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 \[h\].

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

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.

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.

Explain—negative energy, requires calculus concepts.

Exercise 1: 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?

A hydroelectric power system using pumped-storage.

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.

\[F = \frac{Gm_1 m_2}{r^2}\].

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 \[r\] 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 \[r=\infty\], 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 \[5\cdot 10^7~\]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 \[r\] 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:

\[U_g(r) = -\frac{G m_1 m_2}{r}\]

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.

Gravitational potential energy wells of Pluto and Charon

Exercise 2: 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?

What is power?

Much like energy, the word power is something we hear a lot. In everyday life it has a wide range of meanings. In physics however, it has a very specific meaning. It is a measure of the rate at which work is done (or similarly, at which energy is transferred).

The ability to accurately measure power was one of the key abilities which allowed early engineers to develop the steam engines which drove the Industrial Revolution. It continues to be essential for understanding how to best make use of the energy resources which drive the modern world.

How do we measure power?

The standard unit used to measure power is the watt which has the symbol \[\text{W}\]. The unit is named after the Scottish inventor and industrialist James Watt. You have probably come across the watt often in everyday life. The power output of electrical equipment such as light bulbs or stereos is typically advertised in watts.

By definition, one watt is equal to one joule of work done per second. So if \[P\] represents power in watts, \[\Delta E\] is the change in energy (number of joules) and \[\Delta t\] is the time taken in seconds then:

\[P = \frac{\Delta E}{\Delta t}\]

There is also another unit of power which is still widely used: the horsepower. This is usually given the symbol hp and has its origins in the 17th century where it referred to the power of a typical horse when being used to turn a capstan. Since then, a metric horsepower has been defined as the power required to lift a \[75~\text{kg}\] mass through a distance of 1 meter in 1 second. So how much power is this in watts?

Well, we know that when being lifted against gravity, a mass acquires gravitational potential energy \[E_p = m\cdot g\cdot h\]. So putting in the numbers we have:

\[\frac{75~\mathrm{kg} \cdot ~ 9.807 ~\mathrm{m/s^2} \cdot 1~\mathrm{m}}{1 ~\mathrm{s}} = 735.5 ~\mathrm{W}\]

Wait. I thought one horsepower was equal to 746 W?

How do we measure varying power?

In many situations where energy resources are being used, the rate of usage varies over time. The typical usage of electricity in a house (see Figure 1) is one such example. We see minimal usage during the day, followed by peaks when meals are prepared and an extended period of higher usage for evening lighting and heating.

There are at least three ways in which power is expressed which are relevant here: Instantaneous power \[P_\text{i}\], average power \[P_\text{avg}\] and peak power \[P_\text{pk}\]. It is important for the electricity company to keep track of all of these. In fact, different energy resources are often brought to bear in addressing each of them.

Instantaneous power is the power measured at a given instant in time. If we consider the equation for power, \[P = \Delta E / \Delta t\], then this is the measurement we get when \[\Delta t\] is extremely small. If you are lucky enough to have a plot of power vs time, the instantaneous power is simply the value you would read from the plot at any given time.

Average power is the power measured over a long period, i.e., when \[\Delta t\] in the equation for power is very large. One way to calculate this is to find the area under the power vs time curve (which gives the total work done) and divide by the total time. This is usually best done with calculus, but it is often possible to estimate it reasonably accurately just using geometry.

Peak power is the maximum value the instantaneous power can have in a particular system over a long period. Car engines and stereo systems are example of systems which have the ability to deliver a peak power which is much higher than their rated average power. However, it is usually only possible to maintain this power for a short time if damage is to be avoided. Nevertheless, in these applications a high peak power might be more important to the driving or listening experience than a high average power.

Figure 1 : Daily electricity usage of a typical house

Exercise 1 : Using figure 1, estimate the instantaneous power at 10 am, the average power for the entire twenty four hour time interval, and peak power.

Exercise 2: One device in which there is a huge difference between peak and average power is known as an ultrashort pulse laser. These are used in physics research and can produce pulses of light which are extremely bright, but for extremely short periods of time. A typical device might produce a pulse of duration \[100~\mathrm{fs}\] (note that \[1 \text{ fs}=10^{-15}~\mathrm{s}\]), with peak power of \[350~\mathrm{kW}\] – that's about the average power demanded by 700 homes! If such a laser produces 1000 pulses per second, what is the average power output?

Can the concept of power help us describe how objects move?

The equation for power connects work done and time. Since we know that work is done by forces, and forces can move objects, we might expect that knowing the power can allow us to learn something about the motion of a body over time.

If we substitute the work done by a force \[W=F\cdot\Delta x \text{ cos}\theta\] into the equation for power \[P=\dfrac{W}{\Delta t}\] we find:

\[P=\frac{F\cdot\Delta x \cdot \cos{\theta}}{\Delta t}\]

If the force is along the direction of motion (as it is in many problems) then \[\cos(\theta)=1\] and the equation can be re-written

\[P=F\cdot v\]

since a change in distance over time is a velocity. Or equivalently,

\[P_i = m\cdot a\cdot v\]

Note that in this equation we have made sure to specify that the power is the instantaneous power, \[P_i\]. This is because we have both acceleration and velocity in the equation and therefore the velocity is changing over time. It only makes sense if we take the velocity at a given instant. Otherwise, we need to use the average velocity, i.e.:

\[P_\mathrm{avg} = m\cdot a \cdot \frac{1}{2}(v_\mathrm{final}+v_\mathrm{initial})\]

This can be a particularly useful result. Suppose a car has a mass of \[1000 \text{ kg}\] and has an advertised power output to the wheels of \[75 \text{ kW}\] (around \[100 \text{ hp}\]). The advertiser claims that it has constant acceleration over the range of \[0–25 \dfrac{\text m}{\text s}\].

Using only this information we can find out the time the car should take under ideal conditions to accelerate from zero to a speed of 25 m/s.

\[P_{avg} = m \cdot a \cdot \frac{1}{2} v_{final}\]

Because acceleration is \[\Delta v / \Delta t\]:

\[\begin{aligned}
P_\mathrm{avg} &= m \cdot (v_{final} / t) \cdot \frac{1}{2} v_\mathrm{final} \\&= \frac{mv_\mathrm{final}^2}{2t}\end{aligned}\]

Which can be rearranged:

\[ \begin{aligned}
t &= \frac{v_\mathrm{final}^2 \cdot m}{2 \cdot P_\mathrm{avg}} \\
&= \frac{(25~\mathrm{m/s})^2 \cdot 1000~\mathrm{kg}}{2\cdot 75000~\mathrm{W}} \\
&= 4.17~\mathrm{s} \end{aligned}\] \

Exercise 2: In the real world, we are unlikely to observe such a rapid acceleration. This is because work is also being done in the opposite direction (negative work) by the force of drag as the car pushes the air aside. Suppose we trust the manufacturer at their specification, but actually observe a time t=8 s. What fraction of the power of the engine is being used to overcome drag during the test?