The motivation for Isaac Newton to discover his laws of motion was to explain the properties of planetary orbits that were obser
ved by Tycho Brahe and analyzed by Johannes Kepler. A good starting point for understanding this (as well as the speed of the space shuttle and the height of geostationary satellites) is the simplest orbit - a circular one. This problem concerns the properties of circular orbits for a satellite orbiting a planet of mass M. Use G for the universal gravitational constant. Find the orbital period T.
We know that , for an object to remain in circular motion , a force towards centre is required which is called centripetal force. In the circular motion of
satellites around planet , this force is provided by the gravitational attraction between satellite and planet.
If M be the mass of planet and m be the mass of satellite, G be gravitational constant and R be the distance between planet and satellite or R be the radius of orbit
Gravitational force = G Mm / R²
If v be the velocity with which satellite is orbiting
centripetal force
= m v² /R
Centripetal force = gravitational attraction
m v² /R = G Mm / R²
v =
Time period = time the satellite takes to make one rotation
Given there are three blocks of masses , and (ref image in attachment)
When all three masses move together at an acceleration a, the force F is given by
F = ( + + ) *a ................(equation 1)
Also it is given that does not move with respect to , which gives tension T is exerted on pulley by only, Hence tension T is
T = *a ..........(equation 2)
There is also also tension exerted by . There are two components here: horizontal due to acceleration a and vertical component due to gravity g. Thus tension is given by
T = ................(equation 3)
From equation 2 and 3, we get
*a =
Squaring both sides we get
* = * (+)
* = ( * )+ ( *)
( - ) * = *
= */( - )
Taking square root on both sides, we get acceleration a
a = *g/()
Hence substituting the value of a in equation 1, we get