Circular Orbits in Gravitational Fields (AQA A Level Physics): Revision Note

Circular orbits in gravitational fields

• Since most planets and satellites have a near circular orbit, the gravitational force $F_G$ between the Sun and another planet provides the centripetal force needed to stay in an orbit

• The gravitational force is centripetal, therefore it is 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 $v$

$F_G=F_{centripetal}$

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

$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$

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

$v^2=\frac{GM}{r}$

• Where:

  • $v$ = linear speed of the mass in orbit (m s^-1)

  • $G$ = Newton's gravitational constant

  • $M$ = mass of the object being orbited (kg)

  • $r$ = orbital radius (m)

• 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

Time period and orbital radius relation

• Since a planet or a satellite is travelling in circular motion when in orbit, its orbital time period $T$ to travel the circumference of the orbit $2\pi r$, the linear speed $v$ is:

$v=\frac{2\pi r}{T}$

• This is a result of the well-known equation, speed = distance / time and first introduced in the circular motion topic

• Substituting the value of the linear speed $v$ from equating the gravitational and centripetal force into the above equation gives:

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

• Squaring out the brackets and rearranging for $T^2$ gives the equation relating the time period $T$ and orbital radius $r$:

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

• Where:

  • $T$ = time period of the orbit (s)

  • $r$ = orbital radius (m)

  • $G$ = Newton's gravitational constant

  • $M$ = mass of the object being orbited (kg)

• The equation shows that the orbital period $T$ is related to the radius $r$ of the orbit

• This is also 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:

$T^2\propto r^3$

Graphical representation of T² ∝ r³

• The relationship between $T$ and $r$ can be shown using a logarithmic plot

• The graph of log($T$) in years against log($r$) in AU (astronomical units) for the planets in our solar system is a straight line graph:

• The graph does not go through the origin since it has a negative y-intercept

• Only the log of both $T$ and $r$ will produce a straight line graph

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