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# 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 ${U}_{g}$. 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 $m$ being lifted through a height $h$ 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, ${F}_{g}$, are parallel. If $g$ 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, ${F}_{g}$, times the vertical distance, $h$, it has moved through. This assumes the gravitational acceleration is constant over the height $h$.
$\begin{array}{rl}{U}_{g}& ={F}_{g}\cdot h\\ & =m\cdot g\cdot h\end{array}$
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 $h=0$. 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.
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.
Exercise 1b (extension): What is the cost of the elevator journey assuming the cost of electricity is $0.10\frac{\mathrm{}}{\text{kW}\cdot \text{hr}}$?
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 ${}^{1}$. The height difference, $h$, 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?

# 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, ${m}_{1}$ and ${m}_{2}$, decreases with separation distance $r$ squared. If $G$ is the gravitational constant,
$F=\frac{G{m}_{1}{m}_{2}}{{r}^{2}}$.
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 $r$ 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 $r$ 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 $r=\mathrm{\infty }$, 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 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 $r$ 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 $r$ is:
${U}_{g}\left(r\right)=-\frac{G{m}_{1}{m}_{2}}{r}$
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, ${U}_{g}$ 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.
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?

## Want to join the conversation?

• In the Exercise 1b (extension): I think that instead of saying $0.10 kW/hr there should be$0.10 kW.hr because units of Power - kW already have over time value in them. And we are paying for energy not for power - therefore energy equals power.time not power/time • I'm confused by the exercise with the gravity wells of Charon and Pluto. It states that for the descent onto Pluto, work needs to be done equal to the GPE in order to land with zero speed.

But I read earlier that if I'm holding a dumbell and lower my arm, I'm performing negative work on the dumbell. The solution says that the spacecraft is performing positive work in this procedure (if I'm reading it right). Shouldn't it be negative?

Edit: To clarify, what's still confusing me is that if the spacecraft is lifting out of one gravity well and then descending into another, I would think that the work of lifting off Charon (the less massive body) would be a positive value, because the spacecraft is thrusting itself up out of the gravity well, and then when it descends to Pluto (more massive), it would being doing negative work since it's now losing the GPE it gained before, and then some, since Pluto's surface would be a deeper gravity well. So the work would be opposite to the first part, and greater in magnitude. Then the net work done would essentially be the difference between the GPE on the surfaces of the two bodies. Am I on the right track?

The solution takes the energy requirement of getting to the point where the two gravitational wells meet, and adds it to the energy in descending, though, where I would expect it to be subtracted. Can you point out where I'm going wrong?

I appreciate the help. This work/energy thing has eluded my understanding for many years. • algebraically
GPE_final - GPE_initial
= work (or energy) "needed" to move an object from initial to final

1. from Charon_surface to Charon_out
= GPE_Charon_out - GPE_Charon_surface
= -0.07 - (-0.17)
= -0.07 + 0.17
= 0.10 GJ
meaning you "need" 0.10GJ to escape Charon

2. from Charon_out to Pluto_surface
= GPE_Pluto_surface - GPE_Charon_out
= -0.74 - (-0.07)
= -0.74 + 0.07
= -0.67 GJ
meaning you "need" -0.67GJ on the way to Pluto
= you "gain" 0.67GJ on the way to Pluto
= you dash toward Pluto surface with 0.67GJ
= you "need" 0.67GJ to offset (-) the gained 0.67GJ to make you land on (rather than crash into) the surface
# if you "add or do work" -0.67GJ to this spacecraft, you actually push it down to crash on the surface 2x harder
## same for the case of a dumbbell, if you do negative work, you're not landing it gently. you're crashing it to the ground on top of -mg

3-1. from Charon_surface to Pluto_surface
= (GPE_Charon_out - GPE_Charon_surface) + -(GPE_Pluto_surface - GPE_Charon_out)
= 0.10 + -(-0.67)
= 0.10 + 0.67
= 0.77 GJ

3-2. or you can do the math at once tweaking 3-1 a bit
= (GPE_Charon_out - GPE_Charon_surface) + -(GPE_Pluto_surface - GPE_Charon_out)
= GPE_Charon_out - GPE_Charon_surface - GPE_Pluto_surface + GPE_Charon_out
= 2GPE_Charon_out - GPE_Charon_surface - GPE_Pluto_surface
= 2(-0.07) - (-0.17) - (-0.74)
= -0.14 + 0.17 + 0.74
= 0.03 + 0.74
= 0.77 GJ

3-3 don't apply the uppermost euqation directly to Charon and Pluto though
from Charon_surface to Pluto_surface
= GPE_Pluto_surface - GPE_Charon_surface
= -0.74 - (-0.17)
= -0.74 + 0.17
= -0.57 #it's ignoring 1) work for escaping Charon and 2) work for offsetting what's gained on the way to Pluto

in short
a. to escape Charon, you need + work
b. to land on Pluto, you need + work to offset - work gained on the way to Pluto (if you do - work, you actually crash it to the surface harder)
c. in sum you need two + works to escape and then land on Pluto without crashing
• I am not able to understand the concept of efficiency in the second problem.
How and from where does 1.25 come into picture? Please explain • If I lift object I'm doing positive work on the object and gravity is doing negative work on it. then the net work should be zero. Then why is it gaining GPE?

Whats wrong with my thinking? • In Exercise 1A at the end, why would you multiply by 1/.25? Wouldn't you just multiply by .25 to find the potential energy? • Hi,
In exercise 1b: 147kJ * 2.77 * 10^-5 $/kJ = 0.0041$
so the answer should be 0.4 cent, not 41 cents.     