Motion in Gravitational Fields

In short: Every mass attracts every other mass with a force $F = \frac{GM m}{r ^{2}}$ , so doubling the distance quarters the force. For a satellite in a circular orbit, gravity is the centripetal force, which sets its speed: $v = \frac{GM}{r}$ , so higher orbits are slower.

Newton's law of universal gravitation

Every mass attracts every other mass. The force between two masses $M$ and $m$ , with centres a distance $r$ apart, is:

$F = \frac{GM m}{r ^{2}}$

$G = 6.67 \times 1 0^{- 11}$ N m² kg⁻² (the gravitational constant)
$r$ is measured centre to centre, not from the surface.

The two masses pull on each other with equal and opposite forces (Newton's third law). Earth pulls on you exactly as hard as you pull on Earth.

This is an inverse square law:

Distance	Force
$r$	$F$
$2 r$	$F /4$
$3 r$	$F /9$

Gravitational field strength

The gravitational field strength $g$ is the force on each kilogram:

$g = \frac{GM}{r ^{2}}$

At Earth's surface, $g = 9.8$ N/kg. At twice Earth's radius from the centre, it is a quarter of that: 2.45 N/kg.

Orbits: gravity is the centripetal force

For a satellite in a circular orbit, gravity provides the centripetal force:

$\frac{GM m}{r ^{2}} = \frac{m v ^{2}}{r} \Rightarrow v = \frac{GM}{r}$

The satellite's own mass cancels out. A tiny satellite and a huge one at the same height move at the same speed.
Higher orbits are slower and take longer.

Using $v = \frac{2 π r}{T}$ , the period is $T = 2 π \frac{r ^{3}}{GM}$ .

Kepler's three laws

Johannes Kepler found three rules by studying careful records of the planets.

Orbits are ellipses, with the Sun at one focus.
A line from the Sun to a planet sweeps out equal areas in equal times. So a planet moves fastest when it is closest to the Sun.
For every planet around the same star, $\frac{r ^{3}}{T ^{2}}$ is the same:

$\frac{r ^{3}}{T ^{2}} = \frac{GM}{4 π ^{2}}$

This third law lets you find the mass of a star or planet from the orbits around it.

Types of satellite orbits

	Low Earth orbit (LEO)	Geostationary orbit
Height	about 200 to 2000 km	about 36 000 km
Period	about 90 minutes	24 hours (one Earth day)
Speed	faster (about 7.8 km/s)	slower (about 3.1 km/s)
Uses	photos, weather, ISS	TV, communications
Notes	close to the ground, so detailed images	stays above one spot on the equator

A geostationary satellite turns with Earth, so a dish on the ground can point at it all the time.

Energy in a gravitational field

Far from a planet, we take gravitational potential energy as zero. Closer in, it is negative:

$U = - \frac{GM m}{r}$

The closer you are, the more negative $U$ is.
Moving away means $U$ increases (gets closer to zero). This takes work.
Near the surface, $Δ U = m g Δ h$ is a good shortcut, because $g$ hardly changes over small heights.

Escape velocity

The escape velocity is the launch speed that lets an object get infinitely far away and just stop. Its total energy is then zero:

$\frac{1}{2} m v_{esc}^{2} - \frac{GM m}{r} = 0 \Rightarrow v_{esc} = \frac{2 GM}{r}$

For Earth's surface, this is about 11.2 km/s. It does not depend on the mass of the object being launched.

Gravitational Fields

Newton's law

F = GMm / r²
Inverse square
Equal and opposite forces

Field strength

g = GM / r²
9.8 N/kg at Earth's surface

Orbits

Gravity = centripetal force
v = √(GM / r)
Higher is slower

Kepler

Ellipses
Equal areas, equal times
r³ / T² constant

Satellites

Low Earth orbit
Geostationary, 24 h

Energy

U = −GMm / r
E = −GMm / 2r
Escape v = √(2GM / r)

Year 12Total energy of an orbit

For a circular orbit, the kinetic energy is $K = \frac{1}{2} m v^{2} = \frac{GM m}{2 r}$ . The total energy is:

$E = K + U = \frac{GM m}{2 r} - \frac{GM m}{r} = - \frac{GM m}{2 r}$

So $K = - \frac{1}{2} U$ and $E = \frac{1}{2} U$ . The total energy is negative: the satellite is bound to the planet.

Changing orbit. Moving to a higher orbit makes $E$ less negative, so it needs energy from rockets. In the new orbit, $U$ is higher but $K$ is lower: the satellite moves more slowly.

Orbital decay. In low orbit, thin air drags on a satellite. It loses total energy and drops lower. Strangely, it speeds up, because $v = GM / r$ grows as $r$ shrinks. Its potential energy falls by twice as much as its kinetic energy rises.

Area under a graph. On a graph of gravitational force against distance, the area between two distances is the work done, which equals the change in potential energy.

Only gravity acts. A satellite in orbit has just one force on it, gravity, which acts as the centripetal force. There is no forward "thrust" force needed to keep it moving.

Escape from an orbit. Since $v_{orbit} = \frac{GM}{r}$ and $v_{esc} = \frac{2 GM}{r}$ , the escape velocity from any orbit radius is $2$ times the orbital speed there. A satellite orbiting at 2100 m/s needs about 2970 m/s to escape.

Work between two distances. $W = Δ U = GM m (\frac{1}{r _{1}} - \frac{1}{r _{2}})$ . Moving from $r$ to $3 r$ takes $\frac{2}{3}$ of the energy needed to escape from $r$ ; moving on from $3 r$ to $9 r$ takes a third as much again as the first step, because the bracket $\frac{1}{3 r} - \frac{1}{9 r} = \frac{2}{9 r}$ is a third of $\frac{1}{r} - \frac{1}{3 r} = \frac{2}{3 r}$ .

Graphs. A graph of $g$ against $\frac{1}{r ^{2}}$ is a straight line through the origin with gradient $GM$ , so $M = \frac{gradient}{G}$ .

Where the pulls cancel. Between Earth (mass $81 M$ ) and the Moon (mass $M$ ), the pulls are equal where $\frac{81 M}{x ^{2}} = \frac{M}{( d - x ) ^{2}}$ . Taking square roots, $9 (d - x) = x$ , so $x = \frac{9}{10} d$ from Earth's centre.

Quick summary

$F = \frac{GM m}{r ^{2}}$ , with $r$ centre to centre. Doubling $r$ quarters the force.
$g = \frac{GM}{r ^{2}}$ is the force per kilogram.
In a circular orbit, gravity is the centripetal force, so $v = \frac{GM}{r}$ . Higher orbits are slower.
Kepler: ellipses, equal areas in equal times, and $\frac{r ^{3}}{T ^{2}}$ constant for one central body.
Geostationary satellites orbit above the equator once every 24 hours.
$U = - \frac{GM m}{r}$ , and escape velocity is $\frac{2 GM}{r}$ .

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Which law says the gravitational force between two masses falls as 1/r²?
1. A Kepler's first law
2. B Newton's law of universal gravitation
3. C Newton's first law
4. D Kepler's second law
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B: Newton's law of universal gravitation
F = GMm/r² is Newton's law of universal gravitation.
The distance between two moons doubles. The gravitational force between them becomes:
1. A Twice as big
2. B Half as big
3. C One quarter as big
4. D Four times as big
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C: One quarter as big
F is proportional to 1/r². (1/2)² = 1/4.
Earth pulls on the Moon with a force F. The Moon pulls on Earth with a force of:
1. A Much less than F
2. B Exactly F
3. C Much more than F
4. D Zero
Show answer
B: Exactly F
Newton's third law: the forces are equal in size and opposite in direction.