Gravitation • Topic 2 of 3

Gravitational PE & Escape Velocity

Gravitational potential energy (PE) is the energy a body possesses because of its position in a gravitational field. Near the Earth's surface we use the familiar $U=mgh$, but that is only an approximation valid for small heights where $g$ is nearly constant. For larger distances we must account for the fact that gravity weakens with distance.

Taking the reference point at infinity (where the gravitational force is zero), the work done by gravity in bringing a mass $m$ from infinity to a distance $r$ from a mass $M$ gives the potential energy:

  • $U=-\frac{GMm}{r}$.
  • The negative sign shows the field is attractive — the system has less energy than when the masses were infinitely far apart. PE is maximum (zero) at infinity and becomes more negative as the bodies come closer.

Gravitational potential ($V$) is the potential energy per unit mass at a point: it is the work done in bringing a unit mass from infinity to that point.

  • $V=-\frac{GM}{r}$, with SI unit $\text{J/kg}$.
  • Relationship: $U=mV$. Like PE, the potential is negative and tends to zero at infinity.

Escape velocity ($v_e$) is the minimum speed with which a body must be projected from the surface of a planet so that it just escapes the planet's gravitational pull and never returns. To escape, the body's kinetic energy must at least equal the magnitude of its gravitational potential energy at the surface. Setting $\frac{1}{2}mv_e^2=\frac{GMm}{R}$ gives:

  • $v_e=\sqrt{\frac{2GM}{R}}$.
  • Using $g=\frac{GM}{R^2}$, this can be written as $v_e=\sqrt{2gR}$.
  • Escape velocity is independent of the mass and direction of projection of the body — it depends only on the planet.

For the Earth, $v_e\approx11.2\ \text{km/s}$. Notice the neat link to orbital motion: $v_e=\sqrt{2}\times v_o$, where $v_o$ is the orbital velocity close to the surface. This is why a planet with a high escape velocity (like Earth) can retain a thick atmosphere, while the Moon, with $v_e\approx2.4\ \text{km/s}$, has effectively lost its atmosphere.

Escape velocity from a planet and gravitational potential energyM, Rv_ev_e = sqrt(2GM/R) = sqrt(2gR)U = 0 at infinityU = -GMm/r (well)
1
Worked Example
Calculate the gravitational potential energy of a 2 kg body placed at the surface of the Earth. ($M=6\times10^{24}\ \text{kg}$, $R=6.4\times10^{6}\ \text{m}$, $G=6.67\times10^{-11}$ SI units.)
Solution
  1. Step 1: Use $U=-\frac{GMm}{R}$.
  2. Step 2: Substitute: $U=-\frac{6.67\times10^{-11}\times6\times10^{24}\times2}{6.4\times10^{6}}$.
  3. Step 3: Compute the numerator $=8.0\times10^{14}$, so $U=-\frac{8.0\times10^{14}}{6.4\times10^{6}}=-1.25\times10^{8}\ \text{J}$.

Answer: $U\approx-1.25\times10^{8}\ \text{J}$.

2
Worked Example
Find the gravitational potential at the surface of the Earth. ($M=6\times10^{24}\ \text{kg}$, $R=6.4\times10^{6}\ \text{m}$.)
Solution
  1. Step 1: Use $V=-\frac{GM}{R}$.
  2. Step 2: Substitute: $V=-\frac{6.67\times10^{-11}\times6\times10^{24}}{6.4\times10^{6}}$.
  3. Step 3: Compute: $V=-\frac{4.0\times10^{14}}{6.4\times10^{6}}=-6.25\times10^{7}\ \text{J/kg}$.

Answer: $V\approx-6.25\times10^{7}\ \text{J/kg}$.

3
Worked Example
Calculate the escape velocity from the Earth's surface using $g=9.8\ \text{m/s}^2$ and $R=6.4\times10^{6}\ \text{m}$.
Solution
  1. Step 1: Use $v_e=\sqrt{2gR}$.
  2. Step 2: Substitute: $v_e=\sqrt{2\times9.8\times6.4\times10^{6}}=\sqrt{1.254\times10^{8}}$.
  3. Step 3: Compute: $v_e\approx1.12\times10^{4}\ \text{m/s}=11.2\ \text{km/s}$.

Answer: $v_e\approx11.2\ \text{km/s}$.

4
Worked Example
The escape velocity from a planet is $v_e$. If both the mass and the radius of the planet are doubled, what is the new escape velocity?
Solution
  1. Step 1: $v_e=\sqrt{\frac{2GM}{R}}$, so $v_e\propto\sqrt{\frac{M}{R}}$.
  2. Step 2: New value: $v_e'\propto\sqrt{\frac{2M}{2R}}=\sqrt{\frac{M}{R}}$.
  3. Step 3: The ratio $\frac{M}{R}$ is unchanged, so $v_e'=v_e$.

Answer: The escape velocity remains $v_e$ (unchanged).

5
Worked Example
How much work must be done to take a 100 kg satellite from the Earth's surface to infinity? (Take $g=9.8\ \text{m/s}^2$, $R=6.4\times10^{6}\ \text{m}$.)
Solution
  1. Step 1: PE at the surface is $U=-\frac{GMm}{R}=-mgR$ (using $GM=gR^2$); PE at infinity is $0$.
  2. Step 2: Work done $=U_\infty-U_{surface}=0-(-mgR)=mgR$.
  3. Step 3: Substitute: $W=100\times9.8\times6.4\times10^{6}=6.27\times10^{9}\ \text{J}$.

Answer: $W\approx6.27\times10^{9}\ \text{J}$.

6
Worked Example
The escape velocity of the Earth is 11.2 km/s. What would be the escape velocity from a planet whose mass is 4 times and radius is 2 times that of the Earth?
Solution
  1. Step 1: $v_e=\sqrt{\frac{2GM}{R}}$, so $\frac{v_e'}{v_e}=\sqrt{\frac{M'/R'}{M/R}}$.
  2. Step 2: Here $M'=4M$ and $R'=2R$, so $\frac{M'/R'}{M/R}=\frac{4M/2R}{M/R}=2$.
  3. Step 3: Therefore $v_e'=\sqrt{2}\times11.2=15.84\ \text{km/s}$.

Answer: $v_e'\approx15.8\ \text{km/s}$.

Key Points

  • Gravitational PE of a two-body system: $U=-\frac{GMm}{r}$, taking PE as zero at infinity; the negative sign shows attraction.
  • Gravitational potential is PE per unit mass: $V=-\frac{GM}{r}$ (unit J/kg), and $U=mV$.
  • Escape velocity: $v_e=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$, found by equating KE to the surface PE.
  • $v_e$ is independent of the mass and the direction of projection of the body; for Earth $v_e\approx11.2$ km/s.
  • Escape and orbital velocities are linked by $v_e=\sqrt{2}\,v_o$.
Tap an option to check your answer0 / 4
Q1.The gravitational potential energy of a mass $m$ at distance $r$ from mass $M$ is:
Explanation: Taking PE zero at infinity, $U=-\frac{GMm}{r}$; the negative sign reflects the attractive field.
Q2.The escape velocity from the Earth's surface is approximately:
Explanation: $v_e=\sqrt{2gR}\approx\sqrt{2\times9.8\times6.4\times10^6}\approx11.2\ \text{km/s}$.
Q3.Escape velocity is independent of:
Explanation: $v_e=\sqrt{2gR}$ has no dependence on the projectile's mass or its launch direction.
Q4.The SI unit of gravitational potential is:
Explanation: Potential is energy per unit mass, $V=-\frac{GM}{r}$, measured in J/kg.