Gravitational potential energy

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 $U_g$U, start subscript, g, end subscript. 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 $m$m being lifted through a height $h$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, $F_g$F, start subscript, g, end subscript, are parallel. If $g$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, $F_g$F, start subscript, g, end subscript, times the vertical distance, $h$h, it has moved through. This assumes the gravitational acceleration is constant over the height $h$h.

$\begin{aligned}U_g &= F_g\cdot h \\ &= m\cdot g \cdot h\end{aligned}$

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 $h=0$h, equals, 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.

$U(x) = \int \mathbf{F}(x) \cdot \mathrm{d}x$U, left parenthesis, x, right parenthesis, equals, integral, F, left parenthesis, x, right parenthesis, dot, d, x

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

$\Delta U_\mathrm{person}$delta, U, start subscript, p, e, r, s, o, n, end subscript

$\begin{aligned} \Delta U_\mathrm{person} &= m g h \\ &= (75~\mathrm{kg})\cdot (9.81~\mathrm{m/s^2})\cdot (50~\mathrm{m}) \\ &\simeq 3.68\cdot 10^4~\mathrm{J} \\ &\simeq 36.8~\mathrm{kJ} \end{aligned}$

$E$E

$\begin{aligned}E &= \frac{1}{0.25} \Delta U_\mathrm{person}\\ &= \frac{1}{0.25} 36.8~\mathrm{kJ}\\ &= 147~\mathrm{kJ}\end{aligned}$

Exercise 1b (extension): What is the cost of the elevator journey assuming the cost of electricity is $0.10 \dfrac{\$}{\text{kW}\cdot \text{hr}}$0, point, 10, start fraction, dollar sign, divided by, start text, k, W, end text, dot, start text, h, r, end text, end fraction?

$\text{1 kW for 1 hour} = (1~\mathrm{kW})\cdot(60\cdot 60~\mathrm{s})= 3600~\mathrm{kJ}$start text, 1, space, k, W, space, f, o, r, space, 1, space, h, o, u, r, end text, equals, left parenthesis, 1, space, k, W, right parenthesis, dot, left parenthesis, 60, dot, 60, space, s, right parenthesis, equals, 3600, space, k, J

$\frac{0.1\$}{3600~\mathrm{kJ}}= 2.\overline{77}\cdot 10^{-5}~\mathrm{\$/kJ}$start fraction, 0, point, 1, dollar sign, divided by, 3600, space, k, J, end fraction, equals, 2, point, start overline, 77, end overline, dot, 10, start superscript, minus, 5, end superscript, space, dollar sign, slash, k, J

$(2.\overline{77}\cdot 10^{-5}~\mathrm{\$/kJ})\cdot (147~\mathrm{kJ}) = 0.41~\text{cents}$left parenthesis, 2, point, start overline, 77, end overline, dot, 10, start superscript, minus, 5, end superscript, space, dollar sign, slash, k, J, right parenthesis, dot, left parenthesis, 147, space, k, J, right parenthesis, equals, 0, point, 41, space, start text, c, e, n, t, s, end text

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 $^1$start superscript, 1, end superscript. The height difference, $h$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?

A hydroelectric power system using pumped-storage.

A hydroelectric power system using pumped-storage.

[Show solution.]

$P$P$t$t

$\begin{aligned} E &= P\cdot t \\ &= (3\cdot 10^9~\mathrm{W}) \cdot (30 \cdot 60~\mathrm{s}) \\ &= 5.4\cdot 10^{12} ~\mathrm{J} \end{aligned}$

$\frac{1}{0.8}=1.25$start fraction, 1, divided by, 0, point, 8, end fraction, equals, 1, point, 25$U_g=1.25\cdot (5.4\cdot 10^{12}~\mathrm{J}) = 6.75\cdot 10^{12}~\mathrm{J}$U, start subscript, g, end subscript, equals, 1, point, 25, dot, left parenthesis, 5, point, 4, dot, 10, start superscript, 12, end superscript, space, J, right parenthesis, equals, 6, point, 75, dot, 10, start superscript, 12, end superscript, space, J

$\begin{aligned} m &= \frac{U_g}{g\cdot h} \\ &= \frac{6.75\cdot 10^{12}~\mathrm{J}}{(9.81~\mathrm{m/s^2})\cdot(380~\mathrm{m})} \\ &\simeq 1.81\cdot 10^9~\mathrm{kg}\end{aligned}$

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, $m_1$m, start subscript, 1, end subscript and $m_2$m, start subscript, 2, end subscript, decreases with separation distance $r$r squared. If $G$G is the gravitational constant,

$F = \frac{Gm_1 m_2}{r^2}$F, equals, start fraction, G, m, start subscript, 1, end subscript, m, start subscript, 2, end subscript, divided by, r, squared, end fraction.

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 $r$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 $r$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 $r=\infty$r, equals, infinity, 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 $5\cdot 10^7~$5, dot, 10, start superscript, 7, end superscript, spacemeters 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 $r$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 $r$r is:

$U_g(r) = -\frac{G m_1 m_2}{r}$U, start subscript, g, end subscript, left parenthesis, r, right parenthesis, equals, minus, start fraction, G, m, start subscript, 1, end subscript, m, start subscript, 2, end subscript, divided by, r, end fraction

[Show proof with calculus.]

$U(\infty) = 0$U, left parenthesis, infinity, right parenthesis, equals, 0$r$r$W = -\int_\infty^r \vec{F}\cdot \mathrm{d} \vec{r}$W, equals, minus, integral, start subscript, infinity, end subscript, start superscript, r, end superscript, F, with, vector, on top, dot, d, r, with, vector, on top

$\begin{aligned} U_g(r) &= - \int_\infty^r \frac{-Gm_1 m_2}{x^2} \cdot \mathrm{d}x \\ &= \left[ \frac{-G m_1 m_2}{x} \right]^r_\infty \\ &= -\frac{Gm_1 m_2}{r}\end{aligned}$

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, $U_g$U, start subscript, g, end subscript 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.