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

Published by Patrick Mutisya · 8 days ago

Cambridge A-Level Physics 9702 – Gravitational Potential

Gravitational Potential

Objective

Use the expression \$\phi = -\frac{GM}{r}\$ to calculate the gravitational potential at a distance \$r\$ from a point mass \$M\$.

Key Concepts

  • Gravitational field (\$\mathbf{g}\$): The force per unit mass, \$\mathbf{g} = -\frac{GM}{r^{2}}\hat{r}\$.

  • Gravitational potential (\$\phi\$): The work done per unit mass in bringing a test mass from infinity to a point in the field, measured in joules per kilogram (J kg⁻¹).

  • Reference point: By convention, \$\phi = 0\$ at \$r \to \infty\$.

Derivation of \$\phi = -GM/r\$

  1. Start from the definition of potential difference:

    \$\Delta\phi = -\int{r1}^{r_2}\mathbf{g}\cdot d\mathbf{r}.\$

  2. Insert the expression for the gravitational field of a point mass:

    \$\mathbf{g} = -\frac{GM}{r^{2}}\hat{r}.\$

    Since \$\mathbf{g}\$ and \$d\mathbf{r}\$ are collinear,

    \$\$\Delta\phi = -\int{r1}^{r_2}\left(-\frac{GM}{r^{2}}\right)dr

    = GM\int{r1}^{r_2}\frac{dr}{r^{2}}.\$\$

  3. Integrate:

    \$\$GM\int{r1}^{r2}\frac{dr}{r^{2}} = GM\left[-\frac{1}{r}\right]{r1}^{r2}

    = GM\left(\frac{1}{r1}-\frac{1}{r2}\right).\$\$

  4. Choose the reference point at infinity (\$r_2\to\infty\$, \$\phi(\infty)=0\$):

    \$\phi(r) - 0 = -GM\left(\frac{1}{r}\right),\$

    so

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

Properties of Gravitational Potential

  • Scalar quantity – direction is not required.

  • Negative everywhere for an attractive field (since work must be done against the field to move outward).

  • Magnitude decreases with increasing \$r\$ (approaches zero as \$r\to\infty\$).

  • Potential energy of a mass \$m\$ at distance \$r\$ is \$U = m\phi = -\frac{GMm}{r}\$.

Example Calculations

SituationGivenCalculationResult \$\phi\$ (J kg⁻¹)
Near Earth's surface (approximate)\$M_{\oplus}=5.97\times10^{24}\,\text{kg}\$, \$r=6.37\times10^{6}\,\text{m}\$\$\phi = -\frac{(6.674\times10^{-11})(5.97\times10^{24})}{6.37\times10^{6}}\$\$-6.25\times10^{7}\$
At altitude \$h=400\,\text{km}\$ (ISS orbit)\$r = R_{\oplus}+h = 6.77\times10^{6}\,\text{m}\$\$\phi = -\frac{GM_{\oplus}}{r}\$\$-5.88\times10^{7}\$
Near the Sun's surface\$M_{\odot}=1.99\times10^{30}\,\text{kg}\$, \$r=6.96\times10^{8}\,\text{m}\$\$\phi = -\frac{GM_{\odot}}{r}\$\$-1.91\times10^{9}\$

Common Mistakes

  1. Using \$+\frac{GM}{r}\$ instead of the negative sign.

  2. Taking the reference point at the Earth's surface rather than at infinity; this changes the zero level and leads to inconsistent results.

  3. Confusing potential (\$\phi\$) with field strength (\$g\$); remember \$\mathbf{g} = -\nabla\phi\$.

Practice Questions

  1. Calculate the gravitational potential at a distance of \$2.0\times10^{7}\,\text{m}\$ from a planet of mass \$8.0\times10^{24}\,\text{kg}\$.

  2. A satellite of mass \$500\,\text{kg}\$ is in a circular orbit \$10\,000\,\text{km}\$ above the Earth's surface. Determine its gravitational potential energy.

  3. Show that the work required to move a \$1\,\text{kg}\$ test mass from \$r=R{\oplus}\$ to \$r=2R{\oplus}\$ equals \$GM{\oplus}\left(\frac{1}{R{\oplus}}-\frac{1}{2R_{\oplus}}\right)\$ joules.

Suggested diagram: Radial lines from a point mass showing \$r\$, the direction of \$\mathbf{g}\$ (inward), and a test mass being moved from infinity to \$r\$.

Summary

The gravitational potential of a point mass \$M\$ at a distance \$r\$ is given by \$\phi = -\frac{GM}{r}.\$ This simple expression follows directly from the definition of potential as the negative integral of the field, with the conventional zero at infinity. It is a scalar quantity, negative for attractive forces, and forms the basis for calculating gravitational potential energy, escape velocity, and orbital mechanics in A‑Level physics.