What is gravitational potential energy?

Learn what gravitational potential energy means and how to calculate it.

What is gravitational potential energy?

We all know instinctively that a heavy weight raised above someone's head represents a potentially dangerous situation. The weight may be well secured, so it is not necessarily dangerous. Our concern is that whatever is providing the force to secure the weight against gravity might fail. To use correct physics terminology, we are concerned about the gravitational potential energy of the weight.
All conservative forces have potential energy associated with them. The force of gravity is no exception. Gravitational potential energy is usually given the symbol UgU_g. It represents the potential an object has to do work as a result of being located at a particular position in a gravitational field.
Consider an object of mass mm being lifted through a height hh against the force of gravity as shown below. The object is lifted vertically by a pulley and rope, so the force due to lifting the box and the force due to gravity, FgF_g, are parallel. If gg is the magnitude of the gravitational acceleration, we can find the work done by the force on the weight by multiplying the magnitude of the force of gravity, FgF_g, times the vertical distance, hh, it has moved through. This assumes the gravitational acceleration is constant over the height hh.
Ug=Fgh=mgh\begin{aligned}U_g &= F_g\cdot h \\ &= m\cdot g \cdot h\end{aligned}
A weight lifted vertically to acquire gravitational potential energy.
A weight lifted vertically to acquire gravitational potential energy.
If the force were to be removed, the object would fall back down to the ground and the gravitational potential energy would be transferred to kinetic energy of the falling object. Our article on conservation of energy includes some example problems that are solved through an understanding of how gravitational potential energy is converted to other forms.
What is interesting about gravitational potential energy is that the zero is chosen arbitrarily. In other words, we are free to choose any vertical level as the location where h=0h=0. For simple mechanics problems, a convenient zero point would be at the floor of the laboratory or at the surface of a table. In principle however, we could choose any reference point—sometimes called a datum. The gravitational potential energy could even be negative if the object were to pass below the zero point. This doesn't present a problem, though; we just have to be sure that the same zero point is used consistently in the calculation.
The possibility of potential energy having a negative value is a result of the arbitrary constant that arises when performing the indefinite integral U(x)=F(x)dxU(x) = \int \mathbf{F}(x) \cdot \mathrm{d}x.
It is a consequence of the coordinate system chosen and quite distinct from the negative energy one encounters in quantum mechanics when explaining things like the Casimir effect.
Exercise 1a: How much electrical energy would be used by an elevator lifting a 75 kg person through a height of 50 m if the elevator system has an overall efficiency of 25%? Assume the mass of the empty elevator car is properly balanced by a counterweight.
Elevator system
Elevator system
Electrical energy from the grid is transferred into gravitational potential energy of the person and heat due to friction within the elevator system. Because the elevator car is counterweighted, there is no change in the gravitational potential energy of the elevator-counterweight system.
Using the equation for gravitational potential energy, we first find the change in gravitational potential energy of the person ΔUperson\Delta U_\mathrm{person}:
ΔUperson=mgh=(75 kg)(9.81 m/s2)(50 m)3.68104 J36.8 kJ\begin{aligned} \Delta U_\mathrm{person} &= m g h \\ &= (75~\mathrm{kg})\cdot (9.81~\mathrm{m/s^2})\cdot (50~\mathrm{m}) \\ &\simeq 3.68\cdot 10^4~\mathrm{J} \\ &\simeq 36.8~\mathrm{kJ} \end{aligned}
We are told that the system has an overall efficiency of 25%. This means that 25% of the electric energy used by the motor is transferred to useful work—gravitational potential energy in this case—and the remaining 75% is lost to the environment. So the total electrical energy usage EE is
E=10.25ΔUperson=10.2536.8 kJ=147 kJ\begin{aligned}E &= \frac{1}{0.25} \Delta U_\mathrm{person}\\ &= \frac{1}{0.25} 36.8~\mathrm{kJ}\\ &= 147~\mathrm{kJ}\end{aligned}
Exercise 1b (extension): What is the cost of the elevator journey assuming the cost of electricity is 0.10$kWhr0.10 \dfrac{\$}{\text{kW}\cdot \text{hr}}?
We know from the definition of power that the total energy is the power multiplied by the duration:
1 kW for 1 hour=(1 kW)(6060 s)=3600 kJ\text{1 kW for 1 hour} = (1~\mathrm{kW})\cdot(60\cdot 60~\mathrm{s})= 3600~\mathrm{kJ}.
So the electricity cost is 0.1$3600 kJ=2.77105 $/kJ\frac{0.1\$}{3600~\mathrm{kJ}}= 2.\overline{77}\cdot 10^{-5}~\mathrm{\$/kJ}.
And the cost of the journey is (2.77105 $/kJ)(147 kJ)=0.41 cents(2.\overline{77}\cdot 10^{-5}~\mathrm{\$/kJ})\cdot (147~\mathrm{kJ}) = 0.41~\text{cents}.
Exercise 2: Gravitational potential energy is one of very few forms of energy that can be used for practical energy storage at a very large scale. Very large scale energy storage is required for storing excess electrical energy from wind and solar energy resources so that it can be transferred to the electricity grid at times of peak demand. This can be achieved with pumped-storage hydroelectric systems. The image below shows an example of such a system. Water is pumped into an upper reservoir using excess energy to drive a motor which operates a turbine pump. When energy demand is high, the flow is reversed. The pump becomes a generator driven by the gravitational potential energy of the water in the upper reservoir. The water can be released very rapidly to accommodate the peak power needs of a whole city or even many cities.
The Bath County Pumped Storage Station is the world’s largest pumped-storage hydroelectric system. It serves 60 million people and has a generation capacity of around 3 GW 1^1. The height difference, hh, of the system is 380 m. Assume the system has an overall energy efficiency of 80%. What volume of water from the upper reservoir would need to flow through the turbine in a 30 minute period if a city is being provided with 3 GW of power for this time?
A hydroelectric power system using pumped-storage.
A hydroelectric power system using pumped-storage.
First we need to find the energy required, given the known power PP and duration tt:
E=Pt=(3109 W)(3060 s)=5.41012 J\begin{aligned} E &= P\cdot t \\ &= (3\cdot 10^9~\mathrm{W}) \cdot (30 \cdot 60~\mathrm{s}) \\ &= 5.4\cdot 10^{12} ~\mathrm{J} \end{aligned}
Given that the process is 80% efficient, we need more gravitational potential energy than this by a factor of 10.8=1.25\frac{1}{0.8}=1.25, so Ug=1.25(5.41012 J)=6.751012 JU_g=1.25\cdot (5.4\cdot 10^{12}~\mathrm{J}) = 6.75\cdot 10^{12}~\mathrm{J}.
The mass of water required can then be found from the equation for gravitational potential energy:
m=Uggh=6.751012 J(9.81 m/s2)(380 m)1.81109 kg\begin{aligned} m &= \frac{U_g}{g\cdot h} \\ &= \frac{6.75\cdot 10^{12}~\mathrm{J}}{(9.81~\mathrm{m/s^2})\cdot(380~\mathrm{m})} \\ &\simeq 1.81\cdot 10^9~\mathrm{kg}\end{aligned}
By the magic of the metric system—1 L of water has a mass of 1 kg—this is 1.81 GL, giga-liters, of water. As a point of reference, an olympic size swimming pool contains 2.5 ML, mega-liters, of water, so 1810 ML in 30 minutes equates to about 24 olympic size swimming pool of water falling per minute.

What if the gravitational field is not uniform?

If the problem involves large distances, we can no longer assume that the gravitational field is uniform. If we recall Newton's law of gravitation, the attractive force between two masses, m1m_1 and m2m_2, decreases with separation distance rr squared. If GG is the gravitational constant,
F=Gm1m2r2F = \frac{Gm_1 m_2}{r^2}.
When dealing with gravitational potential energy over large distances, we typically make a choice for the location of our zero point which may seem counterintuitive. We place the zero point of gravitational potential energy at a distance rr of infinity. This makes all values of the gravitational potential energy negative.
It turns out that it makes sense to do this because as the distance rr becomes large, the gravitational force tends rapidly towards zero. When you are close to a planet you are effectively bound to the planet by gravity and need a lot of energy to escape. Strictly you have escaped only when r=r=\infty, but because of the inverse-square relationship, we can reach an asymptote where gravitational potential energy becomes very close to zero. For a spacecraft leaving earth, this can be said to occur at a height of about 5107 5\cdot 10^7~meters above the surface which is about four times the Earth's diameter. At that height, the acceleration due to gravity has decreased to about 1% of the surface value.
If we recall that work done is a force times a distance then we can see that multiplying the force of gravity, above, by a distance rr cancels out the squared in the denominator. If we make our zero of potential energy at infinity, then it shouldn't be too much of a surprise that the gravitational potential energy as a function of rr is:
Ug(r)=Gm1m2rU_g(r) = -\frac{G m_1 m_2}{r}
If we define U()=0U(\infty) = 0, then the work done in moving a particle from infinity to a distance rr away from a gravitating object is W=rFdrW = -\int_\infty^r \vec{F}\cdot \mathrm{d} \vec{r}.
The simplest path over which to evaluate the integral is the straight path along the x-axis.
Ug(r)=rGm1m2x2dx=[Gm1m2x]r=Gm1m2r\begin{aligned} U_g(r) &= - \int_\infty^r \frac{-Gm_1 m_2}{x^2} \cdot \mathrm{d}x \\ &= \left[ \frac{-G m_1 m_2}{x} \right]^r_\infty \\ &= -\frac{Gm_1 m_2}{r}\end{aligned}
This formulation is very convenient for describing the energy requirements for traveling between different bodies in the solar system. We can imagine coming in to land on a planet. As we come closer to the planet, we gain kinetic energy. Because energy is conserved, we lose gravitational potential energy to account for this—in other words, UgU_g becomes more negative.
This picture leads to the concept of a gravity well which you need to "climb out of" in order to transfer from one planetary body to another. The image below shows a depiction of the gravity wells of Pluto and its moon Charon, calibrated for a 1,000 kg spacecraft.
Gravitational potential energy wells of Pluto and Charon
Gravitational potential energy wells of Pluto and Charon
Exercise 3: Based on the plot shown in the image above, how much work needs to be done against gravity on a journey beginning at rest on the surface of Charon and arriving at the surface of Pluto with zero speed?
To leave the surface of Charon, we must first climb out of Charon's gravity well. Luckily we don't need to escape the whole system but go just far enough that we have enough energy to avoid falling back to Charon should our rocket shut off. This occurs when the red and blue lines cross.
Reading from the plot this requires (.07)(0.175)=0.105 GJ(-.07)-(-0.175) = 0.105~\mathrm{GJ}.
At this point, we could simply fall into Pluto, but then all our gravitational potential energy would be transferred to kinetic energy, and we would crash into Pluto's surface. We want to land with zero speed, so we need to do work exactly equal to our gravitational potential energy to achieve this. Reading from the plot again we find this requires (0.07)(0.737)=0.667 GJ(-0.07) - (-0.737) = 0.667~\mathrm{GJ}.
So, overall, we require a total energy of 0.105+0.667=0.772 GJ0.105 + 0.667 = 0.772~\mathrm{GJ}. This does not include any energy needed to account for velocity differences between the two objects.