Orbits of planets and satellites (A-level only)

Welcome to the physics of orbits! Whether we are looking at moons orbiting planets, planets orbiting the Sun, or artificial satellites orbiting Earth, the physical principles are exactly the same.

What you'll learn in this topic:

How to link the orbital speed and period of a satellite to its radius.
How to derive the famous relationship $T2∝r3T^2 \propto r^3$ and prove it using logarithmic graphs.
How the kinetic, potential, and total energy of a satellite change with its orbit.
How to calculate the escape velocity of a planet.
The differences between geostationary orbits and Low Earth Orbits (LEO).

The physics of a circular orbit

To keep an object moving in a circle, a centripetal force is required. For a satellite orbiting a planet (or a planet orbiting a star), gravity is the centripetal force.

Orbiting satellite diagram

Let's say a satellite of mass $m$ is orbiting a planet of mass $M$ at a distance $r$ from the planet's centre, moving at a constant speed $v$ .

The gravitational force between them is given by Newton's law of gravitation:

\frac{GMm}{r^2}

The centripetal force required to maintain this circular orbit is:

\frac{mv^2}{r}

Because gravity provides this centripetal force, we equate the two:

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

By cancelling $m$ and one factor of $r$ , we can rearrange this to find the orbital speed $v$ :

v^2 = \frac{GM}{r} \implies v = \sqrt{\frac{GM}{r}}

Key Idea

Orbital speed is independent of mass

Notice that the mass of the satellite ( $m$ ) cancelled out. This means that at a given radius, a massive space station and a tiny bolt must travel at exactly the same speed to maintain their orbit! The orbital speed only depends on the mass of the central object ( $M$ ) and the orbital radius ( $r$ ).

Deriving Kepler's Third Law ( $T2∝r3T^2 \propto r^3$ )

The orbital period $T$ is the time taken for one complete orbit. Since the satellite travels the circumference of the orbit ( $2πr2\pi r$ ) at speed $v$ , we know:

\frac{2\pi r}{T}

Substitute this expression for $v$ into our earlier equation $GMmr2=mv2r\frac{GMm}{r^2} = \frac{mv^2}{r}$ :

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

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

Rearranging to make $T^2$ the subject gives the relationship:

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

Because $4$ , $π\pi$ , $G$ , and $M$ are all constants for a given central mass, the entire bracket is a constant. Therefore:

T^2 \propto r^3

This is a mathematical statement of Kepler's Third Law.

Example

Calculating orbital radius

A satellite is in orbit around the Earth with a period of 90 minutes. Calculate the radius of its orbit. Data: Mass of Earth $\times 10^{24}$ kg, $\times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ .

First, convert the period $T$ into standard SI units (seconds). $\times 60 = 5400$ s
State the formula linking $T$ and $r$ . $T2=4π2r3GMT^2 = \frac{4\pi^2 r^3}{GM}$
Rearrange the formula to make $r^3$ the subject. $r3=GMT24π2r^3 = \frac{GM T^2}{4\pi^2}$
Substitute the values into the equation. $r3=(6.67×10−11)×(5.97×1024)×540024π2r^3 = \frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times 5400^2}{4\pi^2}$ $m3r^3 = 2.94 \times 10^{20} \text{ m}^3$
Take the cube root to find $r$ . $\sqrt[3]{2.94 \times 10^{20}} = 6.65 \times 10^6 \text{ m}$

Analysing orbits with logarithmic graphs

In exams, you may be asked to verify $T2∝r3T^2 \propto r^3$ using data from several planets. Plotting $T^2$ against $r^3$ gives a straight line through the origin, but the numbers involved are astronomically huge. Instead, we often plot a logarithmic graph.

Starting with:

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

Take the natural logarithm ( $ln⁡\ln$ ) of both sides:

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

Using the laws of logarithms ( $ln⁡(ab)=ln⁡(a)+ln⁡(b)\ln(ab) = \ln(a) + \ln(b)$ and $ln(a^n) = n \ln(a)$ ), we can expand this:

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

Divide by 2 to get it into the form $y = m x + c$ :

\ln(T) = \frac{3}{2} \ln(r) + \frac{1}{2} \ln\left( \frac{4\pi^2}{GM} \right)

If you plot $ln⁡(T)\ln(T)$ on the y-axis against $ln⁡(r)\ln(r)$ on the x-axis, you will get a straight line with a gradient of $1.5$ ( $32\frac{3}{2}$ ). The y-intercept allows you to calculate the mass of the central object $M$ .

Common Mistake

Forgetting the units of your intercept

When taking the y-intercept $12ln⁡(4π2GM)\frac{1}{2} \ln\left( \frac{4\pi^2}{GM} \right)$ to find $M$ , ensure you remember that $T$ must have been in seconds and $r$ in metres for $G$ to work correctly. If the data table gave $T$ in days or $r$ in km, you must convert them before calculating $M$ , or your mass will be completely wrong!

Energy in an orbit

A satellite in orbit possesses both kinetic energy ( $E_k$ ) due to its speed, and gravitational potential energy ( $E_p$ ) due to its position in a gravitational field.

Let's find expressions for both. We already know that $v2=GMrv^2 = \frac{GM}{r}$ . Therefore, the kinetic energy is:

E_k = \frac{1}{2}mv^2 = \frac{1}{2}m \left( \frac{GM}{r} \right) = \frac{GMm}{2r}

The gravitational potential energy is defined as being zero at infinity. Because gravity is attractive, work must be done to move a mass to infinity, meaning $E_p$ everywhere else is negative:

E_p = -\frac{GMm}{r}

The total energy ( $E$ ) of the satellite is the sum of its kinetic and potential energies:

E = E_k + E_p

\frac{GMm}{2r} - \frac{GMm}{r}

-\frac{GMm}{2r}

Tip

The energy shortcut

Look closely at those three formulas! The magnitude of the kinetic energy is exactly half the magnitude of the potential energy. $Ek=−12EpE_k = -\frac{1}{2}E_p$ Total Energy $E = -E_k$

If an orbiting satellite experiences atmospheric drag (air resistance in very low orbits), its total energy decreases (becomes more negative). This means its radius $r$ must decrease. As it falls to a lower radius, its $E_p$ decreases (becomes more negative), but its $E_k$ actually increases. Counter-intuitively, air resistance causes the satellite to speed up as it spirals inwards!

Escape velocity

Definition

Escape velocity

Escape velocity is the minimum speed an unpowered object needs at the surface of a planet to escape its gravitational field and reach infinity.

At the surface of a planet of radius $r$ , the object has potential energy $Ep=−GMmrE_p = -\frac{GMm}{r}$ . To just escape, it must reach infinity where its potential energy is zero, and it will arrive there with zero kinetic energy. Therefore, its total energy at the surface must be exactly zero.

\text{Total Energy} = E_k + E_p = 0

\frac{1}{2}mv^2 - \frac{GMm}{r} = 0

\frac{1}{2}mv^2 = \frac{GMm}{r}

The mass of the object $m$ cancels out (meaning the escape velocity is the same for a rocket or a molecule of gas):

v^2 = \frac{2GM}{r} \implies v = \sqrt{\frac{2GM}{r}}

Synchronous and geostationary orbits

Satellites are launched into different orbits depending on their purpose.

Definition

Synchronous orbit

A synchronous orbit is an orbit where the satellite's orbital period $T$ is exactly equal to the rotational period of the object it is orbiting.

Diagram comparing LEO and Geostationary orbits

For Earth, a synchronous orbit has a period of 24 hours. A geostationary orbit is a very specific type of Earth-synchronous orbit with three strict rules:

It must have a period of exactly 24 hours.
It must orbit in the exact same direction as the Earth's rotation.
It must be perfectly aligned with the equatorial plane (it orbits directly above the equator).

Because of these rules, a geostationary satellite stays exactly above the same point on Earth's surface at all times. This makes them ideal for telecommunications and satellite TV broadcasts, because receivers on Earth (like satellite dishes on houses) can be pointed at a fixed spot in the sky and never need to move.

Using $T2∝r3T^2 \propto r^3$ , a geostationary orbit always has a radius of roughly $00042\,000$ km from the centre of the Earth (or about $00036\,000$ km above the surface).

Low Earth Orbits (LEO)

Low Earth Orbits are much closer to the surface (typically $160$ km to $2000$ km altitude). Because $r$ is small, $T$ is also small. A typical LEO satellite completes an orbit in about 90 to 120 minutes.

LEOs are often polar orbits (passing over the North and South poles). Because the Earth rotates beneath the satellite, a polar LEO satellite will eventually scan the entire surface of the Earth over multiple orbits.

LEO satellites are cheaper to launch (requiring less energy) and closer to the surface, making them ideal for high-resolution military surveillance, Earth mapping, and weather monitoring.

Exam technique

In the exam

Watch out for altitude vs. radius: Questions love to give you the height above the surface (altitude) instead of the orbital radius $r$ . Always add the planet's radius to the altitude to find $r$ before using any formulas.
Remember the time conversions: If you are using $T2∝r3T^2 \propto r^3$ to calculate $M$ , $T$ must be converted into seconds. $\text{ hours} = 24 \times 60 \times 60 = 86400 \text{ s}$ .
Check the signs on energy: If a question asks for the change in potential energy, do "final minus initial" carefully. Since $E_p$ values are negative, $number)\Delta E_p = (- \text{small number}) - (- \text{large number})$ , which yields a positive increase if moving outwards.

Self review

Check yourself

Can you derive $\sqrt{\frac{GM}{r}}$ by equating gravitational and centripetal forces?
How do you prove that plotting $ln⁡(T)\ln(T)$ against $ln⁡(r)\ln(r)$ gives a gradient of $1.5$ ?
If a satellite's orbital radius increases, what happens to its kinetic energy, potential energy, and total energy?
What three conditions must be met for an orbit to be geostationary?

Welcome to the physics of orbits! Whether we are looking at moons orbiting planets, planets orbiting the Sun, or artificial satellites orbiting Earth, the physical principles are exactly the same.

What you'll learn in this topic:

How to link the orbital speed and period of a satellite to its radius.
How to derive the famous relationship $T2∝r3T^2 \propto r^3$ and prove it using logarithmic graphs.
How the kinetic, potential, and total energy of a satellite change with its orbit.
How to calculate the escape velocity of a planet.
The differences between geostationary orbits and Low Earth Orbits (LEO).

The physics of a circular orbit

To keep an object moving in a circle, a centripetal force is required. For a satellite orbiting a planet (or a planet orbiting a star), gravity is the centripetal force.

Orbiting satellite diagram

Let's say a satellite of mass $m$ is orbiting a planet of mass $M$ at a distance $r$ from the planet's centre, moving at a constant speed $v$ .

The gravitational force between them is given by Newton's law of gravitation:

\frac{GMm}{r^2}

The centripetal force required to maintain this circular orbit is:

\frac{mv^2}{r}

Because gravity provides this centripetal force, we equate the two:

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

By cancelling $m$ and one factor of $r$ , we can rearrange this to find the orbital speed $v$ :

v^2 = \frac{GM}{r} \implies v = \sqrt{\frac{GM}{r}}

Key Idea

Orbital speed is independent of mass

Deriving Kepler's Third Law ( $T2∝r3T^2 \propto r^3$ )

The orbital period $T$ is the time taken for one complete orbit. Since the satellite travels the circumference of the orbit ( $2πr2\pi r$ ) at speed $v$ , we know:

\frac{2\pi r}{T}

Substitute this expression for $v$ into our earlier equation $GMmr2=mv2r\frac{GMm}{r^2} = \frac{mv^2}{r}$ :

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

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

Rearranging to make $T^2$ the subject gives the relationship:

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

Because $4$ , $π\pi$ , $G$ , and $M$ are all constants for a given central mass, the entire bracket is a constant. Therefore:

T^2 \propto r^3

This is a mathematical statement of Kepler's Third Law.

Example

Calculating orbital radius

A satellite is in orbit around the Earth with a period of 90 minutes. Calculate the radius of its orbit. Data: Mass of Earth $\times 10^{24}$ kg, $\times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$ .

First, convert the period $T$ into standard SI units (seconds). $\times 60 = 5400$ s
State the formula linking $T$ and $r$ . $T2=4π2r3GMT^2 = \frac{4\pi^2 r^3}{GM}$
Rearrange the formula to make $r^3$ the subject. $r3=GMT24π2r^3 = \frac{GM T^2}{4\pi^2}$
Substitute the values into the equation. $r3=(6.67×10−11)×(5.97×1024)×540024π2r^3 = \frac{(6.67 \times 10^{-11}) \times (5.97 \times 10^{24}) \times 5400^2}{4\pi^2}$ $m3r^3 = 2.94 \times 10^{20} \text{ m}^3$
Take the cube root to find $r$ . $\sqrt[3]{2.94 \times 10^{20}} = 6.65 \times 10^6 \text{ m}$

Analysing orbits with logarithmic graphs

Starting with:

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

Take the natural logarithm ( $ln⁡\ln$ ) of both sides:

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

Using the laws of logarithms ( $ln⁡(ab)=ln⁡(a)+ln⁡(b)\ln(ab) = \ln(a) + \ln(b)$ and $ln(a^n) = n \ln(a)$ ), we can expand this:

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

Divide by 2 to get it into the form $y = m x + c$ :

\ln(T) = \frac{3}{2} \ln(r) + \frac{1}{2} \ln\left( \frac{4\pi^2}{GM} \right)

Common Mistake

Forgetting the units of your intercept

Energy in an orbit

A satellite in orbit possesses both kinetic energy ( $E_k$ ) due to its speed, and gravitational potential energy ( $E_p$ ) due to its position in a gravitational field.

Let's find expressions for both. We already know that $v2=GMrv^2 = \frac{GM}{r}$ . Therefore, the kinetic energy is:

E_k = \frac{1}{2}mv^2 = \frac{1}{2}m \left( \frac{GM}{r} \right) = \frac{GMm}{2r}

The gravitational potential energy is defined as being zero at infinity. Because gravity is attractive, work must be done to move a mass to infinity, meaning $E_p$ everywhere else is negative:

E_p = -\frac{GMm}{r}

The total energy ( $E$ ) of the satellite is the sum of its kinetic and potential energies:

E = E_k + E_p

\frac{GMm}{2r} - \frac{GMm}{r}

-\frac{GMm}{2r}

Tip

The energy shortcut

Look closely at those three formulas! The magnitude of the kinetic energy is exactly half the magnitude of the potential energy. $Ek=−12EpE_k = -\frac{1}{2}E_p$ Total Energy $E = -E_k$

Escape velocity

Definition

Escape velocity

Escape velocity is the minimum speed an unpowered object needs at the surface of a planet to escape its gravitational field and reach infinity.

\text{Total Energy} = E_k + E_p = 0

\frac{1}{2}mv^2 - \frac{GMm}{r} = 0

\frac{1}{2}mv^2 = \frac{GMm}{r}

The mass of the object $m$ cancels out (meaning the escape velocity is the same for a rocket or a molecule of gas):

v^2 = \frac{2GM}{r} \implies v = \sqrt{\frac{2GM}{r}}

Synchronous and geostationary orbits

Satellites are launched into different orbits depending on their purpose.

Definition

Synchronous orbit

A synchronous orbit is an orbit where the satellite's orbital period $T$ is exactly equal to the rotational period of the object it is orbiting.

Diagram comparing LEO and Geostationary orbits

For Earth, a synchronous orbit has a period of 24 hours. A geostationary orbit is a very specific type of Earth-synchronous orbit with three strict rules:

It must have a period of exactly 24 hours.
It must orbit in the exact same direction as the Earth's rotation.
It must be perfectly aligned with the equatorial plane (it orbits directly above the equator).

Using $T2∝r3T^2 \propto r^3$ , a geostationary orbit always has a radius of roughly $00042\,000$ km from the centre of the Earth (or about $00036\,000$ km above the surface).

Low Earth Orbits (LEO)

LEO satellites are cheaper to launch (requiring less energy) and closer to the surface, making them ideal for high-resolution military surveillance, Earth mapping, and weather monitoring.

Exam technique

In the exam

Watch out for altitude vs. radius: Questions love to give you the height above the surface (altitude) instead of the orbital radius $r$ . Always add the planet's radius to the altitude to find $r$ before using any formulas.
Remember the time conversions: If you are using $T2∝r3T^2 \propto r^3$ to calculate $M$ , $T$ must be converted into seconds. $\text{ hours} = 24 \times 60 \times 60 = 86400 \text{ s}$ .
Check the signs on energy: If a question asks for the change in potential energy, do "final minus initial" carefully. Since $E_p$ values are negative, $number)\Delta E_p = (- \text{small number}) - (- \text{large number})$ , which yields a positive increase if moving outwards.

Self review

Check yourself

Can you derive $\sqrt{\frac{GM}{r}}$ by equating gravitational and centripetal forces?
How do you prove that plotting $ln⁡(T)\ln(T)$ against $ln⁡(r)\ln(r)$ gives a gradient of $1.5$ ?
If a satellite's orbital radius increases, what happens to its kinetic energy, potential energy, and total energy?
What three conditions must be met for an orbit to be geostationary?

The physics of a circular orbit

Deriving Kepler's Third Law ( $T2∝r3T^2 \propto r^3$ )

Analysing orbits with logarithmic graphs

Energy in an orbit

Escape velocity

Synchronous and geostationary orbits

Low Earth Orbits (LEO)

Review Orbits of planets and satellites (A-level only) by teaching Genie

Remember key concepts with flashcards

Orbits of planets and satellites (A-level only)

The physics of a circular orbit

Deriving Kepler's Third Law ( $T2∝r3T^2 \propto r^3$ )

Analysing orbits with logarithmic graphs

Energy in an orbit

Escape velocity

Synchronous and geostationary orbits

Low Earth Orbits (LEO)

Review Orbits of planets and satellites (A-level only) by teaching Genie

Remember key concepts with flashcards

Orbits of planets and satellites (A-level only)

The physics of a circular orbit

Deriving Kepler's Third Law (T2∝r3T^2 \propto r^3T2∝r3)

Analysing orbits with logarithmic graphs

Energy in an orbit

Escape velocity

Synchronous and geostationary orbits

Low Earth Orbits (LEO)

Review Orbits of planets and satellites (A-level only) by teaching Genie

Remember key concepts with flashcards

Orbits of planets and satellites (A-level only) Revision Guide

Orbits of planets and satellites (A-level only)

The physics of a circular orbit

Deriving Kepler's Third Law (T2∝r3T^2 \propto r^3T2∝r3)

Analysing orbits with logarithmic graphs

Energy in an orbit

Escape velocity

Synchronous and geostationary orbits

Low Earth Orbits (LEO)

Review Orbits of planets and satellites (A-level only) by teaching Genie

Remember key concepts with flashcards

Orbits of planets and satellites (A-level only) Revision Guide

Deriving Kepler's Third Law ( $T2∝r3T^2 \propto r^3$ )

Deriving Kepler's Third Law ( $T2∝r3T^2 \propto r^3$ )