Derive an expression to show that for satellites in a circular orbit T² ∝ r ³ where T is the period of orbit and r is the radius of the orbit.

For an object to stay in a steady orbit; F=mv2/r where: F is the force on the object towards the centre of the orbit, m is the mass of the object, v is the radial velocity of the object, and r is the radius of the orbit.In the case of a satellite orbiting a planet, all of F is provided by the gravitational force acting on the satellite due to the planet/moon/star. This force is given by Newton's law of gravitation:F = GMm/r2where F is the gravitational force, G is the gravitational constant; 6.67 x 10 -11 Nm2kg-2, M is the mass of the planet/moon/star, m is the mass of the satellite, and r is the distance between the planet/moon/star and the satellite.We can therefore equate these two forces, as F = F, giving;GMm/r2 = mv2/rWe can multiply both sides by r and divide both sides by m to give;GM/r = v2Finally, we need the time period, T, not the velocity, v, therefore we can use v = s/t. In this case, s is the circumference of orbit = 2πr, and t is T, the time period of the orbit. We can write:v = 2πr/TSubstituting this into before gives:GM/r = (2πr/T)2Expanding the brackets, multiplying both sides by T, and multiplying both sides by r gives;GMT2 = 4π2r3

JM
Answered by James M. Physics tutor

6033 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

A student studied how a few parameters of the electromagnetic radiation affects the I-V(current-voltage) curve of photoelectricity. By increasing one parameter he saw that the saturation current has risen. Which parameter it was?


Why do I keep losing marks on 6 markers?


A cart starts at rest and moves freely down a ramp without friction or air resistance and descends 8 meters vertically, what is its speed at the bottom?


what depends if the universe is expanding or not


We're here to help

contact us iconContact ustelephone icon+44 (0) 203 773 6020
Facebook logoInstagram logoLinkedIn logo

MyTutor is part of the IXL family of brands:

© 2026 by IXL Learning