Orbital Motion (Edexcel International A Level Physics)

Revision Note

Test yourself
Katie M

Author

Katie M

Last updated

Orbital Motion

Newton's Law of Gravitation & Orbits

  • Since most planets and satellites have a near circular orbit, the gravitational force FG between the sun or another planet provides the centripetal force needed to stay in an orbit
    • This centripetal force is perpendicular to the velocity 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
    • Equating the gravitational force from Newton's Law of Gravitation to the centripetal force for a planet or satellite in orbit gives:

FG = Fcentripetal

Circular Orbits Equation 1

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

Circular Orbits Equation 2

  • Where:
    • v = linear speed of the mass in orbit (m s1)
    • G = Newton's Gravitational Constant (N m2 kg–2)
    • 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 orbital radius r

Newton's Laws of Motion & Orbits

  • Newton's first law of motion states that a body remain at rest or at constant velocity unless a resultant force acts on it
  • Bodies in orbit do have a resultant force acting on them
    • This is the gravitational force due to the mass M being orbited
    • This force is a centripetal force because it acts towards the centre of M, perpendicular to the velocity of the planet or satellite
  • Therefore, since the direction of a planet or satellite orbiting in circular motion is constantly changing, it must be accelerating
    • This is called centripetal acceleration

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

A satellite in orbit around the Earth travels in circular motion

Time Period & Orbital Radius Relation

  • A planet or a satellite orbits in circular motion
    • Therefore, its orbital time period T is the time taken to travel the circumference of the orbit 2πr
    • This means the linear speed, or orbital speed v is:

Circular Orbits Equation 3

  • This is a result of the well-known equation speed = distance / time
  • Equating the two equations for orbital speed gives:

Circular Orbits Equation 4

  • Squaring out the brackets and rearranging for T2 gives the equation relating the time period T and orbital radius r:

Circular Orbits Equation 5

  • Where:
    • T = time period of the orbit (s)
    • r = orbital radius (m)
    • G = Newton's Gravitational Constant (N m2 kg–2)
    • M = mass of the object being orbited (kg)

  • The equation shows the relationship between the orbital period and the orbital radius for any planet or satellite in orbit
  • It is summarised mathematically as:

Circular Orbits in Gravitational Fields equation 7

Worked example

A binary star system constant 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 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

(i) The angular speed ⍵ of M1

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

Circular Orbits Worked Example Part 2

Examiner Tip

This worked example helps you practise two crucial techniques that are often examined: 

  1. The centripetal force is expressed as fraction numerator m v squared over denominator r end fraction or equivalently as m omega squared r. In our case the angular speed is given in the question, so it is best to use the latter expression m omega squared r when equating the centripetal force to the gravitational force. 
  2. The gravitational force is given as fraction numerator G M m over denominator r squared end fractionbut note that the distance r in this question is given as a sum, R1 + R2You should remember that r is defined as the distance between the centre of masses of M and m, therefore, r = R1 + R2 and so r2 = (R1 + R2)2.

You've read 0 of your 5 free revision notes this week

Sign up now. It’s free!

Join the 100,000+ Students that ❤️ Save My Exams

the (exam) results speak for themselves:

Did this page help you?

Katie M

Author: Katie M

Expertise: Physics

Katie has always been passionate about the sciences, and completed a degree in Astrophysics at Sheffield University. She decided that she wanted to inspire other young people, so moved to Bristol to complete a PGCE in Secondary Science. She particularly loves creating fun and absorbing materials to help students achieve their exam potential.