A satellite circles Earth because gravity pulls it toward the center while it moves sideways. Gravity is the centripetal force. That one idea gives the orbital speed, the period, and why the people inside float.
Why it works
Imagine Newton's cannon on a very tall mountain. Fire a ball sideways and it curves down to the ground. Fire it faster and it lands farther away. At about $7.9\ \text{km/s}$ near Earth's surface, the ground curves away as fast as the ball falls: it keeps falling and never lands. That is an orbit. For a circle of radius $r$ (measured from the planet's center), gravity must supply exactly the centripetal force: $$\frac{GMm}{r^2} = \frac{mv^2}{r} \quad\Rightarrow\quad v = \sqrt{\frac{GM}{r}}.$$ The satellite's own mass cancels.
Key idea
For a circular orbit of radius $r = R + h$ around a planet of mass $M$: $v = \sqrt{\dfrac{GM}{r}}$ and period $T = \dfrac{2\pi r}{v}$. Lower orbits are faster, and take less time. Everything in an orbiting craft falls together, so it feels weightless even though gravity is strong.
Worked examples
Worked example: A space station
A space station orbits $400\ \text{km}$ above Earth. Find its speed and period. Use $G = 6.67 \times 10^{-11}\ \text{N·m}^2\text{/kg}^2$, $M = 5.97 \times 10^{24}\ \text{kg}$, $R = 6.37 \times 10^{6}\ \text{m}$.
- $$r = 6.37 \times 10^{6} + 4.00 \times 10^{5} = 6.77 \times 10^{6}\ \text{m}.$$
- $$v = \sqrt{\frac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{6.77 \times 10^{6}}} = 7.67 \times 10^{3}\ \text{m/s}.$$
- $T = \dfrac{2\pi(6.77 \times 10^{6})}{7670} = 5550\ \text{s} \approx 92$ minutes.
Answer: $7.67\ \text{km/s}$; about $92\ \text{min}$
Worked example: How strong is gravity up there?
What is $g$ at that station's height?
- $$g = \dfrac{GM}{r^2} = \dfrac{(6.67 \times 10^{-11})(5.97 \times 10^{24})}{(6.77 \times 10^{6})^2} = 8.7\ \text{m/s}^2.$$
- That is close to 90% of its value at the surface. The crew float because they and the station fall around Earth together, not because gravity is missing.
Answer: $8.7\ \text{m/s}^2$
Worked example: An elliptical orbit
A satellite on an ellipse is $7.0 \times 10^{6}\ \text{m}$ from Earth's center at its closest, moving at $9.0\ \text{km/s}$. Its farthest point is $2.1 \times 10^{7}\ \text{m}$ away. How fast does it move there?
- Gravity points at Earth's center, so it makes no torque about it, and angular momentum is conserved.
- At both ends the velocity is perpendicular to the radius: $v_p r_p = v_a r_a$.
- $v_a = (9.0\ \text{km/s})\dfrac{7.0 \times 10^{6}}{2.1 \times 10^{7}} = 3.0\ \text{km/s}$: slowest when farthest, which is Kepler's second law.
Answer: $3.0\ \text{km/s}$
Common mistake
Thinking astronauts float because there is no gravity in orbit. At $400\ \text{km}$ up, gravity is still about 90% as strong as on the ground, and it is the very force that keeps the station on its curved path. Without it the station would fly off in a straight line. The astronauts float because they are in free fall along with their ship: nothing has to hold them up, just as you feel light at the top of a drop ride.
Higher orbits are slower and take longer: a satellite about $35{,}800\ \text{km}$ up takes one day to go around, so it stays above one spot on the equator. Fly your own orbits in the Orbit Lab of the Science Lab.