Chandrayaan-3 left Sriharikota on 14 July 2023 and did not head straight for the Moon. For more than two weeks it circled the Earth while ISRO fired its engine five times, each time near its lowest point. Each burn stretched the orbit farther out. Only then did it leave for the Moon. Each burn made the spacecraft faster, yet each larger orbit was slower on average. How can both be true?
Three energies of one orbit
A satellite of mass \(m\) circles a planet of mass \(M\) at radius \(r\). From the last lesson, gravity supplies the centripetal force: \(mv_o^2/r = GMm/r^2\). Multiply both sides by \(r/2\). The left side becomes the kinetic energy, so \(K = \tfrac12 mv_o^2 = GMm/2r\). The potential energy is the usual \(U = -GMm/r\), with its zero at infinity.
Add them. The total energy is \(E = K + U = GMm/2r - GMm/r = -GMm/2r\). Three neat relations follow at every radius. The kinetic energy equals minus the total, \(K = -E\). The potential energy is twice the total, \(U = 2E\). And the total is negative, so every circular orbit is bound, as the last lessons showed it must be.
Move the slider outward. The velocity arrow shrinks: at \(2R\) the satellite moves at \(5.6\,\text{km/s}\), at \(7R\) at only \(3.0\,\text{km/s}\). Yet the total energy rises, from \(-0.25\) to \(-0.07\) in units of \(GMm/R\). A higher orbit holds more energy and moves more slowly. The extra energy has gone into potential energy, which rises twice as fast as the kinetic energy falls.
A satellite on a circular orbit has kinetic energy \(8\,\text{GJ}\). What are its potential energy and its total energy?
Binding energy
To free a satellite completely, its total energy must be raised from \(E\) to zero. The energy needed is \(-E = GMm/2r\). This is the binding energy of the satellite in its orbit. It is half the depth of the potential well at that radius, because the satellite's own kinetic energy already supplies the other half.
A body at rest on the Earth's surface has no kinetic energy to help. Its total energy is \(-GMm/R\), so its binding energy is \(GMm/R = mgR\). Putting it into a grazing orbit just above the ground raises \(E\) to \(-GMm/2R\). So the launch supplies half the energy needed for escape, and a second, equal amount would finish the job.
A \(1000\,\text{kg}\) satellite circles the Earth at \(r = 2R\). Take \(g = 9.8\,\text{m/s}^2\) and \(R = 6.4 \times 10^6\,\text{m}\). Find its total energy and its binding energy. How much energy did the launch give it, ignoring the Earth's spin?
Use \(GM = gR^2\), so \(GMm/R = mgR = 1000 \times 9.8 \times 6.4 \times 10^6 = 6.27 \times 10^{10}\,\text{J}\).
In orbit \(E = -GMm/2r = -GMm/4R = -mgR/4\). On the ground at rest, \(E_0 = -mgR\).
\(E = -6.27 \times 10^{10}/4 = -1.57 \times 10^{10}\,\text{J}\), so the binding energy is \(1.57 \times 10^{10}\,\text{J}\). The launch supplied \(E - E_0 = \tfrac34 mgR = 4.70 \times 10^{10}\,\text{J}\).
In orbit \(K = 1.57 \times 10^{10}\,\text{J}\), \(U = -3.14 \times 10^{10}\,\text{J}\) and \(E = -1.57 \times 10^{10}\,\text{J}\). The launch gave \(4.70 \times 10^{10}\,\text{J}\), three times the energy still needed to free it.
A \(500\,\text{kg}\) satellite circles the Earth at \(r = 4R\). Take \(g = 9.8\,\text{m/s}^2\) and \(R = 6.4 \times 10^6\,\text{m}\). What is its binding energy, in gigajoules?
Raising and lowering an orbit
Move a satellite from a circular orbit of radius \(r_1\) to a larger one of radius \(r_2\). The engines must supply the difference in total energy: \[ \Delta E = -\dfrac{GMm}{2r_2} + \dfrac{GMm}{2r_1} = \dfrac{GMm}{2}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right). \] This is positive when \(r_2 > r_1\). It is exactly half the change in potential energy, because the kinetic energy falls by the other half.
Fig 6.1 also explains a strange fact. The thin air at a few hundred kilometres drags on a low satellite and removes energy, so \(E\) falls. A smaller, more negative \(E\) belongs to a smaller orbit, and a smaller orbit is faster. So drag makes a satellite speed up as it spirals down. Since \(K = -E\), every joule that drag removes from \(E\) adds one joule to \(K\), while \(U\) falls by two joules.
Think of each joule removed by drag as buying two joules of fall. One of them pays for the drag; the other becomes extra speed. The satellite speeds up because it has fallen, just as a stone speeds up on its way down.
How much energy must be supplied to move a \(1000\,\text{kg}\) satellite from a circular orbit at \(2R\) to one at \(4R\), in gigajoules? Take \(g = 9.8\,\text{m/s}^2\) and \(R = 6.4 \times 10^6\,\text{m}\).
Thin air slowly drags on a satellite in a low circular orbit. What happens to its speed as its orbit shrinks?
From orbit to escape
A satellite in orbit already has \(K = GMm/2r\). To escape from that radius its total energy must reach zero, so its kinetic energy must reach \(GMm/r\), exactly twice as much. Twice the kinetic energy means \(\sqrt2\) times the speed. So at any radius the escape speed is \(\sqrt2\) times the orbital speed.
At the Earth's surface this gives \(11.2 = \sqrt2 \times 7.9\,\text{km/s}\), which joins the last two lessons. The extra speed is \(41\,\%\) of the orbital speed. In the figure of Newton's cannon, the \(11.2\,\text{km/s}\) chip is exactly \(\sqrt2\) times the \(7.9\,\text{km/s}\) chip, and the path opens out and never closes.
A spacecraft circles the Earth at \(r = 2R\), where its speed is \(5.6\,\text{km/s}\). To what speed must it be raised, along its orbit, to escape? Give the answer in km/s.
Return to Chandrayaan-3. Each burn near the lowest point made the craft faster, raising its \(K\) and so its \(E\). The higher \(E\) belonged to a larger orbit, which reached farther out. Out there the craft moved slowly, so over a whole lap it was slower on average. The same energy account, \(E = -GMm/2r\) for a circle, tells engineers how much fuel each step costs.
At \(400\,\text{km}\) the air is extremely thin, yet it drags on the space station all the time. Without help the station would lose a couple of kilometres of height in a typical month, speeding up as it sank. Every few weeks a docked cargo craft fires its engine along the direction of motion to raise the orbit again. The station needs this reboost because its speed alone cannot hold it at \(400\,\text{km}\) for ever.
- Thinking that a higher orbit, being slower, has less energy.\(E = -GMm/2r\) rises towards zero as \(r\) grows. Slower means less \(K\), but \(U\) rises twice as much.
- Taking the binding energy of an orbit as \(GMm/r\).The satellite's kinetic energy already supplies half. The binding energy is \(GMm/2r\).
- Multiplying the speed by 2 to escape from an orbit.The kinetic energy doubles, so the speed grows by \(\sqrt2\).