Gravitational fields

Definition

A gravitational field 重力场 is a region where a mass 质量 feels a force 力 from other masses. The gravitational field strength 重力场强度 $g$ at a point is the gravitational force per unit mass on a small test mass 检验质量 placed there:

Unit: $\text{N kg}^{-1}$ (the same as $\text{m s}^{-2}$ — the acceleration of free fall in the field). $g$ is a vector 矢量, pointing the way the force acts — towards the source mass.

Field lines

A gravitational field is drawn with field lines 场线 that point the way the force acts on a test mass:

• around a point mass 质点 or a uniform sphere (treated as a point mass from outside), the field lines are radial 径向, pointing inwards.
• near the Earth's surface over a small area, the field lines are nearly parallel and equally spaced, pointing straight down — a uniform field 匀强场.

Closer lines mean a stronger field.

Field-line spacing shows the field strength — closer lines mean a stronger field

For two point masses $m_{1}, m_{2}$ a distance $r$ apart, the force on each is

Two masses attract along the line joining them

pulling them together along the line joining them. This is Newton's law of gravitation 万有引力定律. The constant $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}$ is the universal gravitational constant 万有引力常量.

Spheres treated as point masses

For a uniform sphere (such as a planet or star), the field at any point outside is the same as that of a point mass equal to the total mass at the centre. So from above the surface, you can treat the Earth as a point mass at its centre. (Points inside a sphere are different, and are not in the syllabus.)

Outside a uniform sphere the field is radial, exactly like a point mass at the centre

Field strength from a point mass

Put the gravitational force on a test mass $m$ at distance $r$ from a point mass $M$ into $g = F/m$:

So $g$ falls off as $1/r^{2}$ as you move away from the source.

Field strength falls off as $1/r^2$ with distance from a point mass

Worked example. Find the gravitational field strength at the Earth's surface. (Earth's mass $M = 6.0 \times 10^{24}\ \text{kg}$, radius $R = 6.4 \times 10^{6}\ \text{m}$, $G = 6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}$.)

Why $g$ is nearly constant near the Earth's surface

The Earth's radius is $R \approx 6.4 \times 10^{6}\ \text{m}$. Rising to height $h$ changes the distance from the centre from $R$ to $R + h$. For $h \ll R$ (any building or mountain), $(R + h)/R \approx 1$, so $g$ barely changes — going from $5\ \text{m}$ to $10\ \text{m}$ high changes $r$ by about one part in a million. In the laboratory, $g$ is effectively constant.

The International Space Station orbits Earth, held in its path by gravity.

Gravity provides the centripetal force that keeps a planet in a circular orbit

Saturn, its rings (countless small orbiting pieces) and its moons are all held in orbit by gravity

For a satellite 卫星 of mass $m$ in a circular orbit 轨道 of radius $r$ around a body of mass $M$, gravity provides the centripetal force 向心力:

Cancel $m$ (the orbital speed does not depend on the satellite's mass):

Worked example. A satellite orbits the Earth in a circular orbit of radius $r = 7.0 \times 10^{6}\ \text{m}$. Find its orbital speed. (For the Earth, $GM = 4.0 \times 10^{14}\ \text{m}^{3}\ \text{s}^{-2}$.)

The period 周期 follows from $T = 2\pi r / v$:

This is Kepler's third law 开普勒第三定律 for circular orbits: $T^{2} \propto r^{3}$. A plot of $T^{2}$ against $r^{3}$ is a straight line through the origin with gradient $4\pi^{2}/(GM)$, so orbital data gives the central mass.

Kepler's third law: $T^2 \propto r^3$, a straight line through the origin

Geostationary orbit

A geostationary 地球同步 satellite:

• stays directly above the same point on the Earth (so a fixed dish always points at it),
• has a period of 24 hours (the same as the Earth's rotation),
• orbits west to east (the same way the Earth turns),
• must be directly above the equator 赤道.

It must have the same angular speed 角速度 as the Earth, in the same direction, in the equatorial plane (or it would drift north–south during the day). From $T = 24\ \text{h}$ and $T^{2} = 4\pi^{2} r^{3}/(GM)$, the radius is $r \approx 4.2 \times 10^{7}\ \text{m}$ (about $3.6 \times 10^{7}\ \text{m}$ above the surface).

Gravitational potential 引力势 $\phi$ at a point is the work done per unit mass in bringing a small test mass from infinity 无穷远 to that point:

Unit: $\text{J kg}^{-1}$.

The potential is taken as zero at infinity. As the test mass falls in towards the source, gravity does the work for you, so $\phi$ is negative everywhere except at infinity. For a point mass $M$ at distance $r$:

$\phi$ is a scalar 标量. For several masses, add the potentials.

The gravitational potential well: $\phi = -GM/r$ is negative, rising to zero at infinity

Gravitational potential energy of two point masses

If a test mass $m$ sits where the potential is $\phi$, the gravitational potential energy 重力势能 of the pair is

Like the potential, $E_{\text{P}}$ is negative and reaches zero only at infinite separation. Closer masses have more negative potential energy (more tightly bound).

Link with $\Delta E_{\text{P}} = mg\Delta h$

For small height changes near the surface, $r$ barely changes, so $\Delta E_{\text{P}} \approx mg\Delta h$. For large changes (a satellite moving to a higher orbit) use $-GMm/r$ at each radius and take the difference:

which is positive (energy must be supplied to raise the satellite).

Escape velocity (from conservation of energy)

To escape from radius $r$ to infinity, an object's kinetic energy 动能 must equal the size of its gravitational potential energy:

At the Earth's surface, the escape velocity 逃逸速度 is $\approx 11\ \text{km s}^{-1}$. It does not depend on the object's mass.

Worked example. Find the escape velocity from the Earth's surface. (For the Earth, $GM = 4.0 \times 10^{14}\ \text{m}^{3}\ \text{s}^{-2}$, $R = 6.4 \times 10^{6}\ \text{m}$.)

• Newton's law of gravitation $F = GMm/r^2$ (inverse-square); field strength $g = GM/r^2$.
• Distinguish gravitational potential ($\phi = -GM/r$, always negative, zero at infinity) from field strength.
• For an orbit set gravity $=$ centripetal force to get $T^2 \propto r^3$; a geostationary orbit has $T = 24\ \text{h}$.