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

Exam code: 9702

Leander Oates

Last updated

Circular orbits in gravitational fields

  • The orbits of most planets and satellites in the Solar System are nearly circular

  • The gravitational force between two bodies provides the centripetal force required to maintain a circular orbit

  • Consider a satellite with mass m orbiting Earth with mass M at a distance r from the centre, travelling with linear speed v

F subscript G space equals space F subscript c i r c end subscript

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

fraction numerator G M m over denominator r squared end fraction space equals space fraction numerator m v squared over denominator r end fraction

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

v squared space equals space fraction numerator G M over denominator r end fraction space space space space space rightwards double arrow space space space space space v space equals space square root of fraction numerator G M over denominator r end fraction end root

  • Where:

    • v = orbital speed of the smaller mass (m s−1)

    • G = Newton's Gravitational Constant

    • M = mass of the larger mass being orbited (kg)

    • r = orbital radius (m)

  • This equation shows that all satellites moving in a circular orbit of fixed radius r will travel at a constant speed

  • The direction of a satellite orbiting in circular motion is constantly changing, which means that the satellite has

    • a velocity which is constantly changing

    • a centripetal acceleration that acts towards the satellite

    • a centripetal force, provided by the gravitational force, which causes the satellite to accelerate

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 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 subscript 1 R subscript 1 omega squared space equals space fraction numerator G M subscript 1 M subscript 2 over denominator open parentheses R subscript 1 plus space R subscript 2 close parentheses squared end fraction

Step 2: M1 cancels on both sides

R subscript 1 omega squared space equals fraction numerator space G M subscript 2 over denominator open parentheses R subscript 1 plus space R subscript 2 close parentheses squared end fraction

Step 3: Rearrange for angular velocity ⍵

omega squared space equals fraction numerator space G M subscript 2 over denominator R subscript 1 open parentheses R subscript 1 plus space R subscript 2 close parentheses squared end fraction

Step 4: Square root both sides

omega space equals space square root of fraction numerator G M subscript 2 over denominator R subscript 1 open parentheses R subscript 1 plus R subscript 2 close parentheses squared end fraction end root

(b) 

Step 1: Angular speed equation with time period T

omega space equals fraction numerator space 2 straight pi over denominator T end fraction

Step 2: Rearrange for T

T space equals fraction numerator space 2 straight pi over denominator straight omega end fraction

Step 3: Substitute in ⍵

T space equals space 2 straight pi space divided by space square root of fraction numerator G M subscript 2 over denominator R subscript 1 open parentheses R subscript 1 plus R subscript 2 close parentheses squared end fraction end root space equals space 2 straight pi square root of fraction numerator R subscript 1 open parentheses R subscript 1 plus R subscript 2 close parentheses squared over denominator G M subscript 2 end fraction end root

Examiner Tips and Tricks

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!

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Leander Oates

Author: Leander Oates

Expertise: Physics Content Creator

Leander graduated with First-class honours in Science and Education from Sheffield Hallam University. She won the prestigious Lord Robert Winston Solomon Lipson Prize in recognition of her dedication to science and teaching excellence. After teaching and tutoring both science and maths students, Leander now brings this passion for helping young people reach their potential to her work at SME.