Soon after sunset, a bright dot sometimes crosses the Indian sky from west to east in about five minutes. It does not blink like an aircraft. It is the International Space Station, lit by the Sun that has already set for you. It circles the Earth about \(400\,\text{km}\) up, and it will come round again in about an hour and a half.
Newton's cannon
Fire a cannonball level from a tall mountain. It curves down and lands some distance away. Fire it faster and it lands farther away. But the Earth is round, so the ground ahead also curves down. At one particular speed the ball falls exactly as fast as the ground drops beneath it. It never comes any closer to the ground, and it circles the whole Earth.
The cannon fires level at \(9\,\text{km/s}\), faster than the speed for a circle, \(7.9\,\text{km/s}\), but slower than the escape speed, \(11.2\,\text{km/s}\). What does the ball do?
In the figure above, each chip gives the speed for a launch just above the ground. The mountain is drawn far too tall, so that it can be seen. At \(4\) and \(7\,\text{km/s}\) the ball lands, farther round the Earth each time. At \(7.9\,\text{km/s}\) it circles. At \(9\,\text{km/s}\) it follows an ellipse and returns to the mountain. At \(11.2\,\text{km/s}\) it escapes. Every path is computed from the law of gravitation alone.
A satellite is exactly such a ball. It is always falling towards the Earth, and always missing it. The Moon is the Earth's natural satellite; the space station and the GSAT communication satellites are artificial ones. None of them needs an engine to stay up. What keeps them up is their speed along the orbit.
The speed of a circular orbit
The chapter on Dynamics of Circular Motion showed that a body moving at speed \(v\) on a circle of radius \(r\) needs a net force. That force is \(mv^2/r\), towards the centre. For a satellite, only gravity acts. So gravity must supply exactly that force: \(GMm/r^2 = mv^2/r\). Here \(r\) is measured from the Earth's centre, and \(r = R + h\) for a height \(h\).
\[ \dfrac{GMm}{r^2} = \dfrac{mv_o^2}{r} \]
Why. the pull of the Earth is the only force, and it must be the centripetal force of the circle.
\[ v_o = \sqrt{\dfrac{GM}{r}} \]
Why. cancel \(m\) and one factor of \(r\), then take the square root. The mass of the satellite has gone.
\[ T = \dfrac{2\pi r}{v_o} = 2\pi\sqrt{\dfrac{r^3}{GM}} \]
Read it. one lap is a distance \(2\pi r\), covered at the steady speed \(v_o\).
Just above the ground, \(r \approx R\), and \(v_o = \sqrt{gR} = \sqrt{9.8 \times 6.4 \times 10^6} = 7.9\,\text{km/s}\). One lap then takes \(T = 2\pi\sqrt{R/g}\), about \(85\) minutes. Higher orbits are slower, since \(v_o\) falls as \(1/\sqrt r\). They also have longer laps, since \(T\) grows as \(r^{3/2}\). The Moon, at \(3.84 \times 10^8\,\text{m}\), takes \(27.3\) days, which is its month.
Two facts follow from the derivation. First, the satellite's mass cancelled. A cube-sat of one kilogram and a space station of four hundred tonnes at the same height move at the same speed. Second, gravity points at the Earth's centre, and the centripetal force must point at the centre of the circle. So the centre of every orbit is the centre of the Earth.
The International Space Station orbits \(400\,\text{km}\) above the ground. Take \(g = 9.8\,\text{m/s}^2\) and \(R = 6.4 \times 10^6\,\text{m}\). Find its speed and the time for one lap.
The orbit radius is \(r = R + h = 6.8 \times 10^6\,\text{m}\), measured from the Earth's centre. Also \(GM = gR^2 = 4.01 \times 10^{14}\,\text{m}^3/\text{s}^2\).
\(v_o = \sqrt{GM/r}\) and \(T = 2\pi r/v_o\).
\(v_o = \sqrt{4.01 \times 10^{14}/6.8 \times 10^6} = 7.68 \times 10^3\,\text{m/s}\). Then \(T = 2\pi \times 6.8 \times 10^6/7680 = 5560\,\text{s}\), about \(93\) minutes.
The station moves at \(7.7\,\text{km/s}\) and circles the Earth in about \(93\) minutes, about \(15\) laps a day. That is why it crosses the sky in a few minutes and comes round again so soon.
A satellite circles the Earth at \(r = 4R\), a height of three Earth radii. Near the ground the period would be \(84.6\) minutes. Find its period in hours.
Could a satellite, with its engines off, circle the Earth directly above the Tropic of Cancer, staying over \(23.5°\,\text{N}\) all the way round?
Floating in orbit
Astronauts on the space station float, and a ball let go beside them stays where it is. This is not because gravity is absent. At \(400\,\text{km}\) the field is \(g(R/r)^2 = 9.8 \times (6.4/6.8)^2 = 8.7\,\text{m/s}^2\), about nine tenths of its value on the ground. A \(60\,\text{kg}\) astronaut is pulled down with about \(520\,\text{N}\).
The astronaut floats because the station falls too. Both have the same acceleration, the local \(g\), towards the Earth's centre. So the floor falls away beneath the astronaut's feet exactly as fast as the astronaut falls. The floor does not push on the feet at all, and a weighing scale on it reads zero. That reading, the normal force, is the apparent weight. In the chapter on Accelerated Objects the same thing happened in a lift whose cable had snapped.
Why does an astronaut float inside a space station orbiting \(400\,\text{km}\) above the Earth?
The satellite that stands still
A dish on a roof in Pune points at one fixed spot in the sky, day and night. The satellite it receives must therefore stay above the same point of the Earth. Such a satellite is geostationary. It needs three conditions. Its orbit must lie over the equator. It must move west to east, as the Earth turns. And its period must equal one turn of the Earth, \(24\) hours.
The first condition follows from the last topic. Every orbit is centred on the Earth's centre, so an orbit that stays over one latitude must be the equator itself. The third condition fixes the height. Only one radius gives a period of one day, found by solving \(T = 2\pi\sqrt{r^3/GM}\) for \(r\).
Find the radius and the height of the geostationary orbit, and the satellite's speed. Take \(T = 24\,\text{h}\), \(g = 9.8\,\text{m/s}^2\) and \(R = 6.4 \times 10^6\,\text{m}\).
\(T = 24 \times 3600 = 86\,400\,\text{s}\) and \(GM = gR^2 = 4.01 \times 10^{14}\,\text{m}^3/\text{s}^2\).
Squaring \(T = 2\pi\sqrt{r^3/GM}\) gives \(r^3 = GMT^2/4\pi^2\).
\(r^3 = 4.01 \times 10^{14} \times (86\,400)^2/39.5 = 7.59 \times 10^{22}\,\text{m}^3\), so \(r = 4.23 \times 10^7\,\text{m}\), about \(6.6R\). The height is \(r - R = 3.59 \times 10^7\,\text{m}\).
The satellite sits about \(3.6 \times 10^7\,\text{m}\), some \(36\,000\,\text{km}\), above the equator. Its speed is \(2\pi r/T = 3.1\,\text{km/s}\).
In Fig 5.2 the dashed line from the dish to the outer satellite points at it all day. So the dish never has to move. The inner satellite runs ahead and passes behind the Earth, out of the dish's view, for part of each day. That is why television and weather satellites, such as India's GSAT and INSAT series, use the geostationary orbit. A dish aimed once stays aimed.
The Earth turns once relative to the stars in \(23\) hours \(56\) minutes, not \(24\) hours. The extra four minutes come from its motion round the Sun. A geostationary orbit uses this shorter period, which lowers its height by about \(80\,\text{km}\), to \(3.586 \times 10^7\,\text{m}\). Both answers round to \(3.6 \times 10^7\,\text{m}\).
A satellite must take \(24\,\text{h}\) for one lap. Taking \(GM = 4.01 \times 10^{14}\,\text{m}^3/\text{s}^2\) and \(R = 6400\,\text{km}\), how high above the surface is it, in km?
Isaac Newton wrote a short treatise on the system of the world in the 1680s; it was printed in 1728. In it he drew a mountain on the Earth with a cannon on its peak. He argued that a projectile fired ever faster would land ever farther away. At a great enough speed it would never land, but go round the Earth and return to the mountain. With that picture Newton joined the falling apple and the orbiting Moon: both are falling, and the Moon simply never lands.
A direct-to-home television dish in India points south and upward, at a satellite over the equator, for example near \(93.5°\,\text{E}\). It is fixed with bolts, because the satellite never moves in the sky. A signal that goes up to the satellite and back down travels about \(72\,000\,\text{km}\), which takes about a quarter of a second. You can hear that delay on a live television interview relayed by satellite.
The hook asked about the bright dot. The space station moves at \(7.7\,\text{km/s}\) and laps the Earth in about \(93\) minutes. So it comes round again in about an hour and a half. Its crew floats because they and the station fall together round the Earth, not because gravity has vanished.
- Putting the height \(h\) in place of \(r\) in \(v_o = \sqrt{GM/r}\).\(r\) is measured from the Earth's centre: \(r = R + h\).
- Thinking a heavier satellite must orbit faster or lower.The satellite's mass cancels. The speed depends only on \(M\) and \(r\).
- Explaining weightlessness by a lack of gravity.Gravity is still about nine tenths of its ground value. Everything aboard falls together, so the apparent weight is zero.