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

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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: $$\Delta\phi = -\int_{r_1}^{r_2}\mathbf{g}\cdot d\mathbf{r}.$$
  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, $$\Delta\phi = -\int_{r_1}^{r_2}\left(-\frac{GM}{r^{2}}\right)dr = GM\int_{r_1}^{r_2}\frac{dr}{r^{2}}.$$
  3. Integrate: $$GM\int_{r_1}^{r_2}\frac{dr}{r^{2}} = GM\left[-\frac{1}{r}\right]_{r_1}^{r_2} = GM\left(\frac{1}{r_1}-\frac{1}{r_2}\right).$$
  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 $\mathbf{g} = -abla\phi$.

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.