5.4.4 - Gravitational potential, energy and escape velocity

Potential starts at infinity

Near Earth's surface you have often used $E_p=mgh$. That formula is useful only for small height changes where $g$ is almost constant and where you have chosen a convenient local zero. For a planet-scale gravitational field, OCR uses a different and more powerful zero:

The unit of gravitational potential is joule per kilogram, $\text{J kg}^{-1}$. It is a scalar, even though the gravitational field strength $g$ is a vector quantity. Potential tells you about energy per kilogram, not force per kilogram.

For a point mass, or outside a spherical mass, the data booklet relationship is:

Here $G$ is the gravitational constant, $M$ is the mass producing the field, and $r$ is the distance from the centre of that mass. The minus sign is not decoration. It says that, compared with infinity, a mass near a planet is in a bound state. As the mass moves further away, $V_g$ increases towards zero.

Worked example: gravitational potential at Earth's surface

Take $M=5.97\times10^{24}\,\text{kg}$, $R=6.37\times10^6\,\text{m}$, and $G=6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$.

That is about $-63\,\text{MJ kg}^{-1}$. The negative sign means each kilogram at Earth's surface would need about $63\,\text{MJ}$ added to reach infinity with no kinetic energy left.

[DIAGRAM: asset_slug: 5_04_4_gravitational_potential_energy_and_escape_velocity__diagram_01; description: Gravitational potential is negative outside an attracting mass and approaches zero at infinity.; caption: Gravitational potential is negative outside an attracting mass and approaches zero at infinity.]

Changes in potential and energy

OCR questions often give two values of gravitational potential and ask what energy change follows. The safest route is:

Then multiply by the moving mass:

Here $m$ is the object placed in the field. $V_g$ is energy per kilogram; $E_p$ is energy for the whole object.

Suppose a satellite moves from a position where $V_g=-63\,\text{MJ kg}^{-1}$ to a higher orbit where $V_g=-56\,\text{MJ kg}^{-1}$.

The potential has increased. It is still negative, but it is less negative than before. For a $750\,\text{kg}$ satellite:

So $5.25\,\text{GJ}$ of gravitational potential energy has been gained. If the satellite also needs to be moving fast enough for an orbit, the kinetic energy is an additional part of the total energy calculation; do not hide it inside $E_p$.

Common sign trap: moving away from Earth increases gravitational potential energy even though the final value may still be negative. Increasing from $-63\,\text{MJ kg}^{-1}$ to $-56\,\text{MJ kg}^{-1}$ is a positive change.

Work from a force-distance graph

The gravitational force changes with distance, so $W=Fx$ with one constant $F$ is not enough for large movements. For a point mass or outside a spherical mass, the magnitude of the gravitational force is:

On a graph of force magnitude $F$ against distance $r$, the area under the curve gives the work done against gravity by an external force when the object is moved slowly outward. If the object moves from $r_1$ to $r_2$, that positive work is equal to the increase in gravitational potential energy.

For the inverse-square curve:

That matches the change in potential energy:

Worked example: moving a spacecraft outward

A $500\,\text{kg}$ spacecraft is moved slowly from $r_1=7.0\times10^6\,\text{m}$ to $r_2=8.0\times10^6\,\text{m}$ from Earth's centre. Use $GM=3.98\times10^{14}\,\text{m}^3\text{s}^{-2}$.