Circular Orbits in Gravitational Fields

Since most planets and satellites have a near circular orbit, the gravitational force F_G between the sun or another planet provides the centripetal force needed to stay in an orbit
Both the gravitational force and centripetal force are perpendicular to the direction of travel of the planet
Consider a satellite with mass m orbiting Earth with mass M at a distance r from the centre travelling with linear speed v

$F_{G} = F_{c i r c}$

Equating the gravitational force to the centripetal force for a planet or satellite in orbit gives:

$\frac{G M m}{r^{2}} = \frac{m v^{2}}{r}$

The mass of the satellite m will cancel out on both sides to give:

$v^{2} = \frac{G M}{r}$

This means that all satellites, whatever their mass, will travel at the same speed v in a particular orbit radius r
Recall that since the direction of a planet orbiting in circular motion is constantly changing, it has centripetal acceleration

A Satellite in Orbit Around the Earth

Circular motion satellite, downloadable AS & A Level Physics revision notes

A satellite in orbit around the Earth travels in circular motion

Kepler’s Third Law of Planetary Motion

For the orbital time period T to travel the circumference of the orbit 2πr, the linear speed v can be written as

$v = \frac{2 π r}{T}$

This is a result of the well-known equation, speed = distance / time
Substituting the value of the linear speed v into the above equation:

$v^{2} = {(\frac{2 π r}{T})}^{2} = \frac{G M}{r}$

Rearranging leads to Kepler’s third law equation:

$T^{2} = \frac{4 π^{2} r^{3}}{G M}$

The equation shows that the orbital period T is related to the radius r of the orbit. This is known as Kepler’s third law:

For planets or satellites in a circular orbit about the same central body, the square of the time period is proportional to the cube of the radius of the orbit

Kepler’s third law can be summarised as:

$T^{2} \propto r^{3}$

Maths Tip

The ∝ symbol means ‘proportional to’
Find out more about proportional relationships between two variables in the “proportional relationships ” section of the A Level Maths revision notes

Worked example

A binary star system consists of two stars orbiting about a fixed point B. The star of mass M₁ has a circular orbit of radius R₁ and mass M₂ has a radius of R₂. Both have a linear speed v and an angular speed ⍵ about B. Worked example - circular orbits in g fields, downloadable AS & A Level Physics revision notes

State the following formula, in terms of G, M₂, R₁ and R₂

(a) The angular speed ⍵ of M₁

(b) The time period T for each star in terms of angular speed ⍵

Answer:

(a)

Step 1: Equating the centripetal force of mass M₁ to the gravitational force between M₁ and M₂

$M_{1} R_{1} ω^{2} = | \frac{G M_{1} M_{2}}{{(R_{1} + R_{2})}^{2}}$

Step 2: M₁ cancels on both sides

$R_{1} ω^{2} = \frac{G M_{2}}{{(R_{1} + R_{2})}^{2}}$

Step 3: Rearrange for angular velocity ⍵

$ω^{2} = \frac{G M_{2}}{R_{1} {(R_{1} + R_{2})}^{2}}$

Step 4: Square root both sides

$ω = \sqrt{\frac{G M_{2}}{R_{1} {(R_{1} + R_{2})}^{2}}}$

(b)

Step 1: Angular speed equation with time period T

$ω = \frac{2 π}{T}$

Step 2: Rearrange for T

$T = \frac{2 π}{ω}$

Step 3: Substitute in ⍵

$T = 2 π \div \sqrt{\frac{G M_{2}}{R_{1} {(R_{1} + R_{2})}^{2}}} = 2 π \sqrt{\frac{R_{1} {(R_{1} + R_{2})}^{2}}{{GM}_{2}}}$

Exam Tip

Many of the calculations in the Gravitation questions depend on the equations for circular motion. Be sure to revisit these and understand how to use them!

Circular Orbits in Gravitational Fields (CIE A Level Physics)

Revision Note