Gravitational field of a point mass
Why the Moon doesn't fall
- The Moon is always falling toward Earth — but it also moves sideways fast.
- So it keeps missing: it is in orbit.
- Gravity provides exactly the centripetal force needed.
Field of a point mass
- At distance $r$ from a mass $M$: $g = \dfrac{GM}{r^{2}}$ — falling off as $\dfrac{1}{r^{2}}$.
- Near Earth's surface $r \approx R$ is huge, so climbing a building barely changes $g$.
The gravitational field strength at distance $r$ from a point mass $M$ is:
From $g = \dfrac{F}{m}$ with $F = \dfrac{GMm}{r^{2}}$, the test mass cancels: $g = \dfrac{GM}{r^{2}}$.
Climbing a tall building changes the value of g by a large amount.
Earth's radius is ~6400 km, so a few metres of height barely changes $r$ — $g$ is effectively constant.
Orbital speed
- Gravity = centripetal force: $\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}$.
- The mass $m$ cancels: $v = \sqrt{\dfrac{GM}{r}}$ — independent of the satellite's mass.
A satellite's orbital speed does not depend on its own mass.
The mass $m$ cancels in $\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}$, leaving $v = \sqrt{\dfrac{GM}{r}}$.
Kepler's third law
- From $T = \dfrac{2\pi r}{v}$: $T^{2} = \dfrac{4\pi^{2}}{GM}\,r^{3}$, so $T^{2} \propto r^{3}$.
- A graph of $T^{2}$ against $r^{3}$ is a straight line — its gradient gives the central mass.
Kepler's third law (for circular orbits) states:
$T^{2} = \dfrac{4\pi^{2}}{GM}\,r^{3}$, so the square of the period is proportional to the cube of the radius.
If an orbit radius increases by a factor of 4, the period increases by a factor of:
$T \propto r^{3/2}$, so $T$ scales by $4^{3/2} = 8$.
Geostationary orbit
- Period 24 hours, directly above the equator, orbiting west to east.
- It stays fixed above one point — perfect for a TV dish. Radius $\approx 4.2 \times 10^{7}\ \text{m}$.
Select all the features of a geostationary satellite.
It matches Earth's rotation (24 h, west to east) above the equator, so it stays fixed above one point. A polar orbit does not.
You've got it
- field of a point mass: $g = \dfrac{GM}{r^{2}}$ (so $g$ is nearly constant near the surface)
- orbit: $v = \sqrt{\dfrac{GM}{r}}$ (independent of satellite mass); $T^{2} \propto r^{3}$
- geostationary: 24 h, above the equator, west to east