3.7.2.3 Gravitational fields
Understanding of definition of gravitational potential, including zero value at infinity.
Understanding of gravitational potential difference.
Work done in moving mass m given by ΔW = mΔV
Equipotential surfaces.
Idea that no work is done when moving along an equipotential surface.
V in a radial field given by V = −GM/r
Significance of the negative sign.
Graphical representations of variations of g and V with r. V related to g by: g = −ΔV/Δr; ΔV from area under graph of g against r.
Gravitational potential energy vs gravitational potential
Why is cycling up a hill so much harder than cycling on flat ground? We have all experienced the difficulty of trying to pedal uphill — even in a bike’s lowest gear it can be very hard work! When doing this we are converting chemical energy stored in the glycogen molecules in our body into heat, kinetic energy, and gravitational potential energy. If we have a super-light bike, our lower mass will make it easier, as the lower weight means we have to do less work, but also less gravitational potential energy is gained, because as you know $E_{P}=mg\Delta h$.
If you stop pedalling at any point you will start to roll downhill, but what if the hill stretches on for a long distance, maybe to the edge of space? Ignoring the obvious problems of lack of oxygen and physical exertion, would it get any easier as you move further and further from the Earth? What if the hill extended into deep space? Would you ever reach a point where you no longer rolled back if you stopped pedalling, even though, as we know, gravity’s reach is infinite?
As height above the surface increases, our gravitational potential increases. If two cyclists with different masses are climbing side by side, the one with the greater mass will gain more gravitational potential energy, but as they are both moving through the same gravitational field, it can be useful to compare their change in potential energy independent of mass in the same way we compare the potential difference across a resistor independent of the amount of charge flowing through it, $V=W/Q$.
This is called gravitational potential ($V$), and is defined as the work done per unit mass in bringing a small test mass from infinity to that point. This means gravitational potential is energy per kilogram, so $V=W/m$, not the total energy of an object, but energy per unit mass. This removes the dependence on the mass of the object, just like $g=F/m$ removed it for field strength. Potential is an energy quantity, not a force. Its unit is $\units{J kg^{-1}}$, not $\units{N}$ or $\units{N kg^{-1}}$, but it is not potential energy either. Potential energy $E_{p}=mV$ depends on the mass placed in the field; potential $V$ does not.
Defining V
Sticking with the cycling analogy, as we cycle further up the hill away from Earth, the gravitational force weakens. Gravity becomes negligible at very large distances, so eventually we stop gaining significant potential energy with each extra metre climbed. We say that potential reaches zero at this point, and we define this to be at infinity. This is not a physical fact, it is a convention. Infinity is the most useful zero because it is the same for all masses and all fields; we could choose a different point, since potential is always relative (we could define zero at Earth’s surface, and this is sometimes done in mechanics) but infinity gives a universal reference.
As you can see from Figure 2 above, as the test mass falls towards Earth, its gravitational potential becomes more and more negative, but why is this? Why does it not increase positively? Well, let’s imagine we place an object at infinity, where $V=0$ by definition and no gravitational force acts. If it now moves towards a source mass, gravity will pull the object inward. The object accelerates, and as it does, it gains kinetic energy: gravity does positive work on the object. Where did this energy come from? The potential energy decreased as the object moved towards Earth, and since we started at zero, the potential energy must now be below zero. All points closer than infinity have $V<0$; the closer to the source mass, the more negative $V$ becomes, and to move back outward requires energy input. We often describe being close to the source mass as being in a potential well, and to move out of it we have to climb a potential ‘hill’. The larger the source mass, the deeper the potential well, and the more work needs to be done to climb out of it.
Worth remembering
The negative sign in $V=-GM/r$ isn’t just a mathematical detail — it’s telling you something physical. A negative potential means the mass sits in a potential well relative to infinity, and energy must be put in to remove it. The size of that negative number tells you how deep the well is: a more negative $V$ means it would take more energy to escape from that point to infinity.
Work done moving a mass
Looking at the definition again, we can derive an equation for gravitational potential. Potential is the work done per unit mass in moving a test mass from infinity to a point, so: $$V=-\frac{W}{m}\rightarrow V=-\frac{Fs}{m}\rightarrow V=-\frac{G m M}{r^{2}m}\times s$$
One of the mass terms cancels, and since $r$ and $s$ both represent a length in this approximation, they cancel too, leaving us with: $$V= -\frac{GM}{r}$$ This derivation does take a few mathematical shortcuts, and we should strictly integrate to explain it fully, but that is beyond the scope of A level physics.
So a satellite in a low Earth orbit with an altitude of $1000 \units{km}$ would have a gravitational potential of: $$V=-\frac{6.67\times10^{-11}\times6.0\times 10^{24}}{(6370+1000)\times 10^{3} }$$ $$V\approx -54,300,000 \units{J kg^{-1}} \text{ or }-54\units{MJ kg^{-1}}$$ If the satellite increases in height it will have to do work against the gravitational field, and again from the definition, the work done will be $\Delta W = m \Delta V$. So if a satellite with a mass of $3000 \units{kg}$ changes potential from $-54 \units{MJ kg^{-1}}$ to $-50 \units{MJ kg^{-1}}$, it will have had $\Delta W= 3000 \times ((-50)-(-54))\times 10^6 = 12 \units{GJ}$ of work done on it. In fact, this would also be the change in gravitational potential energy of the satellite.
We can use the equation for gravitational potential to derive the equation for gravitational potential energy that you already know from mechanics. When an object changes in height from $r_A$ to $r_B$, its change in GPE is given by:
$$\begin{aligned} \Delta GPE &= m \Delta V=m(V_B-V_A) \\ \\ &= \frac{-GMm}{r_B}-\frac{-GMm}{r_A} \\ \\ &=GMm \left(\frac{1}{r_A}-\frac{1}{r_B}\right)= GMm \left(\frac{r_B-r_A}{r_B r_A} \right) \end{aligned}$$ If $r_A$ and $r_B$ are similar, then the product $r_B r_A$ is approximately equal to $r^2$, and the difference $r_B-r_A$ is $\Delta h$, both of which can be substituted into the equation:
$$\Delta GPE=\frac{GMm \Delta h}{r^2}$$ We have seen previously that $\frac{GM}{r^2}=g$, so: $$\Delta GPE= mg \Delta h$$
Common exam mistake
It is easy to lose track of the minus signs when rearranging $V_B - V_A$ in this derivation. Write out $V=-GM/r$ for each point in full before substituting, rather than trying to do the sign-juggling in your head — a dropped negative sign here gives you $mg\Delta h$ with the wrong sign, which then makes an energy-conservation question impossible to complete correctly.
Equipotential surfaces
Around a point mass, or an object such as a planet, the gravitational field is radial, and as the field lines spread out, the potential rises towards zero. As we have seen, potential is the work done per unit mass, and since it changes independently of the mass of any object moving through the field, it can be useful to see how the potential varies through space. If we join up points of equal potential we can create a sort of “map” of the potential, which we call equipotential surfaces, such as in the diagram below.
If an object moves along a line of equipotential there is no change in its gravitational potential energy, just like someone walking along a contour line on a hill does not gain or lose gravitational potential energy if they do not increase or decrease their height. Contour lines on a map are almost exactly analogous to equipotential lines, as any two points on the same contour represent the same gravitational potential. When we look at more complex gravitational interactions later in the course, this contour-map analogy becomes even more useful.
Because potential is proportional to $1/r$, the distance between equal changes in potential increases as we move further from the source mass. The change in potential per metre is called the potential gradient. For small changes in height near the Earth’s surface, the magnitude of the potential gradient is $9.81\units{J kg^{-1} m^{-1}}$; for larger changes in distance, or further from the surface, it decreases as $g$ reduces. In general, for small changes, the potential gradient is given by $\frac{\Delta V}{\Delta r}$.
Graphs of g and V against r
We have already seen a graph of $V$ against $r$ at the top of this page, and can see that its shape is that of $-\frac{1}{r}$. A graph of $g$ against $r$ is shown below, and we can see that although the shape is similar, because $g=\frac{GM}{r^{2}}$, it is in fact a $\frac{1}{r^{2}}$ curve.
It is useful to compare these two graphs qualitatively first. For the field strength graph, when the test mass is close to the source mass, $g$ is large, as there are many field lines per unit area, so there is a high force per kilogram. When the test mass is far from the source mass, $g$ is small, but never zero for a source mass with finite $r$. This graph is always positive, because gravity always attracts. It falls as $1/r^2$, so doubling $r$ quarters $g$.
The potential graph, on the other hand, starts with a very negative $V$ when the test mass is close to the source mass (the test mass is deep in the ‘well’.) Far from the mass, $V$ approaches zero from below; it is always negative, only becoming zero at infinity. Its magnitude falls as $1/r$, so less steeply than $g$.
Linking g and V: gradients and areas
The two graphs can also be linked quantitatively. We have already seen that the potential gradient is $\frac{\Delta V}{\Delta r}$, so if we look at the equation for $V$, which is $V=-\frac{GM}{r}$, we can see that the potential gradient could also be written as:
$$\text{potential gradient}=-\frac{GM}{\Delta r^2} $$
Which is, of course, the equation for $g$. Therefore, the gradient of the potential graph is in fact $-g$, or:
$$g=-\frac{\Delta V}{\Delta r}$$ So taking a tangent at a point on a graph of $V$ against $r$ would give you the negative of the gravitational field strength at that point.
Taking the graph of $g$ against $r$, the area under the curve between $r_1$ and $r_2$ equals the magnitude of the change in potential, $\Delta V$, for that range. This is because, as above, $g=-\Delta V/\Delta r$, which means that $\Delta V=-g\times\Delta r$ for small changes, where $g$ does not vary. Summing all these steps over a greater range gives the total area.
Common exam mistake
It is easy to get tangled up trying to track the sign of $g=-\Delta V/\Delta r$ algebraically, since $g$ is always quoted as a positive number even though gravity is attractive. Don’t try to reason through the signs from first principles under exam pressure, instead, just read off the magnitude of the gradient or area from the graph, and separately state the direction of $g$ using what you already know: it always points towards the source mass. This two-step approach (magnitude from the maths, direction from the physics) avoids sign errors almost entirely.
Together, these two relationships mean that the $g$–$r$ and $V$–$r$ graphs are not simply two separate ways of describing the same field — they are mathematically linked, each containing exactly the same information as the other, just presented differently. Given either graph, we could in principle reconstruct the other: differentiate $V$ to find $g$, or sum the area under $g$ to find $V$. This relationship between a quantity and the gradient or area of its graph will reappear throughout this course, from force–displacement graphs in mechanics to charge–voltage graphs for capacitors, so it is well worth getting comfortable with it here.
Worked example
- A space probe is a very large distance from a planet, where the gravitational potential can be taken to be zero. Explain why the gravitational potential becomes negative as the probe approaches the planet.
- At a point a distance $r=1.00\times10^{7}\units{m}$ from the centre of the planet, the gravitational potential is measured to be $V=-2.50\times10^{7}\units{J kg^{-1}}$. Show that the mass of the planet is approximately $3.75\times10^{24}\units{kg}$.
- A satellite of mass $1500\units{kg}$ orbits at the point described in part b), where $V=-2.50\times10^{7}\units{J kg^{-1}}$. It is moved outward to a new position where the potential is $-2.20\times10^{7}\units{J kg^{-1}}$. Calculate the work done in moving the satellite.
Since $V=0$ is defined at infinity, and gravity does positive work on the probe as it is pulled in from infinity towards the planet, the probe’s potential energy decreases as it gets closer. Since potential energy started at zero, it must now be below zero. Because $V$ is potential energy per unit mass, $V$ itself must therefore be negative at every finite distance from the planet.
Rearranging $V=-\frac{GM}{r}$ for $M$: $$M=-\frac{Vr}{G}$$
Substituting in the values given: $$M=-\frac{(-2.50\times10^{7})\times1.00\times10^{7}}{6.67\times10^{-11}}$$ $$M\approx3.75\times10^{24}\units{kg}$$
Using $\Delta W=m\Delta V$: $$\Delta W=1500\times\left[(-2.20\times10^{7})-(-2.50\times10^{7})\right]$$ $$\Delta W=1500\times3.0\times10^{6}$$ $$\Delta W=4.5\times10^{9}\units{J}=4.5\units{GJ}$$
Test yourself
Try the questions below to check your understanding of this topic. Numerical questions use different numbers each time, so you can attempt them more than once.