair of point masses via \$U = -GMm/r\$. Understanding this relationship is essential for solving problems involving orbital mechanics, escape velocity, and energy changes in gravitational fields at A‑Level.