Every few months a PSLV rocket rises from Sriharikota, on the coast of Andhra Pradesh. It lifts a satellite of about a tonne to a height of several hundred kilometres. At that height the Earth still pulls hard, but less hard than at the ground. The formula mgh assumes that the pull never changes. So how much energy does the climb really take?
When mgh is not enough
In the chapter on Potential Energy and Conservation of Energy, raising a mass \(m\) through a height \(h\) stored \(mgh\). That result rested on one fact: the weight \(mg\) stays the same all the way up. Near the ground this is true to a very good approximation. A climb of \(3\,\text{km}\) changes \(g\) by less than one part in a thousand.
In the chapter on Gravitational Force and Field you met Newton's law of gravitation. Two point masses \(M\) and \(m\) a distance \(r\) apart attract each other with a force \(GMm/r^2\), where \(G = 6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2\). At the Earth's surface this pull is the weight, so \(GMm/R^2 = mg\), where \(R = 6.4 \times 10^6\,\text{m}\) is the radius of the Earth. Here \(r\) is measured from the Earth's centre, and \(g = 9.8\,\text{m/s}^2\) in this unit.
The work of a weakening pull
Now move a mass \(m\) straight outward, slowly, from a distance \(r_i\) to a distance \(r_f\). At each point your pull just balances gravity, so you pull with \(GMm/r^2\). This force changes at every step, so the work is not force times distance. As in the chapter on Work, Kinetic Energy and Power, the work is the area under the force-distance graph, an integral.
\[ dW = \dfrac{GMm}{r^2}\,dr \]
Why. over a small step \(dr\) the pull hardly changes, so the work of the slow outward pull is the force times the step.
\[ W = \int_{r_i}^{r_f} \dfrac{GMm}{r^2}\,dr \]
Why. add the small steps from the start \(r_i\) to the end \(r_f\).
\[ W = GMm\left[-\dfrac{1}{r}\right]_{r_i}^{r_f} \]
Why. the integral of \(1/r^2\) is \(-1/r\).
\[ W = GMm\left(\dfrac{1}{r_i} - \dfrac{1}{r_f}\right) \]
Read it. substitute the limits. The result is positive when \(r_f > r_i\): moving outward always takes work.
Gravity is a central force. In the chapter on Potential Energy and Conservation of Energy you saw that every central force is conservative. So this work depends only on \(r_i\) and \(r_f\), not on the route. A satellite lifted along a slanting, curving path needs the same work as one lifted straight up.
Drag the far end outward. The area keeps growing, but more and more slowly, because the pull out there is weak. The lower box shows the work still left beyond \(r\). It is \(GMm/r\), and it shrinks towards zero as \(r\) grows. So even the whole journey to infinity costs a finite amount: \(GMm/R\) from the surface. That finite total is what allows a natural choice of zero, in the next topic.
A \(500\,\text{kg}\) satellite is moved slowly from \(r = 2R\) to \(r = 3R\), where \(R = 6.4 \times 10^6\,\text{m}\). Take \(g = 9.8\,\text{m/s}^2\). How much work is needed, in gigajoules (GJ)?
Choosing the zero
Only a change in potential energy has meaning, so \(U = 0\) may be placed anywhere. Near the ground we chose the floor. For gravity on a large scale, one choice is far better than the rest. We put \(U = 0\) when the two masses are infinitely far apart. There the pull has vanished, and the two masses no longer interact at all.
Take the mass \(m\) from a distance \(r\) out to infinity. The work \(GMm/r\) that you do is stored as potential energy, so \(U(\infty) - U(r) = GMm/r\). With \(U(\infty) = 0\), this gives \(U(r) = -GMm/r\). This is the gravitational potential energy of the pair.
Two point masses \(M\) and \(m\) a distance \(r\) apart have the gravitational potential energy \(U = -GMm/r\), with \(U = 0\) when they are infinitely far apart. The same holds for two uniform spheres, with \(r\) measured between their centres.
Why the value is negative
The minus sign is not an accident of the choice. Gravity always pulls the two masses together, so separating them to infinity always takes positive work. The pair therefore starts with less energy than it has at infinity, and less than zero is negative. A negative \(U\) says that the pair is bound: to part the two masses, energy must be supplied.
As \(r\) grows, \(-GMm/r\) becomes less negative, so \(U\) increases towards zero. A satellite raised to a greater height gains potential energy, just as a brick raised to a higher shelf does. Negative numbers need care here: \(-5\) is greater than \(-10\).
Keep the minus sign when you compare energies. At \(r = 2R\) the energy is \(-GMm/2R\), which is greater than \(-GMm/R\) at the surface. The change on the way up, \(+GMm/2R\), is positive, as it must be for a lift.
The potential energy of a satellite and the Earth is \(U_1\) when the satellite is at \(r = R\) and \(U_2\) when it is at \(r = 2R\). Which statement is correct?
The energy \(-GMm/r\) belongs to the Earth and the satellite together. The chapter on Potential Energy and Conservation of Energy showed the same for a ball and the Earth. The Earth is so much heavier that it hardly moves, so we say "the potential energy of the satellite" for short. For two stars of similar mass both bodies move, and the energy must be given to the pair. Lesson 7 returns to such a pair.
The Earth has a mass of \(6.0 \times 10^{24}\,\text{kg}\) and the Moon \(7.4 \times 10^{22}\,\text{kg}\). Their centres are \(3.84 \times 10^8\,\text{m}\) apart. (a) Find the gravitational potential energy of the pair. (b) How much work would separate them completely?
(a) \(U = -GMm/r = -(6.67 \times 10^{-11})(6.0 \times 10^{24})(7.4 \times 10^{22})/(3.84 \times 10^8) = -7.7 \times 10^{28}\,\text{J}\).
(b) Separating them means taking \(U\) from \(-7.7 \times 10^{28}\,\text{J}\) up to zero. That needs \(+7.7 \times 10^{28}\,\text{J}\) of work, about a hundred million times what all of humanity uses in a year.
How much work is needed to take a \(1\,\text{kg}\) mass slowly from the Earth's surface to infinity, in megajoules? Take \(g = 9.8\,\text{m/s}^2\) and \(R = 6.4 \times 10^6\,\text{m}\).
Where mgh comes from
Lift a mass \(m\) from the surface, \(r = R\), to a height \(h\), \(r = R + h\). The potential energy rises by \[ \Delta U = -\dfrac{GMm}{R + h} + \dfrac{GMm}{R} = \dfrac{GMm\,h}{R(R + h)}. \] Put \(GM = gR^2\), and the rise becomes \(\Delta U = mgh \cdot R/(R + h)\).
When \(h\) is small compared with \(R\), the factor \(R/(R + h)\) is very close to 1, and \(\Delta U = mgh\). For a climb of \(10\,\text{km}\) the factor is \(0.998\). So the old formula is the near-surface form of the new one, not a different law. On a graph of \(U\) against \(r\), the straight line \(mgh\) is the tangent to the curve at the surface.
Look at \(r = 2R\) in Fig 1.1, a height \(h = R\). The curve has risen by half of \(GMm/R\), from \(-1\) to \(-0.5\). The line \(mgh\) has risen by \(mgR = GMm/R\), the whole of it. So at that height \(mgh\) doubles the true value. The formula \(mgh\) always overestimates, because it keeps the full surface pull all the way up.
A PSLV places a \(1000\,\text{kg}\) satellite at a height of \(1600\,\text{km}\). Take \(g = 9.8\,\text{m/s}^2\) and \(R = 6.4 \times 10^6\,\text{m}\). Find the rise in potential energy, and compare it with \(mgh\).
The height is \(h = 1.6 \times 10^6\,\text{m}\), a quarter of \(R\). So \(r\) goes from \(R\) to \(1.25R\).
\(\Delta U = mgh \cdot R/(R + h)\).
\(mgh = 1000 \times 9.8 \times 1.6 \times 10^6 = 1.57 \times 10^{10}\,\text{J}\). The factor is \(R/(R + h) = 6.4/8.0 = 0.8\). So \(\Delta U = 0.8 \times 1.57 \times 10^{10} = 1.25 \times 10^{10}\,\text{J}\).
The climb stores \(1.25 \times 10^{10}\,\text{J}\). The formula \(mgh\) claims \(25\,\%\) more, because at \(1600\,\text{km}\) the pull is only \((6.4/8.0)^2 = 0.64\) of its value at the ground.
A \(100\,\text{kg}\) probe is lifted from the Earth's surface to a height equal to the Earth's radius, \(6.4 \times 10^6\,\text{m}\). Take \(g = 9.8\,\text{m/s}^2\). By how much does its potential energy rise, in gigajoules (GJ)?
For which of these climbs from the ground does \(mgh\) give the rise in potential energy to better than \(1\,\%\)?
The hook asked how much energy the climb really takes. For the satellite in the example, the honest answer is \(1.25 \times 10^{10}\,\text{J}\), not the \(1.57 \times 10^{10}\,\text{J}\) of \(mgh\). The rocket must do more than lift, though. It must also give the satellite a great speed along its orbit, and lesson 6 counts that part of the bill.
A space station orbits at about \(400\,\text{km}\). Lifting each kilogram there raises its potential energy by about \(3.7\,\text{MJ}\). But in orbit each kilogram also moves at about \(7.7\,\text{km/s}\), which is about \(30\,\text{MJ}\) of kinetic energy. So roughly nine tenths of the energy a rocket delivers goes into speed, not height. That is why launch sites near the equator, such as Sriharikota, help: the Earth's own spin supplies a little of the speed.
- Using \(mgh\) for a climb comparable to the Earth's radius.Use \(\Delta U = GMm(1/r_i - 1/r_f)\), or \(mgh \cdot R/(R + h)\) from the surface.
- Measuring \(r\) from the Earth's surface.In \(U = -GMm/r\), \(r\) is measured from the Earth's centre: \(r = R + h\).
- Dropping the minus sign, or thinking that \(-GMm/2R\) is less than \(-GMm/R\).\(U\) rises towards zero as \(r\) grows. A less negative value is a greater value.