Circular Orbits and Geostationary Satellites

A2 · 10 min

A satellite in a circular orbit is an object in permanent free fall: gravity pulls it towards the planet, but it moves sideways fast enough that it keeps missing. Linking Newton's law of gravitation to centripetal force gives the speed and period of any circular orbit, lets us "weigh" planets and stars from the motion of their moons, and explains why geostationary satellites must sit at one particular height above the Equator. Orbit calculations and the derivation of $T^2 \propto r^3$ appear in almost every Paper 4.

Gravity provides the centripetal force

For a satellite of mass $m$ in a circular orbit of radius $r$ around a planet of mass $M$ , the only force acting is the planet's gravitational pull. It points towards the centre of the planet, which is the centre of the orbit, so it is exactly the centripetal force:

Key result

\frac{GMm}{r^{2}} = \frac{mv^{2}}{r} = mr\omega^{2}

Gravitational force $=$ mass $\times$ centripetal acceleration. The satellite's mass $m$ cancels.

Because $m$ cancels, every object at the same orbital radius around the same planet has the same speed and period, whether it is a space station or a spanner dropped by an astronaut.

Orbital speed

Cancelling $m$ and one factor of $r$ :

\frac{GM}{r} = v^{2} \quad\Rightarrow\quad v = \sqrt{\frac{GM}{r}}

Satellites in lower orbits move faster. The International Space Station, a few hundred kilometres up, moves at nearly $8\ \text{km s}^{-1}$ ; the Moon, far away, moves at about $1\ \text{km s}^{-1}$ .

Orbital period: deriving $T^2 \propto r^3$

This derivation is a frequent "show that" question. Write it out in full.

Period of a circular orbit

The gravitational force provides the centripetal force:

\frac{GMm}{r^{2}} = mr\omega^{2}

Substitute $\omega = \dfrac{2\pi}{T}$ :

\frac{GMm}{r^{2}} = mr\left(\frac{2\pi}{T}\right)^{2} = \frac{4\pi^{2}mr}{T^{2}}

Cancel $m$ and rearrange:

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

Since $4\pi^2/(GM)$ is a constant for a given central mass, $T^2 \propto r^3$ . This is Kepler's third law.

Key result

T^{2} = \left(\frac{4\pi^{2}}{GM}\right) r^{3}

A graph of $T^2$ against $r^3$ for all the satellites of one planet is a straight line through the origin with gradient $4\pi^2/(GM)$ . A graph of $\lg T$ against $\lg r$ is a straight line of gradient $1.5$ .

The same equation is the most common way to find the mass of a central body: measure the radius and period of anything orbiting it, then

M = \frac{4\pi^{2}r^{3}}{GT^{2}}

Notice what you can and cannot find this way: the central mass $M$ , never the mass of the orbiting body.

Geostationary orbits

A geostationary satellite stays above the same point on the Earth's surface at all times. That makes it ideal for communications and broadcasting: a ground dish can point at it permanently without tracking.

Definition

A satellite in a geostationary orbit:

remains at the same point above the Earth's surface;
has an orbital period of $24$ hours, the same as the Earth's rotation;
orbits from west to east, in the same direction as the Earth rotates;
is directly above the Equator.

Why each condition is needed:

Period 24 hours: the satellite must turn through the same angle as the ground beneath it in the same time.
West to east: it must rotate in the same sense as the Earth, otherwise it would move across the sky.
Above the Equator: the centre of every orbit must be the centre of the Earth (gravity points there). An orbit inclined to the Equator would carry the satellite north and south of the Equator each day, so only an equatorial orbit can stay above a fixed point.

Because the period is fixed, $T^2 \propto r^3$ means the radius is fixed too: there is only one geostationary orbit radius, about $4.2 \times 10^{7}\ \text{m}$ from the centre of the Earth (about $36\,000\ \text{km}$ above the surface).

Tip

Strictly, the Earth rotates once relative to the stars every $23$ hours $56$ minutes (a sidereal day), and a geostationary satellite has that period. The syllabus and mark schemes use $24$ hours; use $24$ hours unless a question gives you something else. The difference in radius is less than $0.2\%$ .

Geostationary satellites are not the only useful kind. Satellites in low orbits (a few hundred kilometres up) take about $90$ minutes per orbit. They pass over different parts of the Earth on each orbit, giving detailed, close-up images and short signal delays, but a ground station sees each one for only a few minutes. Geostationary satellites are far away, so their images are less detailed and signals take about a quarter of a second for the round trip, but they provide continuous coverage of almost a third of the Earth's surface.

Weightlessness in orbit

Astronauts in an orbiting station are not beyond the reach of gravity: at $400\ \text{km}$ up, the gravitational field strength is still about $90\%$ of its surface value. They feel weightless because they and their spacecraft are in free fall together, both accelerating towards the Earth at the same rate $g$ . There is no contact force between astronaut and floor, and it is the contact force that we feel as "weight".

Orbit calculations

Write "gravitational force provides the centripetal force".
Write $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ or $\dfrac{GMm}{r^2} = mr\omega^2$ , depending on whether speed or period is involved.
Make sure $r$ is the distance from the centre of the planet: $r = R + h$ .
Convert the period to seconds.
Rearrange symbolically first, then substitute.
For ratio problems between two satellites of the same planet, use $T^2 \propto r^3$ or $v \propto 1/\sqrt{r}$ directly.

Worked examples

The International Space Station

The International Space Station orbits $400\ \text{km}$ above the Earth's surface. The Earth has mass $5.97 \times 10^{24}\ \text{kg}$ and radius $6.37 \times 10^{6}\ \text{m}$ . Calculate the orbital speed and the period of the station.

Solution

$r = 6.37 \times 10^{6} + 0.40 \times 10^{6} = 6.77 \times 10^{6}\ \text{m}$ .

Gravitational force provides the centripetal force: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ , so

v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.77 \times 10^{6}}} = 7.67 \times 10^{3}\ \text{m s}^{-1}

T = \frac{2\pi r}{v} = \frac{2\pi \times 6.77 \times 10^{6}}{7.67 \times 10^{3}} = 5.55 \times 10^{3}\ \text{s} \ (92 \text{ minutes})

Radius of a geostationary orbit

Using the data above, show that the radius of a geostationary orbit is about $4.2 \times 10^{7}\ \text{m}$ , and calculate the height above the Earth's surface and the orbital speed.

Solution

$T = 24 \times 3600 = 86\,400\ \text{s}$ .

r^{3} = \frac{GMT^{2}}{4\pi^{2}} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times (86\,400)^{2}}{4\pi^{2}} = 7.53 \times 10^{22}\ \text{m}^3

r = 4.22 \times 10^{7}\ \text{m}

Height above surface: $4.22 \times 10^{7} - 6.37 \times 10^{6} = 3.59 \times 10^{7}\ \text{m}$ .

Speed: $v = 2\pi r/T = 2\pi \times 4.22 \times 10^{7}/86\,400 = 3.07 \times 10^{3}\ \text{m s}^{-1}$ .

Mass of Mars from its moon

Phobos orbits Mars in a circle of radius $9.38 \times 10^{6}\ \text{m}$ with a period of $7.65$ hours. Calculate the mass of Mars.

Solution

$T = 7.65 \times 3600 = 2.754 \times 10^{4}\ \text{s}$ .

M = \frac{4\pi^{2}r^{3}}{GT^{2}} = \frac{4\pi^{2} \times (9.38 \times 10^{6})^{3}}{6.67 \times 10^{-11} \times (2.754 \times 10^{4})^{2}} = 6.44 \times 10^{23}\ \text{kg}

The mass of Phobos is not needed and cannot be found from this data.

Ratio of periods

Two moons orbit the same planet. Moon A has orbital radius $2.0 \times 10^{8}\ \text{m}$ and period $3.0$ days. Moon B has orbital radius $5.0 \times 10^{8}\ \text{m}$ . Calculate the period of moon B and the ratio of their orbital speeds $v_A / v_B$ .

Solution

$T^2 \propto r^3$ for the same central mass:

T_B = T_A\left(\frac{r_B}{r_A}\right)^{3/2} = 3.0 \times (2.5)^{1.5} = 11.9\ \text{days}

$v \propto 1/\sqrt{r}$ :

\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A}} = \sqrt{2.5} = 1.58

The inner moon moves faster and has the shorter period.

Using surface gravity instead of G and M

A question gives only $g = 9.81\ \text{N kg}^{-1}$ at the Earth's surface and $R = 6.37 \times 10^{6}\ \text{m}$ . Find the period of a satellite orbiting at a radius of $2R$ .

Solution

At the surface $g = GM/R^2$ , so $GM = gR^2 = 9.81 \times (6.37 \times 10^{6})^2 = 3.98 \times 10^{14}\ \text{N m}^2\ \text{kg}^{-1}$ .

T = 2\pi\sqrt{\frac{r^{3}}{GM}} = 2\pi\sqrt{\frac{(1.274 \times 10^{7})^{3}}{3.98 \times 10^{14}}} = 1.43 \times 10^{4}\ \text{s} \ (\text{about } 4.0 \text{ hours})

The substitution $GM = gR^2$ is a standard trick worth remembering.

Why a geostationary satellite must be above the Equator

A student suggests placing a geostationary satellite permanently above a city at latitude $40^\circ\text{N}$ . Explain why this is impossible.

Solution

The only force on the satellite is gravity, which acts towards the centre of the Earth. For a circular orbit, the centripetal force must point to the centre of the orbit, so the plane of every orbit must pass through the Earth's centre. A satellite that stayed above latitude $40^\circ\text{N}$ would have to move in a circle centred on a point on the Earth's axis north of the Earth's centre, which gravity cannot provide. An orbit through the Earth's centre that passes over $40^\circ\text{N}$ must also pass equally far south of the Equator during each orbit, so the satellite would not stay above the city.

Watch out

Using height instead of radius. In every orbit equation $r$ is measured from the centre of the planet. Height above surface must have the planet's radius added.

Watch out

Writing "the satellite is weightless because there is no gravity". There is gravity; it provides the centripetal force. The satellite and everything in it are in free fall together.

Watch out

Including the satellite's mass in the answer. The satellite's mass cancels from $GMm/r^2 = mv^2/r$ . If a question gives you the satellite's mass, it is for a later part (such as kinetic energy), not for the speed or period.

Exam tip

"Show that $T^2 = (4\pi^2/GM)r^3$ ": start from "gravitational force provides centripetal force", write both expressions, substitute $\omega = 2\pi/T$ (or $v = 2\pi r/T$ ), and rearrange. Each step is a mark; skipping straight to the result scores nothing.
"State three features of a geostationary orbit": period 24 hours; west to east (same direction as the Earth's rotation); above the Equator. "Stays above the same point" is the definition, so it does not count as an extra feature.
Orbit answers are often required to 2 or 3 significant figures; carry at least 4 through the working, especially when cubing or taking square roots.

Summary

In a circular orbit, gravity provides the centripetal force: $GMm/r^2 = mv^2/r = mr\omega^2$ .
Orbital speed $v = \sqrt{GM/r}$ : lower orbits are faster.
$T^2 = (4\pi^2/GM)r^3$ (Kepler's third law): derive it from the force equation.
The mass of the central body is $M = 4\pi^2r^3/(GT^2)$ ; the orbiting mass cancels.
Geostationary: period 24 hours, west to east, above the Equator, at $r \approx 4.2 \times 10^{7}\ \text{m}$ ; stays above one point on the surface.
$GM = gR^2$ lets you work without $G$ and $M$ .
Astronauts in orbit are weightless because they are in free fall, not because gravity is absent.

Practice questions

Question

State what provides the centripetal force on a satellite and write the equation linking orbital speed to orbital radius.
Calculate the orbital speed and period of a satellite at a height of $2.0 \times 10^{6}\ \text{m}$ above the Earth's surface. ( $M_E = 5.97 \times 10^{24}\ \text{kg}$ , $R_E = 6.37 \times 10^{6}\ \text{m}$ )
Calculate the radius of orbit of a satellite with a period of $90$ minutes, and hence its height above the surface.
Two satellites orbit the Earth, one at four times the orbital radius of the other. Find the ratio of their periods.
The Earth orbits the Sun at $1.50 \times 10^{11}\ \text{m}$ once a year ( $365.25$ days). Calculate the mass of the Sun.
Explain why astronauts on the International Space Station appear weightless even though the gravitational field strength there is about $8.7\ \text{N kg}^{-1}$ .
A satellite of mass $500\ \text{kg}$ is in geostationary orbit at $4.22 \times 10^{7}\ \text{m}$ from the Earth's centre. Calculate its kinetic energy.
(a) Show that, for circular orbits around a planet of mass $M$ , $\lg T = \tfrac{3}{2}\lg r + \tfrac{1}{2}\lg\left(\dfrac{4\pi^2}{GM}\right)$ . (b) For the moons of a planet, a graph of $\lg(T/\text{s})$ against $\lg(r/\text{m})$ has gradient $1.50$ and passes through the point $(8.00,\ 5.20)$ . Determine the mass of the planet.

Answers

The gravitational force of the planet on the satellite. $GMm/r^2 = mv^2/r$ , so $v = \sqrt{GM/r}$ .
$r = 8.37 \times 10^{6}\ \text{m}$ . $v = \sqrt{6.67 \times 10^{-11} \times 5.97 \times 10^{24}/8.37 \times 10^{6}} = 6.90 \times 10^{3}\ \text{m s}^{-1}$ . $T = 2\pi r/v = 7.62 \times 10^{3}\ \text{s}$ (about $127$ minutes).
$T = 5400\ \text{s}$ . $r = (GMT^2/4\pi^2)^{1/3} = 6.65 \times 10^{6}\ \text{m}$ ; height $= 6.65 \times 10^{6} - 6.37 \times 10^{6} = 2.8 \times 10^{5}\ \text{m}$ (about $280\ \text{km}$ ).
$T \propto r^{3/2}$ : ratio $= 4^{1.5} = 8$ . The outer satellite's period is $8$ times longer.
$T = 365.25 \times 86\,400 = 3.156 \times 10^{7}\ \text{s}$ . $M = 4\pi^2r^3/(GT^2) = 4\pi^2 \times (1.50 \times 10^{11})^3/(6.67 \times 10^{-11} \times (3.156 \times 10^{7})^2) = 2.0 \times 10^{30}\ \text{kg}$ .
The station and the astronauts are both in free fall, accelerating towards the Earth at the same rate (gravity provides their centripetal acceleration). There is no contact force between the astronauts and the station, so they feel weightless.
$v^2 = GM/r$ , so $E_K = \tfrac{1}{2}mv^2 = \dfrac{GMm}{2r} = \dfrac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 500}{2 \times 4.22 \times 10^{7}} = 2.36 \times 10^{9}\ \text{J}$ .
(a) $T^2 = (4\pi^2/GM)r^3$ . Take $\lg$ of both sides: $2\lg T = 3\lg r + \lg(4\pi^2/GM)$ , then divide by $2$ . (b) Intercept: $5.20 = 1.50 \times 8.00 + c$ , so $c = -6.80$ . Then $\tfrac{1}{2}\lg(4\pi^2/GM) = -6.80$ , so $4\pi^2/GM = 10^{-13.6} = 2.51 \times 10^{-14}$ . $M = 4\pi^2/(G \times 2.51 \times 10^{-14}) = 2.4 \times 10^{25}\ \text{kg}$ .

Gravity provides the centripetal force

Orbital speed

Orbital period: deriving T2∝r3T^2 \propto r^3T2∝r3

Geostationary orbits

Weightlessness in orbit

Worked examples

Practice questions

Console

Orbital period: deriving $T^2 \propto r^3$