Gravitational Fields and Orbits

You already know that weight is gravity pulling on a mass, and that a satellite is held in a circular orbit by gravity acting as the centripetal force. But how strong is that pull, exactly? How does it change as you climb away from the Earth? And how fast must a satellite travel to stay up? At A-level we answer all three with a single equation Newton wrote down in 1687 — the law that steers every spacecraft, predicts every eclipse, and let astronomers "weigh" the Sun without ever leaving the ground.

The key idea is that gravity is a field: every mass fills the space around it with a pull that any other mass feels, weakening with distance in a very particular way — the inverse-square law. Get that one shape into your bones and orbits, tides, weightless astronauts and geostationary TV satellites all fall out of it.

Newton's law of gravitation

Newton's great leap was that the pull is universal: it is not special to the Earth. Every chunk of matter attracts every other chunk — your body and this page, two apples on a table, the Earth and the Moon — with a force that grows with each mass and shrinks with the square of their separation.

Every point mass attracts every other point mass with a force

F = \dfrac{G\,m_1 m_2}{r^2},

the force is attractive, acting along the straight line joining the two masses;
m_1, m_2 are the two masses (kg) and r the distance between their centres (m);
it is an inverse-square law: double r and the force drops to a \tfrac14, treble it and the force drops to a \tfrac19;
G is the universal gravitational constant, G = 6.67\times10^{-11}\ \text{N}\,\text{m}^2\,\text{kg}^{-2} — the same everywhere in the Universe.

Because G is so tiny, gravity between everyday objects is far too weak to notice — two people standing a metre apart attract each other with a force smaller than the weight of a speck of dust. Gravity only becomes commanding when one of the masses is astronomical: a planet, a star, a whole galaxy.

Worked example — the pull on a satellite

A 1000\ \text{kg} satellite orbits at r = 7.0\times10^{6}\ \text{m} from the Earth's centre. Earth's mass is M = 5.97\times10^{24}\ \text{kg}. Find the gravitational force.

Step 1 — write the law.

F = \dfrac{G M m}{r^2}.

Step 2 — substitute.

F = \dfrac{(6.67\times10^{-11})(5.97\times10^{24})(1000)}{(7.0\times10^{6})^2}.

Step 3 — evaluate. The top is 3.98\times10^{17} and the bottom is 4.9\times10^{13}, so

F \approx 8.1\times10^{3}\ \text{N} = 8100\ \text{N}.

A hefty pull — and it is exactly this force, aimed at the Earth's centre, that bends the satellite's path into a circle.

Gravitational field strength g

Rather than recompute the force for every different mass, we describe the space once. The gravitational field strength g at a point is the force per kilogram that a mass would feel there — a property of the location, not of the test mass:

The field strength at distance r from a point (or spherical) mass M is the force per unit mass,

g = \dfrac{F}{m} = \dfrac{GM}{r^2},

measured in newtons per kilogram (N/kg) — numerically the same as the free-fall acceleration in \text{m}\,\text{s}^{-2};
it is a vector, pointing radially inward, straight towards the centre of M;
it too obeys the inverse-square law, and depends on the central mass M — not on the mass placed in the field;
the weight of any mass m is then just W = mg.

This is why g \approx 9.81\ \text{N/kg} at the Earth's surface. Put the Earth's mass and its radius R = 6.37\times10^{6}\ \text{m} into g = GM/r^2:

g_{\text{surface}} = \dfrac{(6.67\times10^{-11})(5.97\times10^{24})}{(6.37\times10^{6})^2} \approx 9.81\ \text{N/kg}.

The famous 9.81 isn't a magic constant — it is GM/R^2 for this planet, and it falls off with altitude. Because the distance is measured from the centre, climbing a mountain or reaching orbit barely dents it: at the International Space Station's 400\ \text{km} the radius grows only from 6370 to 6770\ \text{km}, so g is still about 8.7\ \text{N/kg} — roughly 90\% of its surface value. Astronauts are not beyond gravity.

See the inverse-square fall

The curve below is g = GM/r^2 for the Earth, with distance measured in Earth-radii from the centre (so r = 1 is the surface, where g \approx 9.81\ \text{N/kg}). Drag the slider to move out from the planet and watch how fast the field collapses: it is already quartered by r = 2 and down to a ninth by r = 3. That steep, never-quite-reaching-zero shape is the signature of every inverse-square law.

Worked example — comparing two worlds

Field strength depends only on M and r, so we can find g anywhere. The Moon has M = 7.35\times10^{22}\ \text{kg} and radius R = 1.74\times10^{6}\ \text{m}. What is g at its surface?

g_{\text{Moon}} = \dfrac{GM}{R^2} = \dfrac{(6.67\times10^{-11})(7.35\times10^{22})}{(1.74\times10^{6})^2} \approx 1.62\ \text{N/kg}.

About one-sixth of Earth's 9.81 — the reason the Apollo astronauts bounded across the surface in slow, floating leaps. Notice we never needed the astronaut's mass: the field is set by the Moon, and each visitor's weight then follows from W = mg.

Orbits: gravity as the centripetal force

For a satellite of mass m in a circular orbit of radius r, gravity supplies exactly the centripetal force needed to keep it turning. Set the gravitational pull equal to the required centripetal force (mv^2/r):

\underbrace{\dfrac{GMm}{r^2}}_{\text{gravity}} = \underbrace{\dfrac{mv^2}{r}}_{\text{centripetal}}.

The satellite's mass m cancels from both sides — a beautiful result: the speed needed to hold an orbit does not depend on how heavy the satellite is, only on the central mass and the radius. Multiply through by r/m:

v^2 = \dfrac{GM}{r} \quad\Longrightarrow\quad v = \sqrt{\dfrac{GM}{r}}.

For a circular orbit of radius r about a central mass M:

the orbital speed is v = \sqrt{\dfrac{GM}{r}} — smaller r means faster;
the period (time for one lap) is T = \dfrac{2\pi r}{v} = 2\pi\sqrt{\dfrac{r^3}{GM}};
squaring gives Kepler's third law, T^2 = \dfrac{4\pi^2}{GM}\,r^3 \quad\Longrightarrow\quad T^2 \propto r^3, so the period grows steeply as the orbit widens.

Worked example — speed and period of a low orbit

Take the same satellite at r = 7.0\times10^{6}\ \text{m} (GM = 3.98\times10^{14} for the Earth).

Speed:

v = \sqrt{\dfrac{3.98\times10^{14}}{7.0\times10^{6}}} = \sqrt{5.69\times10^{7}} \approx 7.5\times10^{3}\ \text{m/s}.

About 7.5\ \text{km/s} — roughly 27 000 km/h. Period:

T = \dfrac{2\pi r}{v} = \dfrac{2\pi (7.0\times10^{6})}{7.5\times10^{3}} \approx 5.8\times10^{3}\ \text{s} \approx 97\ \text{minutes}.

That is why astronauts on the Space Station see a sunrise every hour and a half: they really do lap the whole planet in about an hour and a half.

The geostationary orbit

A geostationary satellite hangs over one fixed point on the equator, so a TV dish can be aimed once and bolted down. For that, its period must match one Earth rotation: T = 24\ \text{hours} = 86\,400\ \text{s}. Rearranging Kepler's third law for the radius:

r = \left(\dfrac{GM\,T^2}{4\pi^2}\right)^{1/3} = \left(\dfrac{(3.98\times10^{14})(86\,400)^2}{4\pi^2}\right)^{1/3} \approx 4.22\times10^{7}\ \text{m}.

That is about 42\,200\ \text{km} from the Earth's centre — some 36\,000\ \text{km} above the surface. There is only one such radius: the geostationary condition fixes the height completely, which is why that ring of space is such prized, crowded real estate.

These are the classic A-level traps — check yourself against all four:

It is inverse-square, not inverse. Doubling the distance quarters the force (1/2^2), it does not halve it. Trebling it leaves a ninth. Always square the distance ratio.
r is measured from the CENTRE. The distance in g = GM/r^2 runs to the centre of the planet, not the surface. That is why g barely changes over a few kilometres of altitude but the equation still uses r \approx 6.4\times10^{6}\ \text{m}.
Field strength depends on M, not on the mass in the field. A feather and a boulder sit in the same field g; they feel different forces (W = mg) but the field itself is set by the central mass alone.
Orbital speed doesn't depend on the satellite's mass. The m cancelled: v = \sqrt{GM/r} uses only the central mass and the radius. A tiny cubesat and a huge space station at the same height orbit at the same speed.

You cannot put a planet on a scale, yet we know the Earth's mass to four figures. The trick is to watch anything orbiting it. Rearranging v = \sqrt{GM/r} gives M = \dfrac{v^2 r}{G} (or, from a measured period, M = \dfrac{4\pi^2 r^3}{G T^2}). Measure the Moon's distance and 27-day period, plug them in, and out drops the Earth's mass — no rocket required.

The same move on the Earth's own orbit — radius 1.5\times10^{11}\ \text{m}, period one year — weighs the Sun at about 2\times10^{30}\ \text{kg}. Astronomers weigh distant stars, black holes and whole galaxies exactly this way: everything that orbits is a set of scales. It even matters for your phone — GPS satellites at 20\,000\ \text{km} feel a weaker field, and both their speed and that altitude shift their onboard clocks by tens of microseconds a day; uncorrected, the position error would grow by kilometres daily.