Circular Orbits in Gravitational Fields

For objects in a circular orbit in a gravitational field the gravitational force is related to the centripetal acceleration it causes
Planets travel around the Sun in orbits that are (approximately) circular and within the Sun's gravitational field
- Objects in a circular orbit travel at a constant speed
The orbit is a circular path, therefore the direction in which the object is travelling will be constantly changing
- A change in direction causes a change in velocity
Acceleration is the rate of change of velocity, therefore if an object is constantly changing direction then its velocity is constantly changing, so an object in orbit is accelerating
A resultant force is needed to cause an acceleration
This resultant centripetal force is due to the gravitational force F_G created by the object being orbitted
The resultant force must act at right angles to the instantaneous velocity of the object to create a circular orbit
- This is always towards the centre of the orbit
- The instantaneous velocity of the object is the velocity at a given time

Gravitational force and instantaneous velocity

Motion in an orbit, downloadable IGCSE & GCSE Physics revision notes

The direction of the instantaneous velocity and the gravitational force at different points of the Earth’s orbit around the sun

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 ⍵

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

Examiner 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