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

6011 Views

See similar Physics A Level tutors

Related Physics A Level answers

All answers ▸

a ball is dropped from rest off a cliff of height 50m, determine the final velocity of the ball assuming no air resistance.


Describe how a stationary wave is formed at a boundary?


How many fission event occur per second if a Uranium 235 Nuclear Reactor outputs 210MW of energy? Average Binding Energy per Nucleon of Uranium 235- 7.6 MeV Average Binding Energy per Nucleon of Products-8.5 MeV


A circuit with a voltage source of 18V, has 3 resistors all connected on parallel, values at 2ohms, 6ohms and 7.5ohms. Find the total circuit resistance, and then subsequently, the total current supplied and power dissipated in the curcuit.


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