In the last module we found, by trial and error in a simulator, that a cannonball at 500 km altitude needs to fly at about 7.62 km/s to enter a circular orbit. Now we'll work out where that number comes from — and the formula that gives the right answer at any altitude.
A circular orbit is a balancing act between two effects. Gravity pulls the orbiting body toward the central one. Inertia — the body's tendency to keep moving in a straight line — pulls it tangentially, away from a curved path. For a circle to be the steady-state result, these two effects have to be perfectly matched at every instant.
The mathematical statement is the equation of uniform circular motion. Any object moving at constant speed $v$ along a circle of radius $r$ has an acceleration directed toward the center of the circle, with magnitude:
$$ a_{\text{centripetal}} = \frac{v^2}{r} $$This isn't a force — it's the acceleration that the body's path requires. A body that goes in a circle must be accelerating inward at exactly this rate; otherwise it would fly off in a straight line. Something has to be supplying that acceleration. For a ball on a string, the string tension does it. For a satellite, gravity does.
To find the speed at which a body orbits in a circle of radius $r$, we set the centripetal acceleration equal to the gravitational acceleration. The gravitational force on the orbiting body is $F = GMm/r^2$, so its acceleration (force divided by its own mass) is $GM/r^2$. Setting that equal to $v^2/r$ and solving for $v$:
Start by setting the gravitational acceleration equal to the centripetal acceleration:
$$ \frac{GM}{r^2} = \frac{v^2}{r} $$Multiply both sides by $r$:
$$ \frac{GM}{r} = v^2 $$And take the square root:
$$ \boxed{\, v = \sqrt{\frac{GM}{r}} \,} $$That's it. The orbital speed at radius $r$ is fully determined by the mass of the body you're orbiting and your distance from its center. Notice what's not in this equation: the orbiting body's own mass $m$ cancelled out in the second step. A bowling ball and a feather, given the right speed at the right altitude, orbit identically. This is the same equivalence-of-falling that Galileo demonstrated when he (probably didn't actually) drop things off the Leaning Tower of Pisa, dressed up for a different geometry.
Plugging in numbers for our 500-km mountain in the previous module: $r = R_\oplus + 500\ \mathrm{km} = 6{,}871\ \mathrm{km}$, and $GM_\oplus = 3.986 \times 10^{14}\ \mathrm{m^3/s^2}$. The square root works out to $7{,}616\ \mathrm{m/s}$, or 7.62 km/s — exactly the number Newton's cannonball converged on.
If we know the speed and the size of the circle, the period is just distance divided by speed: one trip around is $2\pi r$, and we cover it at $v = \sqrt{GM/r}$. So:
Period is circumference over speed:
$$ T = \frac{2\pi r}{v} = \frac{2\pi r}{\sqrt{GM/r}} $$Simplify by pulling the $r$ inside the square root:
$$ \boxed{\, T = 2\pi \sqrt{\frac{r^3}{GM}} \,} $$This is the third of Kepler's laws in disguise. Squaring both sides gives $T^2 = (4\pi^2/GM)\, r^3$, the relation $T^2 \propto r^3$ that Kepler had pulled out of Tycho's data three quarters of a century before Newton showed why it had to be true. Notice that — like the speed equation — it depends only on the mass of what you're orbiting and your distance from it. The orbit doesn't care what's orbiting.
The interactive below lets you pick an altitude above Earth's surface and see the resulting circular-orbit speed and period. The presets are real satellites — drag the slider to find them, or click to jump.
A few things are worth noticing as you slide the altitude up. First, the relationship between altitude and speed is fairly mild. Doubling the orbital radius from 6,800 km to 13,600 km only drops the orbital speed by a factor of $\sqrt{2}$ — about 30%. Higher orbits are a little slower than lower ones, but not dramatically so.
The period, on the other hand, climbs sharply. Quadrupling the radius makes the period eight times longer, because $T \propto r^{3/2}$. The ISS at 408 km circles Earth in about 92 minutes; geostationary satellites at 35,786 km take a full 24 hours.
That last number is not an accident. Geostationary orbit is defined as the altitude at which the orbital period equals one sidereal day — about 23 hours, 56 minutes — so a satellite there hovers above a fixed point on the equator. From the ground it appears stationary, which is why TV dishes can be bolted in place rather than tracking. To find that altitude, set $T = 86{,}164\ \mathrm{seconds}$ in the period equation and solve for $r$:
$$ r = \left( \frac{GM \, T^2}{4\pi^2} \right)^{1/3} $$The arithmetic gives $r \approx 42{,}164\ \mathrm{km}$ from Earth's center, or 35,786 km of altitude. The presets above will land you exactly there. Satview shows the geostationary belt as a striking ring of satellites all turning together at that altitude.
Real orbits are almost never perfect circles. Every satellite has at least a little eccentricity, and the math we just derived strictly applies only when $e = 0$. But circular orbits are a useful starting point for two reasons. The first is that the math is clean enough to derive from scratch in two lines, which is what we just did. The second is that the results carry forward: in module 3 we'll meet a more general expression — the vis-viva equation — that tells you the speed at any point on any orbit, not just circular ones. The circular formula falls out of vis-viva as the special case $r = a$.