Why doesn’t an orbiting satellite fall to Earth?

It is falling continuously, but its sideways speed means the surface curves away as fast as it falls.

Why do higher satellites move more slowly?

Gravity is weaker further out, so less centripetal force is available, and v = GM/r decreases.

Why must a geostationary orbit be above the equator?

The orbit must be centred on Earth’s centre. Only an equatorial orbit keeps the satellite above a fixed point as Earth rotates.

Why is total orbital energy negative?

The satellite is bound: it would need extra energy to escape to infinity, where total energy would be zero.

Back to course

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Orbits and Satellites

Newton imagined a cannon on a mountain firing ever faster. Eventually the ball falls towards Earth but keeps missing it, because the surface curves away just as fast as it drops. That is an $\textbf{orbit}$ — continuous free fall — and one equation from circular motion describes it completely.

40 min Topic 12 · Gravitational Fields

What you'll be able to do

Equate gravitational and centripetal force to derive orbital speed
Derive and use Kepler’s third law, $T^{2} = \dfrac{4\pi^{2}}{GM}r^{3}$
Describe the properties and uses of geostationary and low Earth orbits
Analyse the kinetic, potential and total energy of an orbiting satellite

Orbital speed

For a satellite of mass $m$ in a circular orbit of radius $r$ around a mass $M$ , the gravitational force provides the centripetal force.

The satellite’s mass cancels, so all satellites at the same radius move at the same speed. Larger orbits are slower.

\dfrac{GMm}{r^{2}} = \dfrac{mv^{2}}{r}

Gravity provides centripetal force.

v = \sqrt{\dfrac{GM}{r}}

Orbital speed.

Example 1Find the orbital speed of the ISS at radius

6.78\times10^{6}

m. Take

GM = 3.98\times10^{14}\ \mathrm{m^{3}\,s^{-2}}

for Earth.

v = \sqrt{\dfrac{3.98\times10^{14}}{6.78\times10^{6}}} = \sqrt{5.87\times10^{7}}

Answer

7.66\times10^{3}\ \mathrm{m\,s^{-1}}

Tip — Compare with escape velocity: $v_{\text{esc}} = \sqrt{2GM/r}$ is always $\sqrt{2}$ times the orbital speed at the same radius.

Kepler’s third law

Substituting $v = 2\pi r/T$ into the orbital speed equation gives a relation between period and radius.

$\textbf{Kepler’s third law}$ : the square of the orbital period is proportional to the cube of the orbit radius. The constant $4\pi^{2}/GM$ depends only on the central mass.

That makes orbits a way to $\textbf{weigh}$ astronomical bodies: measure the period and radius of any moon or planet, and you have the mass of what it orbits.

T^{2} = \dfrac{4\pi^{2}}{GM}r^{3}

Kepler’s third law.

Example 1The Moon orbits Earth with period 27.3 days at radius

3.84\times10^{8}

m. Use this to find Earth’s mass.

T = 27.3 \times 86\,400 = 2.359\times10^{6}

M = \dfrac{4\pi^{2}r^{3}}{GT^{2}} = \dfrac{4\pi^{2} \times (3.84\times10^{8})^{3}}{6.67\times10^{-11} \times (2.359\times10^{6})^{2}}

Answer

6.02\times10^{24}\ \mathrm{kg}

Geostationary and low orbits

A $\textbf{geostationary}$ satellite stays above the same point on Earth’s surface. It must orbit above the equator, in the same direction as Earth rotates, with a period of 24 hours (strictly, one sidereal day of about 23 h 56 min).

Kepler’s law fixes its radius at about $4.2\times10^{7}$ m, some 36 000 km above the surface. Ground dishes can point at it permanently, making it ideal for communications and weather monitoring.

$\textbf{Low Earth orbits}$ , a few hundred kilometres up, have periods of about 90 minutes. Satellites there can pass over the whole planet as Earth turns beneath them, and they are close enough for high-resolution imaging.

Example 1Find the radius of a geostationary orbit, using

T = 8.62\times10^{4}

r^{3} = \dfrac{GMT^{2}}{4\pi^{2}} = \dfrac{3.98\times10^{14} \times (8.62\times10^{4})^{2}}{4\pi^{2}}

r^{3} = 7.49\times10^{22}

Answer

r = 4.22\times10^{7}\ \mathrm{m}

Tip — The three conditions for geostationary orbit — equatorial, same direction as rotation, period one day — are all needed for full marks.

Energy of an orbiting satellite

Using $v^{2} = GM/r$ , a satellite’s kinetic energy is $E_{k} = \tfrac12 mv^{2} = \dfrac{GMm}{2r}$ .

Its potential energy is $E_{p} = -\dfrac{GMm}{r}$ . So the total energy is $E = -\dfrac{GMm}{2r}$ — negative, because the satellite is bound to the planet.

Moving to a higher orbit makes the total energy less negative, so energy must be supplied. But kinetic energy falls, because the satellite moves more slowly in a larger orbit; potential energy rises by twice as much as kinetic energy falls.

E_{\text{total}} = -\dfrac{GMm}{2r}

Total energy in a circular orbit.

Equation recap

v = \sqrt{GM/r}

Orbital speed.

T^{2} = 4\pi^{2}r^{3}/GM

Kepler’s third law.

E = -GMm/2r

Total orbital energy.

Common mistakes to avoid

Using altitude as the orbit radius.

Add the planet’s radius.

Saying heavier satellites must orbit faster.

Mass cancels; speed depends only on

M

and

r

Leaving

T

in hours or days.

Convert to seconds for SI calculations.

Saying any 24-hour orbit is geostationary.

It must also be equatorial and in the direction of Earth’s rotation.

Key takeaways

Gravity provides the centripetal force: $GMm/r^{2} = mv^{2}/r$ .
Orbital speed $v = \sqrt{GM/r}$ , independent of satellite mass.
Kepler’s third law: $T^{2} = 4\pi^{2}r^{3}/GM$ .
Geostationary: equatorial, rotating with Earth, 24-hour period, $r \approx 4.2\times10^{7}$ m.
Total orbital energy $-GMm/2r$ ; higher orbits need energy but are slower.

Test yourself

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