Gravitational Potential
- The gravitational potential energy (G.P.E) is the energy an object has when lifted off the ground given by the familiar equation:
$
\text { G.P.E }=\mathbf{m g} \mathbf{\Delta h}
$
- The G.P.E on the surface of the Earth is taken to be 0
- This means work is done to lift the object
- However, outside the Earth’s surface, G.P.E can be defined as:
- The energy an object possess due to its position in a gravitational field
- The gravitational potential at a point is the gravitational potential energy per unit mass at that point
- Therefore, the gravitational potential is defined as:
- The work done per unit mass in bringing a test mass from infinity to a defined point
- Calculating Gravitational Potential
- The equation for gravitational potential $\phi$ is defined by the mass $M$ and distance $r$ :
$
\phi=-\frac{G M}{r}
$
- Where:
- $\phi=$ gravitational potential $\left(\mathrm{Jg}^{-1}\right)$
- $\mathrm{G}=$ Newton’s gravitational constant
- $M=$ mass of the body producing the gravitational field $(\mathrm{kg})$
- $r=$ distance from the centre of the mass to the point mass $(m)$
- The gravitational potential is negative near an isolated mass, such as a planet, because the potential when $r$ is at infinity is defined as 0
- Gravitational forces are always attractive so as $r$ decreases, positive work is done by the mass when moving from infinity to that point
- When a mass is closer to a planet, its gravitational potential becomes smaller (more negative)
- As a mass moves away from a planet, its gravitational potential becomes larger (less negative) until it reaches 0 at infinity
- This means when the distance $r$ because very large, the gravitational force tends rapidly towards 0 at a point further away from a planet
Gravitational potential increases and decreases depending on whether the object is travelling towards or against the field lines from infinity
Gravitational Potential Energy Between Two Point Masses
- The gravitational potential energy (G.P.E) at point in a gravitational field is defined as:
The work done in bringing a mass from infinity to that point - The equation for G.P.E of two point masses $m$ and $M$ at a distance $r$ is:
$
\text { G.P.E }=-\frac{G M m}{r}
$
- The change in G.P.E is given by:
$
\Delta G . P . E=m g \Delta h
$
- Where:
- $\mathrm{m}=$ mass of the object $(\mathrm{kg})$
- $\phi=$ gravitational potential at that point $\left(\mathrm{Jg}^{-1}\right)$
- $\Delta \mathrm{h}=$ change in height $(\mathrm{m})$
- Recall that at infinity, $\phi=0$ and therefore G.P.E $=0$
- It is more useful to find the change in G.P.E e.g. a satellite lifted into space from the Earth’s surface
- The change in G.P.E from for an object of mass $m$ at a distance $r_1$ from the centre of mass $M$, to a distance of $r_2$ further away is:
$
\Delta \text { G.P.E }=-\frac{G M m}{r_2}-\left(-\frac{G M m}{r_1}\right)=G M m\left(\frac{1}{r_1}-\frac{1}{r_2}\right)
$
- Change in gravitational potential energy between two points
- The change in potential $\Delta \phi$ is the same, without the mass of the object $m$ :
$
\Delta \phi=-\frac{G M}{r_2}-\left(-\frac{G M}{r_1}\right)=\mathrm{GM}\left(\frac{1}{r_1}-\frac{1}{r_2}\right)
$
Change in gravitational potential between two points
Gravitational potential energy increases as a satellite leaves the surface of the Moon
Maths tip
- Multiplying two negative numbers equals a positive number, for example:
$
-\left(-\frac{G M}{r}\right)=+\frac{G M}{r}
$
Worked example: Gravitational potential energy
A spacecraft of mass $300 \mathrm{~kg}$ leaves the surface of Mars to an altitude of $700 \mathrm{~km}$. Calculate the change in gravitational potential energy of the spacecraft. Radius of Mars $=3400 \mathrm{~km}$ Mass of Mars $=6.40 \times 10^{23} \mathrm{~kg}$
Answer/Explanation
Step 1:
Difference in gravitational potential energy equation
$
\Delta \text { G.P.E }=G M m\left(\frac{1}{r_1}-\frac{1}{r_2}\right)
$
Step 2:
Determine values for $r_1$ and $r_2$
$r_1$ is the radius of Mars $=3400 \mathrm{~km}=3400 \times 10^3 \mathrm{~m}$
$r_2$ is the radius + altitude $=3400+700=4100 \mathrm{~km}=4100 \times 10^3 \mathrm{~m}$
Step 3: $\quad$ Substitute in values
$
\begin{gathered}
\Delta \text { G.P.E }=6.67 \times 10^{-11} \times 6.40 \times 10^{23} \times 300 \times\left(\frac{1}{3400 \times 10^3}-\frac{1}{4100 \times 10^3}\right) \\
\text { DG.P.E }=643.076 \times 10^6=640 \text { MJ (2 s.f.) }
\end{gathered}
$
Exam Tip
Make sure to not confuse the $\Delta G$.P.E equation with $\Delta \mathbf{G} . \mathbf{P} . \mathbf{m}=\mathbf{m g} \Delta \mathbf{h}$
The above equation is only relevant for an object lifted in a uniform gravitational field (close to the Earth’s surface). The new equation for G.P.E will not include $g$, because this varies for different planets and is no longer a constant (decreases by $1 / r^2$ ) outside the surface of a planet.