Why Don’t Satellites Fall Down?: Going Further 9–12

Falling and missing

You met the idea in Grade 10: a ball fired sideways from a high mountain falls, but if it is fast enough the round Earth curves away beneath it as fast as it falls, and it never lands (Orbiting is falling, with Newton’s cannon to play with). Everything inside a falling spacecraft falls with it, so the astronauts float Grade 10: Weightlessness. And in Grade 9 you met the three kinds of orbit, low, medium and geostationary Grade 9: Orbits of artificial satellites. Here we go further, and put numbers on it.

Start with one second of the Space Station’s flight. It moves at about 7.67 km/s, so in one second it goes 7,670 m sideways. At its height, 400 km up, gravity is weaker than on the ground: you worked out in Grade 10 that g there is 8.69 m/s² Grade 10: g decreases with height. Anything dropped there falls

$s = \tfrac{1}{2} g t^2 = \tfrac{1}{2} \times 8.69\ \text{m/s}^2 \times (1\ \text{s})^2 = 4.35\ \text{m}$

in the first second. Now the Earth. It is round, so a straight line drawn level from the Space Station rises away from the Earth’s surface. Over 7,670 m, how far does the curved Earth drop below that straight line? The answer (you can check it with Pythagoras’ theorem, below) is 4.34 m. The station falls 4.35 m, and the ground curves away by 4.34 m: the same, to within the rounding of the numbers. It falls, but gets no closer to the ground. That is an orbit, in numbers.

Figure 33.1 Five minutes of the Space Station’s orbit, drawn to scale. With no gravity it would fly off along the straight line (hollow dots, one a minute) and be about 380 km higher than its orbit after five minutes. Gravity bends its path instead (blue dots): in those five minutes it has fallen about 380 km away from the straight line, and it is still 400 km above the ground, because the ground has curved away beneath it just as much.

Why doesn’t the Space Station need its engines to stay up?

Pick an answer to see why.

Gravity at 400 km is still 8.69 m/s², about 89 % of what it is on the ground; that pull is exactly what bends the station’s path round the Earth. There is almost no air up there, so nothing holds it up and almost nothing slows it down. Once a satellite is moving fast enough, it needs no engine to keep going round (the Space Station does fire its engines now and then, but only to make up for the tiny drag of the thin air).

How fast must a satellite go?

Going round in a circle needs a pull

Whirl a stone round on a string. The stone wants to go straight (Newton’s first law), and the string keeps pulling it inwards, towards your hand, so its path bends into a circle. Let go, and it flies off in a straight line, along the direction it was moving at that moment. A bus going round a bend on a hill road is the same: the road’s grip on the tyres pulls the bus towards the inside of the bend, and you, sliding on your seat, feel thrown outwards because your body tries to go straight on.

So anything moving in a circle is being pulled towards the center all the time. Its speed may stay the same, but its direction keeps changing, so it is accelerating (Grade 9: acceleration is any change in velocity, and velocity includes direction). This acceleration towards the center of a circle is called centripetal acceleration (from Latin for “center-seeking”). How big must it be?

Look at Figure 33.1 again, for a very short time t. The station, at distance r from the Earth’s center, moves a distance vt sideways along the straight line. The straight line, the line from the station to the center, and the line from the center to the end of the straight line make a right-angled triangle. The end of the straight line is a little farther from the center than the orbit: call the extra distance d. That is how far the station must fall in time t to stay on its orbit. The end of the straight line is r + d from the center, so by Pythagoras:

$(r + d)^2 = r^2 + (vt)^2 \quad\Rightarrow\quad 2rd + d^2 = v^2t^2$

For a short time, d is tiny next to r (4 m next to 6,771 km), so d² is far smaller still and can be left out:

$d = \frac{v^2 t^2}{2r}$

But we also know how far something falls from rest in time t with acceleration a: $d = \tfrac{1}{2} a t^2$ Grade 10: Equations of motion. For the orbit to work, the two must be equal:

$\tfrac{1}{2} a t^2 = \frac{v^2 t^2}{2r} \quad\Rightarrow\quad a = \frac{v^2}{r}$

Any object going round a circle of radius r at speed v must have an acceleration of v²/r towards the center, and so, by Newton’s second law, a force mv²/r towards the center. That force is called the centripetal force.

Gravity does the pulling

For a satellite, gravity is the only force on it, so gravity must give exactly the acceleration v²/r. You know from Grade 10 how strong gravity is at a distance r from the Earth’s center: $g = \dfrac{GM}{r^2}$ , where M is the Earth’s mass and G = 6.67 × 10⁻¹¹ N m²/kg² Grade 10: Newton’s universal law of gravitation. So for a circular orbit:

$\frac{v^2}{r} = \frac{GM}{r^2} \quad\Rightarrow\quad v = \sqrt{\frac{GM}{r}}$

This is the orbital speed: the speed a satellite must have to go round in a circle at distance r from the center of the planet it orbits. Three things to notice:

r is measured from the Earth’s center: the Earth’s radius plus the height above the ground.
The farther out, the slower the orbit. Gravity is weaker there, so less speed is enough.
The satellite’s own mass is nowhere in it. It canceled: a heavier satellite is pulled harder, but is harder to turn by just as much. A 1.3 kg CubeSat and the 420-tonne Space Station at the same height go at the same speed. So does an astronaut, and a drop of her tea, which is why they float beside the station instead of falling to its floor.

A 1,000 kg satellite and a 1 kg satellite are put into circular orbits at the same height. Which needs the greater speed?

Pick an answer to see why.

A story from science

The rocket that fell short

Munal, the satellite lost on 12 January 2026, would not have been Nepal’s first in orbit. The first, NepaliSat-1, a 1.3 kg cube 10 cm on each side built by two Nepali engineers studying in Japan, went up in April 2019 on a cargo flight to the Space Station, and was pushed out from the station on 17 June 2019 into its own orbit, about 400 km up. It already had the station’s 7.7 km/s, so it needed no rocket of its own.

Munal’s ride went wrong in the last part of the climb. India’s space agency said the rocket’s third stage was disturbed while it burned, the rocket strayed from its path, and it never reached orbital speed. Without that speed, everything it carried simply fell back. (You can read more about NepaliSat-1 in Grade 9: NepaliSat-1.)

How long does one orbit take?

Once you know the speed, the time for one orbit, its period T, follows: once round the circle is a distance 2πr, so

$T = \frac{2\pi r}{v}$

Putting in $v = \sqrt{GM/r}$ gives the period from the distance alone:

$T = 2\pi \sqrt{\frac{r^3}{GM}}$

Higher orbits take longer for two reasons at once: the way round is longer, and the satellite goes more slowly.

Move the satellite up and down, from just above the air to beyond the satellites that carry TV. Everything is drawn to scale, seen from above the North Pole, and the view zooms out as the orbit grows. Watch the speed and the time for one orbit, and watch the red line that turns with the Earth, pointing towards Nepal. Can you find the height where the satellite keeps up with it?

Height above the ground: 400 km

Speed: 7.67 km/s
One orbit takes: 1 h 32 min
Gravity up there: 8.69 N/kg

The height of the Space Station: one lap in about an hour and a half, nearly 16 laps a day. From Nepal it crosses the sky in a few minutes.

Why geostationary satellites are 36,000 km up

A dish on a roof in Bhaktapur stays pointed at one spot in the southern sky for years. The satellite it looks at must go round once in exactly the time the Earth takes to turn once, 23 hours 56 minutes 4 seconds (Grade 9 explains why that, and not 24 hours: Geostationary Earth Orbit), in the same direction, above the equator. Only one distance from the Earth gives that period. Nobody chose it: the law of gravitation did.

Worked example

Finding the geostationary height

Find the radius of an orbit whose period is 23 h 56 min 4 s, and its height above the equator (the Earth’s equatorial radius is 6,378 km).

Answer

T = 23 × 3,600 + 56 × 60 + 4 = 86,164 s.
Rearrange $T = 2\pi\sqrt{r^3/GM}$ for r: square both sides, $T^2 = 4\pi^2 r^3 / GM$ , so
$r^3 = \frac{GMT^2}{4\pi^2} = \frac{3.98 \times 10^{14} \times (86{,}164)^2}{4 \times (3.14)^2} = \frac{3.98 \times 10^{14} \times 7.424 \times 10^{9}}{39.4} = 7.50 \times 10^{22}\ \text{m}^3$
$r = \sqrt[3]{7.50 \times 10^{22}} = 4.22 \times 10^{7}\ \text{m} = 42{,}200\ \text{km}$
Height = 42,200 − 6,378 ≈ 35,800 km above the equator. (With more exact values of G, M and π, the answer is 35,786 km, the figure you met in Grade 9.) Its speed is $2\pi r / T$ = 2 × 3.14 × 4.22 × 10⁷ ÷ 86,164 = 3.08 × 10³ m/s, about 3.1 km/s.

A dark star field full of short slanting streaks, with six boxes marked on it; below, the six boxes enlarged, each showing one or two sharp dots labeled with satellite names such as Thuraya 2, Turksat 3A and Intelsat 12 — **Figure 33.2** A long exposure of the sky around the celestial equator, with the camera held still on the ground. The stars, which the turning Earth carries past, leave short streaks; the geostationary satellites, which turn with the Earth, stay as sharp dots (enlarged below, with their names). Photo: Marcoaliaslama, CC BY-SA 3.0

A company wants a satellite that stays above Nepal’s longitude all the time, but at only 10,000 km up, so its signal is stronger. Why won’t it work?

Pick an answer to see why.

Weighing the Earth, the Sun and Jupiter

Turn the period formula round, and it does something remarkable: it weighs whatever is in the middle. Squaring $T = 2\pi\sqrt{r^3/GM}$ and rearranging,

$M = \frac{4\pi^2 r^3}{G T^2}$

Measure how far a moon is from its planet, and how long it takes to go round, and you have the planet’s mass. Nobody has ever put the Sun on a balance; this is how we know its mass.

A story from science

The apple and the Moon

In 1665–66 a plague closed Cambridge University, and the young Isaac Newton went home to his family’s farm. There, he later wrote, he began to think that the pull that makes an apple fall might reach out as far as the Moon, getting weaker with distance, as one over the distance squared. He could test it. The Moon is about 60 Earth radii from the Earth’s center, so its pull there should be 60 × 60 = 3,600 times weaker than at the ground:

$\frac{9.8\ \text{m/s}^2}{3600} = 0.00272\ \text{m/s}^2$

And the Moon’s actual acceleration, from its orbit (r = 3.844 × 10⁸ m, T = 27.32 days = 2.36 × 10⁶ s), is

$a = \frac{v^2}{r} = \frac{4\pi^2 r}{T^2} = \frac{39.5 \times 3.844 \times 10^{8}}{(2.36 \times 10^{6})^2} = 0.00272\ \text{m/s}^2$

The same. The Moon falls towards the Earth by about 1.4 mm every second, while moving about 1 km sideways, and the apple and the Moon obey one law. Newton wrote that he found the two “answer pretty nearly”. He published the full theory in 1687, in his book the Principia.

Worked example

Weighing the Earth with the Moon

The Moon goes round the Earth once in 27.32 days, at an average distance of 3.844 × 10⁸ m. What mass does this give for the Earth?

Answer

T = 27.32 × 86,400 s = 2.360 × 10⁶ s.
$M = \frac{4\pi^2 r^3}{G T^2} = \frac{39.5 \times (3.844 \times 10^{8})^3}{6.67 \times 10^{-11} \times (2.360 \times 10^{6})^2} = \frac{39.5 \times 5.68 \times 10^{25}}{6.67 \times 10^{-11} \times 5.57 \times 10^{12}}$
$M = \frac{2.24 \times 10^{27}}{3.72 \times 10^{2}} = 6.0 \times 10^{24}\ \text{kg}$
Very close to the Earth’s mass, 5.97 × 10²⁴ kg. (The small difference is real: the Moon is not tiny next to the Earth, a little over 1 % of its mass, and strictly the formula gives the two masses added together.)

Try it

Weighing Jupiter

Jupiter’s moon Io goes round Jupiter once every 1.769 days, at 4.217 × 10⁸ m from Jupiter’s center. Find Jupiter’s mass, and compare it with the Earth’s.

Answer

T = 1.769 × 86,400 = 1.528 × 10⁵ s.
$M = \frac{39.5 \times (4.217 \times 10^{8})^3}{6.67 \times 10^{-11} \times (1.528 \times 10^{5})^2} = \frac{39.5 \times 7.50 \times 10^{25}}{6.67 \times 10^{-11} \times 2.335 \times 10^{10}} = 1.90 \times 10^{27}\ \text{kg}$
That is 1.90 × 10²⁷ ÷ 5.97 × 10²⁴ = about 318 times the Earth’s mass. Galileo found Io in 1610; once Newton’s law was known, Io and its sister moons weighed Jupiter.

Kepler’s three laws

The orbits so far have been circles. Real orbits are only nearly circles, and the rules they follow were found from the sky, half a century before Newton explained them.

A story from science

Eight minutes of arc

For twenty years the Danish nobleman Tycho Brahe measured the positions of the planets night after night, without a telescope (it hadn’t been invented), with giant instruments of his own design, more accurately than anyone before him. In 1600 a poor German mathematician, Johannes Kepler, came to work for him in Prague. Tycho died the next year, and Kepler inherited his records.

Kepler set out to find the shape of Mars’s orbit. Everyone, for two thousand years, had assumed the planets moved in circles, or circles on circles. After years of calculation by hand, Kepler’s best circle still disagreed with Tycho’s measurements by 8 minutes of arc, about a quarter of the width of the full Moon. Anyone else might have blamed the measurements. Kepler trusted Tycho, who had never been that far wrong, and threw the circle away. He wrote that those eight minutes alone had led the way to the reform of the whole of astronomy. The orbit, he found, was an ellipse. He published this in 1609. His third law took nine more years: he found it on 15 May 1618 and published it the next year.

A painted portrait of a man with a mustache and a pointed beard and a wide white lace collar, holding a pair of dividers over a globe — **Figure 33.3** Johannes Kepler (1571–1630). Photo: Anonymous Unknown author, Public domain

The first law: ellipses

Kepler’s first law: each planet moves in an ellipse, with the Sun at one focus.

An ellipse is a stretched circle. It has two special points inside it, its foci (one focus, two foci), and for every point on the ellipse, the distances to the two foci add up to the same total. Half the ellipse’s longest width is its semi-major axis, a; for a planet it is the average of its nearest and farthest distances from the Sun. How stretched an ellipse is, is its eccentricity, e: 0 for a circle, close to 1 for a long, thin one.

Figure 33.4 Drawing an ellipse with two pins and a loop of string. However the pencil moves, the string stays the same length, so the distances from the pencil to the two pins always add up to the same total. Each pin is at a focus.

Activity

Draw an orbit

You need: a piece of cardboard or a thick folder, a sheet of paper, two drawing pins, about 30 cm of thread, and a pencil.

Tie the thread into a loop about 12 cm long when stretched out flat (doubled, it is 24 cm round).
Pin the paper to the cardboard, and push the two pins in 6 cm apart, near the middle.
Put the loop round both pins, put the pencil inside it, and pull it tight. Keeping the thread tight, draw all the way round.
Now move the pins closer together, 3 cm, then 1 cm, and draw again with the same loop. Then try them very close: 4 mm apart.
Mark one pin “Sun”. Where on your ellipse would a planet be nearest the Sun? Farthest?

What you should see

Each drawing is an ellipse. The farther apart the pins, the more stretched it is; with the pins together, it is a circle. With the pins 6 cm apart and a loop of 24 cm, the ellipse is 18 cm long (a = 9 cm) and its eccentricity is 3 ÷ 9 = 0.33, more stretched than any planet’s orbit (Mercury’s, the most stretched, is 0.21). With the pins 4 mm apart (e ≈ 0.017), you have drawn the shape of the Earth’s orbit, and you can’t tell it from a circle. The planet is nearest the Sun at the end of the long axis closest to the “Sun” pin, and farthest at the other end. The other pin marks nothing at all: in a real orbit, the second focus is empty space.

Most planets’ orbits are very nearly circles: the Earth’s eccentricity is 0.017, so its distance from the Sun changes by only about 3 %, from 147.1 million km in early January to 152.1 million km in early July. Comets have long, thin orbits: Halley’s comet (e = 0.967) swings in to 0.59 times the Earth’s distance from the Sun, inside the orbit of Venus, and out to 35 times, beyond Neptune.

The second law: equal areas in equal times

Kepler’s second law: the line from the Sun to a planet sweeps out equal areas in equal times.

On a stretched orbit, a planet near the Sun is on a short line, so it must move fast to sweep out its share of area; far away, the line is long, and a slow crawl sweeps out just as much. So a planet speeds up as it falls in towards the Sun, and slows down as it climbs away, like a ball thrown up a hill and rolling back.

Each colored slice is the area the line from the Sun sweeps out in one twelfth of the orbit’s time. Stretch the orbit and watch the planet and the slices. Every orbit here has the same semi-major axis.

How stretched (eccentricity e): 0.50

Nearest ÷ farthest distance: 0.33
Fastest ÷ slowest speed: 3.00
Time for one orbit: the same for every e

Stretched: close to the Sun the planet races along, far away it crawls. Each slice still takes a twelfth of the orbit, and has the same area.

Take a very short time at the nearest point, and the same short time at the farthest. Each slice swept out is then a thin triangle, of height r and base v × (the time), so equal areas mean the same value of r × v:

$r_1 v_1 = r_2 v_2$

where r₁ and v₁ are the distance and speed at the nearest point and r₂ and v₂ at the farthest. (This is true only at those two points, where the planet moves at right angles to the line from the Sun.)

A comet with a bright head and a long, straight, faint tail stretching across a starry sky — **Figure 33.5** Halley’s comet on 8 March 1986, photographed from Easter Island. It will next come close to the Sun in 2061. Photo: NASA/W. Liller, Public domain

In which month does the Earth move fastest along its orbit?

Pick an answer to see why.

The third law: the period and the size of the orbit

Kepler’s third law: the square of a planet’s period is proportional to the cube of its semi-major axis, $T^2 \propto a^3$ .

You already have it. For a circular orbit, squaring $T = 2\pi\sqrt{r^3/GM}$ gives

$T^2 = \frac{4\pi^2}{GM} \, r^3$

and $4\pi^2/GM$ is the same for every planet of the Sun. Newton showed in the Principia that this is still true for an ellipse, with the semi-major axis a in place of r. With T in years and a in astronomical units (AU; 1 AU is the Earth’s average distance from the Sun, 1.496 × 10⁸ km), the Earth has T = 1 and a = 1, so for every planet T² = a³. Check it:

Planet	a (AU)	T (years)	a³	T²
Mercury	0.387	0.241	0.058	0.058
Venus	0.723	0.615	0.378	0.378
Earth	1.000	1.000	1.000	1.000
Mars	1.524	1.881	3.54	3.54
Jupiter	5.20	11.86	141	141
Saturn	9.55	29.46	871	868
Uranus	19.22	84.0	7,100	7,060
Neptune	30.11	164.8	27,300	27,200

(Distances and periods from NASA JPL’s planetary data; the outer planets are off by a fraction of a percent because they tug on one another a little.)

The same law works for the Earth’s satellites, with the Earth’s mass in it in place of the Sun’s, and with an ellipse’s semi-major axis in place of the radius.

Worked example

Sputnik’s ellipse

Sputnik 1, the first satellite, went round on an ellipse that came down to 215 km above the ground and went up to 939 km. Find its period.

Answer

The long axis of the ellipse runs from the lowest point, through the Earth’s center, to the highest point: 2a = (6,371 + 215) + (6,371 + 939) = 13,896 km, so a = 6,948 km = 6.948 × 10⁶ m.
$T = 2\pi\sqrt{\frac{a^3}{GM}} = 6.28 \times \sqrt{\frac{3.354 \times 10^{20}}{3.98 \times 10^{14}}} = 6.28 \times \sqrt{8.43 \times 10^{5}} = 6.28 \times 918 = 5{,}765\ \text{s}$
5,765 s ÷ 60 = 96.1 minutes. The measured period in October 1957 was 96.2 minutes.

Two satellites have the same semi-major axis. One has a circular orbit, the other a stretched ellipse. Which takes longer to go round?

Pick an answer to see why.

Changing orbit

Mostly speed, not height

Getting a satellite into orbit takes energy, and you can count it as you did in Kinetic energy: the energy of motion and Gravitational potential energy: the energy of height.

Worked example

Lifting it, or speeding it up?

Each kilogram of the Space Station has to be raised 400 km and given a speed of 7.67 km/s. a. How much kinetic energy does each kilogram need? b. Estimate the gravitational potential energy each kilogram gains, with GPE = mgh and g = 9.8 m/s². c. Which is bigger, and by how much?

Answer

a.KE = ½mv² = ½ × 1 kg × (7,670 m/s)² = 2.94 × 10⁷ J, 29 MJ.
b.GPE = mgh = 1 kg × 9.8 m/s² × 400,000 m = 3.9 × 10⁶ J, 3.9 MJ. (This is a slight overestimate: g falls to 8.69 m/s² on the way up. Done properly, the answer is 3.7 MJ.)
c.The kinetic energy is about 8 times bigger (29 MJ ÷ 3.7 MJ). Getting into orbit is mostly about speed, not height. That is why the rockets that carry satellites turn over sideways soon after launch, and spend most of their fuel speeding up along the horizon.

Going up by speeding up

Communications satellites are rarely launched straight to 35,786 km. The rocket first puts them in a low orbit, then sends them up a transfer orbit: an ellipse that just touches the low orbit at its nearest point and the geostationary orbit at its farthest. To get on to it, the satellite fires its engine to push itself forwards, and speeds up. It then climbs, slowing down as it climbs, like a ball thrown upwards. At the top it is going too slowly to stay up there, so it fires its engine to push forwards once more, up to the geostationary speed.

Figure 33.6 From a low orbit to geostationary orbit, to scale. Speeding up at the low orbit sends the satellite on a stretched ellipse up to 35,786 km. It arrives about 5¼ hours later going slowly, and speeds up again to stay on the circle. The two speeds at the ends of the transfer orbit obey Kepler’s second law: 10.2 × 6,578 = 1.6 × 42,164, near enough.

Slowly falling for real

Even at 400 km there is a trace of air, a few trillionths as dense as at the ground, and the Space Station plows through it at 7.67 km/s. The drag slowly takes away its energy, and it sinks lower, where the air is thicker. Left alone, it would come down in a year or two, so every few weeks or months a spacecraft docked to it fires its engines to push it back up.

Small satellites have no engines to do that. NepaliSat-1, pushed out from the Space Station at 400 km in June 2019, sank lower and lower and burned up in the air on 4 October 2021. Nepal PQ-1, a 250 g satellite only 5 cm on a side, built by the Nepali company ORION Space with AMSAT-Nepal, was launched in January 2022 and burned up in February 2024. Satellites in geostationary orbit, far above the air, can stay up for millions of years; when one’s working life is over, its last fuel is used to move it a few hundred kilometers higher, out of everyone’s way, into a graveyard orbit.

A satellite in a low orbit loses a little energy to air drag every orbit. Over the months, it

Pick an answer to see why.

A starry night sky over a lake, with a long, straight, thin streak of light running from high in the sky down towards the glowing horizon — **Figure 33.7** The Space Station crossing the evening sky over a lake in Florida in October 2016, in two long exposures joined together. The stars, turning slowly with the sky, make short dashes; the station, falling round the Earth at 7.7 km/s, draws a long line. You can watch it cross Nepal’s sky too: our Where Is the Space Station? map shows where it is now, and when it will next pass over Kathmandu. Photo: Michael Seeley, CC BY 2.0

Predicting an orbit, step by step

The formulas in this topic are for circles and ellipses, with one body pulling one other. Real spacecraft feel the Earth, the Moon, the Sun, the drag of the air and their own engines all at once, and no neat formula fits. So how do space agencies know exactly where a spacecraft will be next week? They let a computer apply Newton’s law over and over, in small steps of time. Each step is simple enough to do with a calculator:

Work out the pull of gravity on the satellite where it is now, GM/r² towards the Earth’s center.
Change its velocity: the pull changes it by (pull × time step), in the direction of the pull.
Move the satellite: with its new velocity, it moves (velocity × time step) in a straight line.
Go back to step 1, from the new position.

This is a computational model: the orbit is never written down as a formula; it is built up from the law, one step at a time. The smaller the steps, the closer the answer, and the more steps a computer must do.

A satellite starts 6,371 km up, moving sideways too slowly for a circle, so its exact orbit (dashed) is an ellipse that dips to 835 km. Work out its path step by step, first with big steps, then with smaller ones. Watch how far the path is from the exact orbit, and the speed near the Earth.

Step:

Steps: 0
Time: 0 min
Speed: 4.75 km/s
Missed the start by: –

Press One step. The computer works out the pull, changes the velocity, then moves the satellite in a straight line for one step.

Think about it

What the stepping shows

Play with the widget. With 20-minute steps, the computed satellite misses its starting point by more than 2,000 km; with 30-second steps, by almost nothing. Why do big steps go wrong? And which of Kepler’s laws can you see in the 30-second path?

Answer

In each step the computer pretends that the pull stays the same, in size and direction, for the whole step, and that the satellite moves in a straight line. In a short step that is nearly true. In a long step, the satellite moves so far that the real pull has changed a lot by the end of it, especially near the Earth, where the pull is strong and changes quickly; the errors build up from step to step. With small steps, the path closes on itself into an ellipse with the Earth at one focus (the first law), and the satellite covers much more ground per step near the Earth than far away, speeding up from 4.75 km/s at the top to about 8.4 km/s at the bottom (the second law). Newton’s law alone, applied over and over, gives Kepler’s laws.

In short

A satellite is falling all the time. It stays up because it moves sideways so fast that the Earth curves away beneath it as fast as it falls.
Moving in a circle needs an acceleration v²/r towards the center (centripetal acceleration). For a satellite, gravity provides it: GM/r² = v²/r, so the orbital speed is $v = \sqrt{GM/r}$ , with r measured from the Earth’s center. The satellite’s mass doesn’t matter.
The period is $T = 2\pi r/v = 2\pi\sqrt{r^3/GM}$ : about 92 minutes at 400 km, 12 hours for GPS, and exactly one turn of the Earth, 23 h 56 min, at 35,786 km above the equator, the geostationary height.
Turned round, $M = 4\pi^2 r^3/GT^2$ weighs a planet or star from the orbit of anything going round it.
Kepler’s laws: orbits are ellipses with the Sun (or planet) at one focus; the line to it sweeps equal areas in equal times, so an orbiting body is fastest when nearest; and T² ∝ a³. All three follow from Newton’s law of gravitation.
Getting into orbit is mostly about speed: the kinetic energy needed for low orbit is about 8 times the energy needed to lift the satellite there. Speeding up raises the other side of an orbit; air drag slowly brings low satellites down.
Computers predict orbits by applying Newton’s law in small steps of time; smaller steps give more accurate predictions.

Questions

Satellite 1 orbits 400 km up, satellite 2 orbits 4,000 km up. Compared with satellite 1, satellite 2

Pick an answer to see why.

A planet is found going round a distant star in a circular orbit. Which measurements are enough to work out the star’s mass?

Pick an answer to see why.

a.

Explain, using the idea of falling, why a satellite in a circular orbit needs no engine to keep going round, but a satellite 300 km up does need one now and then.

Answer

In orbit, the only force needed is gravity, which pulls the satellite towards the Earth; it is moving sideways so fast that as it falls, the curved surface of the Earth drops away beneath it just as fast, so it never gets closer. In a circular orbit, gravity always acts at right angles to the motion, so it changes the satellite’s direction but not its speed, and with no air there is nothing to slow it: it keeps going by Newton’s first law. But at 300 km there is still a trace of air. Its drag slowly takes away the satellite’s energy, and the satellite sinks lower, into thicker air, and would eventually burn up. An engine is needed now and then to push it back up.

b.

A satellite is to orbit 1,000 km above the Earth. (GM for the Earth = 3.98 × 10¹⁴ N m²/kg, R = 6,371 km.) a. Find its orbital speed. b. Find its period. c. How many times does it go round in a day?

Answer

a. r = 6,371 + 1,000 = 7,371 km = 7.371 × 10⁶ m.
$v = \sqrt{GM/r} = \sqrt{3.98 \times 10^{14} \div 7.371 \times 10^{6}} = \sqrt{5.40 \times 10^{7}}$ = 7.35 × 10³ m/s (7.35 km/s).

b. $T = 2\pi r / v = 2 \times 3.14 \times 7.371 \times 10^{6} \div 7.35 \times 10^{3}$ = 6.30 × 10³ s = 105 minutes.
c.In 1,440 minutes it goes round 1,440 ÷ 105 = 13.7 times.

c.

Use Kepler’s third law. a. Jupiter is 5.20 AU from the Sun. Find its period in years. b. A dwarf planet takes 248 years to go round the Sun. What is the semi-major axis of its orbit? (This is Pluto.)

Answer

a.T² = a³ = 5.20³ = 140.6, so T = √140.6 = 11.9 years.
b.a³ = T² = 248² = 61,500, so a = ∛61,500 = 39.5 AU. (Pluto’s orbit is quite stretched, e = 0.25: it comes in to about 30 AU, a little inside Neptune’s orbit, and goes out to about 49 AU.)

d.

The Moon is 3.844 × 10⁸ m from the Earth and takes 27.3 days to go round it. A geostationary satellite is 4.22 × 10⁷ m from the Earth’s center. Use Kepler’s third law (T² ∝ r³) and the Moon’s orbit to predict the satellite’s period, without using G or the Earth’s mass.

Hint

Since T² ÷ r³ is the same for both, $T_2 = T_1 \times (r_2 / r_1)^{3/2}$ .

Answer

$\frac{T_2^2}{r_2^3} = \frac{T_1^2}{r_1^3} \quad\Rightarrow\quad T_2 = T_1 \left( \frac{r_2}{r_1} \right)^{3/2} = 27.3\ \text{days} \times \left( \frac{4.22 \times 10^{7}}{3.844 \times 10^{8}} \right)^{3/2}$
$T_2 = 27.3 \times (0.1098)^{1.5} = 27.3 \times 0.03638 = 0.993\ \text{days}$
About 0.99 days, 23.8 hours: one turn of the Earth, as it should be. (The small difference comes from rounding, and because the Moon’s mass adds a little to the pull in its own orbit.)

e.

In early January the Earth is 147.1 million km from the Sun and moves at 30.3 km/s. In early July it is 152.1 million km away. a. Use Kepler’s second law to find its speed in early July. b. Explain why it is slower then.

Answer

a. At the nearest and farthest points, r₁v₁ = r₂v₂:
$v_2 = \frac{r_1 v_1}{r_2} = \frac{147.1 \times 30.3}{152.1} = 29.3\ \text{km/s}$
b. Between January and July the Earth is moving away from the Sun, against the Sun’s pull, which slows it down, just as a ball thrown upwards slows as it climbs. Between July and January it falls back in towards the Sun and speeds up again. Kepler’s second law describes this: the line from the Sun is longer in July, so the Earth needs to move less fast to sweep out the same area each day.

f.

Nepal’s government wants to choose an orbit for two satellites. One must take sharp photos of Nepal’s glacial lakes, to warn of floods; the other must relay television to all of Nepal, all day. For each, choose a height and say what its period would be, giving the reasons for each choice. What does each choice cost?

Answer

Glacial lakes: a low Earth orbit, a few hundred km up (say 500–600 km): close, so its camera can see small details. Its period is about 95 minutes. The cost: it races over Nepal in a few minutes at a time, so it sees any one lake only now and then (once every few days on a typical orbit), and drag will bring it down within years unless it has fuel to push itself back up. (Most such satellites use a sun-synchronous orbit over the poles, which passes over each place at the same time of day.)

Television: a geostationary orbit, 35,786 km above the equator, with a period of 23 h 56 min, the same as the Earth’s turn. It stays in the same place in Nepal’s sky, so fixed dishes on every roof can point at it, and from that height it can see the whole country, and nearly half the Earth, at once. The cost: it is about 60 to 70 times farther away than the low satellite, so the signal is weaker and takes about a quarter of a second to go up and back down; and getting there takes a much bigger rocket.

g.

Galileo found four moons of Jupiter in 1610. The innermost, Io, goes round in 1.77 days at 422,000 km from Jupiter’s center; the outermost, Callisto, at 1,883,000 km. a. Predict Callisto’s period. b. Its measured period is 16.7 days. What does the agreement show?

Answer

a. By Kepler’s third law for Jupiter’s moons,
$T_2 = T_1 \left( \frac{r_2}{r_1} \right)^{3/2} = 1.77 \times \left( \frac{1{,}883{,}000}{422{,}000} \right)^{3/2} = 1.77 \times (4.462)^{1.5} = 1.77 \times 9.43 = 16.7\ \text{days}$
b. Kepler found his third law for the planets going round the Sun. That it predicts the periods of moons going round Jupiter just as well shows that the same law of gravity, with a different mass in the middle, rules both: it is a universal law, not a special rule for the Sun’s family.

Where to go next

Why Is Gravity So Weak? The pull that holds satellites up is the weakest force there is; yet it rules the sky. Compare it with the electric force, and see what a gravitational field is.
How Long Will the Sun Shine? Weighing stars by their orbits, as you weighed the Sun here, is how astronomers found that heavier stars live far shorter lives.
How Do We Know the Universe Is Expanding? Gravity also pulls on the whole universe. Whether the galaxies, like a rocket, can escape each other’s pull is the question at the end of Grade 10’s future of the universe.