Orbits: a fall that never ends.

Step 1 of 6

Falling and missing

Picture a cannon on top of a high mountain, firing a ball horizontally. As in projectile motion, the ball falls while it moves forward, and lands some way off. Fire it faster, and it lands farther away. But the Earth is round: fast enough, and the ground curves away beneath the ball as quickly as it falls, so it never lands. That is an orbit: a fall that keeps missing.

In the Scene the small body is $1\%$⁠ of the big one, like a small satellite beside a planet. The arrow on the small body shows the gravity on it: always towards the big body, and at right angles to the motion, just like the string in the earlier lessons. Here gravity is the centripetal force.

Step 2 of 6

How much gravity there is up there

The International Space Station orbits the Earth at a height of about 400 km. To know how much weaker gravity is there, you need the distance from the Earth’s centre, not from the ground: $r = R + h$⁠.

Step 3 of 6

Orbital speed

On a circular orbit, gravity is exactly the force needed to move on the circle:

$G\,\frac{M m}{r^2} = \frac{m v^2}{r} \quad\Longrightarrow\quad v = \sqrt{\frac{G M}{r}}$$G\,\frac{M m}{r^2} =$$\frac{m v^2}{r} \quad\Longrightarrow\quad v =$$\sqrt{\frac{G M}{r}}$⁠

Here $M$⁠ is the Earth’s mass and $m$⁠ the satellite’s. Any slower, and the satellite falls inward; any faster, and it swings out on a stretched orbit. This speed is set by $M$⁠ and $r$⁠ alone.

Step 4 of 6

Kepler’s third law

The period is the circumference divided by the speed, $T = \dfrac{2\pi r}{v}$$T =$$\dfrac{2\pi r}{v}$⁠. Put in $v$⁠ from the last step and you get

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

The square of the period is proportional to the cube of the distance. Kepler found this in measurements of the planets, about seventy years before Newton explained why. Strictly, the denominator is $G(M + m)$⁠, and $r$⁠ is the distance between the two bodies, as the Scene works it out; when the small body is negligible, that is very nearly $GM$⁠.

The Scene’s clock is sped up enormously: an orbit here takes seconds, not hours or years. But the ratios between periods are exact. The small body is now $4$⁠ away, and one orbit takes about 3.3 s.

Step 5 of 6

On the same orbit

The formula $v = \sqrt{\dfrac{G M}{r}}$$v =$$\sqrt{\dfrac{G M}{r}}$⁠ does not contain the satellite’s mass. It appeared on both sides of the equation, and cancelled.

Step 6 of 6

Twice the period

The small body is now $6$⁠ away, and one orbit takes about 6.1 s in the Scene. When you press Check, you will see how long an orbit takes at the distance you chose. Use Kepler’s law to work it out first, rather than guessing.