Gravitation • Topic 3 of 3

Satellites & Kepler's Laws

A satellite is any body that revolves around a planet under the influence of the planet's gravity. The Moon is Earth's natural satellite; communication and weather satellites are artificial ones. For a satellite in a stable circular orbit, the gravitational pull provides exactly the centripetal force needed to keep it moving in a circle.

Orbital velocity ($v_o$) is the speed a satellite must have to stay in orbit at radius $r$ from the centre of the planet. Equating gravity to the centripetal force, $\frac{GMm}{r^2}=\frac{mv_o^2}{r}$, gives:

  • $v_o=\sqrt{\frac{GM}{r}}$. For an orbit close to the surface ($r\approx R$), $v_o=\sqrt{gR}\approx7.9\ \text{km/s}$.
  • The orbital velocity is independent of the satellite's mass and decreases for higher orbits.

Time period ($T$) is the time for one complete revolution: $T=\frac{2\pi r}{v_o}=2\pi\sqrt{\frac{r^3}{GM}}$. The period grows with the orbital radius.

Energy of a satellite. A satellite has both kinetic and potential energy:

  • Kinetic energy: $KE=\frac{1}{2}mv_o^2=\frac{GMm}{2r}$.
  • Potential energy: $PE=-\frac{GMm}{r}$.
  • Total energy: $E=KE+PE=-\frac{GMm}{2r}$. The total energy is negative, confirming that the satellite is bound to the planet.

Geostationary satellites. A geostationary satellite has an orbital period of exactly 24 hours, so it appears fixed over one point on the equator. It orbits in the equatorial plane, from west to east, at a height of about 36,000 km above the surface. These are used for communication and television broadcasting.

Kepler's laws of planetary motion describe how planets move around the Sun, and they apply equally to satellites:

  • First law (law of orbits): every planet moves in an ellipse with the Sun at one focus.
  • Second law (law of areas): the line joining a planet to the Sun sweeps out equal areas in equal times, so a planet moves faster when nearer the Sun. This is a consequence of the conservation of angular momentum.
  • Third law (law of periods): the square of the period is proportional to the cube of the semi-major axis: $T^2\propto r^3$, i.e. $\frac{T^2}{r^3}=\text{constant}$ for all planets around the Sun.
Satellite in circular orbit and Kepler's law of equal areasMv_ov_o = sqrt(GM/r)SunEqual areas in equal times
Solved Examples
1
Worked Example
A satellite orbits the Earth at a height equal to the Earth's radius. Find its orbital velocity. (Take $g=9.8\ \text{m/s}^2$, $R=6.4\times10^{6}\ \text{m}$.)
Solution
  1. Step 1: Orbit radius $r=R+R=2R$. Use $v_o=\sqrt{\frac{GM}{r}}=\sqrt{\frac{gR^2}{2R}}=\sqrt{\frac{gR}{2}}$.
  2. Step 2: Substitute: $v_o=\sqrt{\frac{9.8\times6.4\times10^{6}}{2}}=\sqrt{3.136\times10^{7}}$.
  3. Step 3: Compute: $v_o\approx5.6\times10^{3}\ \text{m/s}=5.6\ \text{km/s}$.

Answer: $v_o\approx5.6\ \text{km/s}$.

2
Worked Example
Find the orbital velocity of a satellite in an orbit very close to the Earth's surface. (Take $g=9.8\ \text{m/s}^2$, $R=6.4\times10^{6}\ \text{m}$.)
Solution
  1. Step 1: For $r\approx R$, $v_o=\sqrt{gR}$.
  2. Step 2: Substitute: $v_o=\sqrt{9.8\times6.4\times10^{6}}=\sqrt{6.272\times10^{7}}$.
  3. Step 3: Compute: $v_o\approx7.92\times10^{3}\ \text{m/s}\approx7.9\ \text{km/s}$.

Answer: $v_o\approx7.9\ \text{km/s}$.

3
Worked Example
The time period of a satellite in a circular orbit of radius $r$ is $T$. What will be the period in an orbit of radius $4r$?
Solution
  1. Step 1: By Kepler's third law $T^2\propto r^3$, so $\frac{T'^2}{T^2}=\left(\frac{4r}{r}\right)^3=64$.
  2. Step 2: Take the square root: $\frac{T'}{T}=\sqrt{64}=8$.
  3. Step 3: Therefore $T'=8T$.

Answer: $T'=8T$.

4
Worked Example
Find the total energy of a 200 kg satellite orbiting at radius $r=8\times10^{6}\ \text{m}$. ($M=6\times10^{24}\ \text{kg}$, $G=6.67\times10^{-11}$ SI units.)
Solution
  1. Step 1: Total energy $E=-\frac{GMm}{2r}$.
  2. Step 2: Substitute: $E=-\frac{6.67\times10^{-11}\times6\times10^{24}\times200}{2\times8\times10^{6}}$.
  3. Step 3: Numerator $=8.0\times10^{16}$, denominator $=1.6\times10^{7}$, so $E=-5.0\times10^{9}\ \text{J}$.

Answer: $E\approx-5.0\times10^{9}\ \text{J}$ (negative, so the satellite is bound).

5
Worked Example
A planet revolves around the Sun in an orbit of mean radius 4 times that of the Earth. Find its period of revolution in Earth years.
Solution
  1. Step 1: By Kepler's third law $\frac{T^2}{r^3}=$ constant, so $\frac{T_p^2}{T_e^2}=\left(\frac{r_p}{r_e}\right)^3$.
  2. Step 2: With $r_p=4r_e$ and $T_e=1$ year: $T_p^2=4^3=64$.
  3. Step 3: Therefore $T_p=\sqrt{64}=8$ years.

Answer: $T_p=8$ Earth years.

6
Worked Example
State why a geostationary satellite must orbit at a particular height, and give its approximate period and height.
Solution
  1. Step 1: To appear stationary above one point, its period must match the Earth's rotation: $T=24\ \text{hours}$.
  2. Step 2: By $T=2\pi\sqrt{\frac{r^3}{GM}}$, a fixed $T$ fixes the radius $r$, and hence the height above the surface.
  3. Step 3: This gives an orbit radius of about $42{,}000\ \text{km}$, i.e. a height of roughly $36{,}000\ \text{km}$, in the equatorial plane moving west to east.

Answer: Period $=24$ hours; height $\approx36{,}000\ \text{km}$ above the equator.

Key Points

  • Orbital velocity: $v_o=\sqrt{\frac{GM}{r}}$; near the surface $v_o=\sqrt{gR}\approx7.9$ km/s, independent of the satellite's mass.
  • Time period: $T=2\pi\sqrt{\frac{r^3}{GM}}$; larger orbits have longer periods.
  • Satellite energy: $KE=\frac{GMm}{2r}$, $PE=-\frac{GMm}{r}$, total $E=-\frac{GMm}{2r}$ (negative ⇒ bound).
  • A geostationary satellite has a 24-hour period, orbits over the equator at about 36,000 km height.
  • Kepler's laws: orbits are ellipses (focus at Sun); equal areas in equal times; $T^2\propto r^3$.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The orbital velocity of a satellite at radius $r$ around a planet of mass $M$ is:
Explanation: Equating gravity to centripetal force gives $v_o=\sqrt{\frac{GM}{r}}$.
Q2.Kepler's third law states that:
Explanation: The square of the period is proportional to the cube of the orbit's semi-major axis: $T^2\propto r^3$.
Q3.The total energy of a satellite in a stable circular orbit is:
Explanation: $E=-\frac{GMm}{2r}$ is negative, which is why the satellite stays bound to the planet.
Q4.A geostationary satellite has a time period of:
Explanation: Its 24-hour period matches the Earth's rotation, so it stays fixed above one point on the equator.
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