What you'll learn

• What gravitational potential means, and why it is negative near a mass.
• How to use $V_g=-GM/r$ and calculate changes in gravitational potential.
• How a force–distance graph links to work done and gravitational potential energy.
• How energy conservation gives escape velocity, including for atoms in atmospheres.

The starting point: force, work and the “zero” point

You already know that two masses attract with gravitational force:

$$F=\frac{GMm}{r^2}$$

where $M$ is the mass creating the field, $m$ is the test mass, $r$ is the separation of their centres, and $G$ is the gravitational constant.

For a constant force, work done is force multiplied by distance moved in the direction of the force. In a gravitational field, the force changes with distance, so we often use energy and graphs instead.

The key convention in this topic is:

gravitational potential is zero at infinity.

That means we choose a point infinitely far from the mass as the place where the gravitational effect has “run out”.

Gravitational potential

Because gravity is attractive, an object released from infinity would naturally fall inwards and gain kinetic energy. To bring it in slowly, an external agent would have to hold it back, so the work done by that external agent is negative.

That is why gravitational potential near a mass is negative.

Potential around a point or spherical mass

For a point mass, or for a point outside a spherical mass, the gravitational potential at distance $r$ from the centre is:

$$V_g=-\frac{GM}{r}$$

where:

• $V_g$ is gravitational potential in J kg⁻¹
• $G=6.67\times 10^{-11}{N\; m}^2{kg}^{-2}$
• $M$ is the mass producing the field in kg
• $r$ is the distance from the centre of the mass in m

For a spherical planet or star, you can treat the mass as if it were concentrated at its centre, provided you are outside the object.

Changes in gravitational potential

A change in potential is found from:

$$\Delta V_g=V_{g,2}-V_{g,1}$$

Using $V_g=-GM/r$:

$$\Delta V_g=-\frac{GM}{r_2}-\left(-\frac{GM}{r_1}\right)$$

so:

$$\Delta V_g=GM\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$$

If an object moves outwards, then $r_2>r_1$, so $\Delta V_g$ is positive. The potential becomes less negative.

Force–distance graphs and work done

A graph of gravitational force against distance is not a straight line, because:

$$F=\frac{GMm}{r^2}$$

The force gets much larger at small distances and weaker as distance increases.

For any force–distance graph, the area under the graph gives the work done. This is the graphical version of work done.

For gravity, if you plot the magnitude of the attractive force against $r$, the area under the graph from $r_1$ to $r_2$ represents the work needed to move the mass slowly outwards between those distances.

If the final distance is infinity, the area from $r$ to infinity is the energy needed to escape from that position:

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

Gravitational potential energy

Since gravitational potential is energy per unit mass:

$$E=m{V}_g$$

Using $V_g=-GM/r$:

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

The unit is the joule, J.

Like gravitational potential, gravitational potential energy is negative near an isolated mass because zero is defined at infinity.

The change in gravitational potential energy is:

$$\Delta E=m\Delta V_g$$

or, between two distances:

$$\Delta E=GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$$

for an outward move from $r_1$ to $r_2$.

Escape velocity

To escape, the object must have enough kinetic energy to raise its gravitational potential energy from a negative value to zero.

At distance $r$ from mass $M$:

$$\frac{1}{2}m{v}_{esc}^2=\frac{GMm}{r}$$

The mass $m$ of the escaping object cancels:

$$v_{esc}=\sqrt{\frac{2GM}{r}}$$

At the surface of a planet, use $r=R$, where $R$ is the planet’s radius:

$$v_{esc}=\sqrt{\frac{2GM}{R}}$$

This idea helps scientists predict whether planets can keep atmospheres. If atoms or molecules in the upper atmosphere often have speeds comparable to the escape velocity, some can escape into space over time.