Gravitational field of a point mass
| English | Chinese | Pinyin |
|---|---|---|
| orbit | 轨道 | guǐ dào |
| orbital speed | 轨道速度 | guǐ dào sù dù |
| satellite | 卫星 | wèi xīng |
| Kepler's third law | 开普勒第三定律 | kāi pǔ lēi dì sān dìng lǜ |
| geostationary | 地球同步 | dì qiú tóng bù |
| equator | 赤道 | chì dào |
| free fall | 自由落体 | zì yóu luò tǐ |
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 轨道.
- 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 $g$.
Field strength drops off as 1/r²: doubling the distance quarters the pull
Field-line spacing shows the field strength — closer lines mean a stronger field
Field of a point mass
g ∝ M/r²
Gravitational field strength obeys the inverse-square law — halve the distance and it quadruples.
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 轨道速度
- Gravity = centripetal force: $\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}$.
- The mass $m$ cancels: $v = \sqrt{\dfrac{GM}{r}}$ — independent of the satellite 卫星'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 开普勒第三定律
- 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.
Plotting T² against r³ gives a straight line through the origin — its gradient fixes the central mass M
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 $4^{3/2} = 8$.
Worked example: a satellite above Mars
A satellite orbits Mars (mass $6.4 \times 10^{23}\ \text{kg}$, radius $3.4 \times 10^{6}\ \text{m}$) at a height of $1.7 \times 10^{6}\ \text{m}$ above the surface. Find its speed and period.
- Orbit radius, from the centre: $r = 3.4 \times 10^{6} + 1.7 \times 10^{6} = 5.1 \times 10^{6}\ \text{m}$.
- Speed: $v = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{6.67 \times 10^{-11} \times 6.4 \times 10^{23}}{5.1 \times 10^{6}}} = 2.9 \times 10^{3}\ \dfrac{\text{m}}{\text{s}}$.
- Period: $T = \dfrac{2\pi r}{v} = \dfrac{2\pi \times 5.1 \times 10^{6}}{2.9 \times 10^{3}} = 1.1 \times 10^{4}\ \text{s}$, about three hours.
- Check: the satellite's own mass never appeared — every satellite at this height moves at this speed. Using the height alone for $r$ gives a speed that is far too high.
A satellite orbits the Earth (mass $6.0 \times 10^{24}\ \text{kg}$) at a radius of $8.0 \times 10^{6}\ \text{m}$ from the centre. What is its orbital speed, in km/s?
$v = \sqrt{\dfrac{GM}{r}} = \sqrt{\dfrac{6.67 \times 10^{-11} \times 6.0 \times 10^{24}}{8.0 \times 10^{6}}} = 7.1 \times 10^{3}\ \dfrac{\text{m}}{\text{s}} = 7.1\ \dfrac{\text{km}}{\text{s}}$.
Geostationary 地球同步 orbit
- Period 24 hours, directly above the equator 赤道, orbiting west to east — the same direction as the Earth turns.
- It stays fixed above one point — perfect for a TV dish. From $r^{3} = \dfrac{GMT^{2}}{4\pi^{2}}$ with $T = 86\,400\ \text{s}$, the radius is $\approx 4.2 \times 10^{7}\ \text{m}$, about $3.6 \times 10^{7}\ \text{m}$ above the surface.
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.
Synchronous orbits elsewhere
- Any planet has its own synchronous orbit: the period equals the planet's rotation period, and the orbit lies above the equator, moving with the rotation.
- Mars turns once in about $25$ hours, so a satellite with a $25$-hour period above its equator stays over one Martian spot.
- Change the period and the radius follows from Kepler's law: a longer period means a larger orbit.
In every orbit formula $r$ is measured from the centre of the planet — add the planet's radius to a height. The satellite's mass cancels, so "a heavier satellite must go faster" is wrong. A geostationary orbit must be over the equator: any other orbit crosses the equator twice a day and drifts north and south as seen from the ground.
A satellite is $600\ \text{km}$ above the surface of a planet of radius $6400\ \text{km}$. In the orbit equations, $r$ = ____ km.
$r$ is measured from the centre: $6400 + 600 = 7000\ \text{km}$.
Not weightless — falling
- At the Space Station's height of $400\ \text{km}$, $g = 9.81 \times \left(\dfrac{6370}{6770}\right)^{2} = 8.7\ \dfrac{\text{N}}{\text{kg}}$ — nearly the surface value.
- Astronauts float because they and the station are in free fall 自由落体 together, not because gravity has vanished.
Earth's radius is $6370\ \text{km}$ and its surface field strength is $9.81\ \dfrac{\text{N}}{\text{kg}}$. What is the field strength at a height of $400\ \text{km}$, in N/kg?
$g \propto \dfrac{1}{r^{2}}$, so $g = 9.81 \times \left(\dfrac{6370}{6770}\right)^{2} = 8.7\ \dfrac{\text{N}}{\text{kg}}$ — the astronauts are not weightless, they are falling.
You've got it
- field of a point mass: $g = \dfrac{GM}{r^{2}}$ (so $g$ is nearly constant near the surface)
- orbit: $v = \sqrt{\dfrac{GM}{r}}$ with $r$ from the centre (independent of satellite mass); $T^{2} \propto r^{3}$
- geostationary: 24 h, above the equator, west to east; a synchronous orbit matches any planet's rotation