as the orbit of the planet around the Sun is elliptical. But this circular orbit assumption is justifiable because planet’s orbit is very close to being circular and there is only a very small deviation from the circular shape. Figure . Point mass orbiting in a circular orbit.
O a= - r v Here v is the velocity and r, the distance of the planet from the center of the orbit (Figure . ). The velocity in terms of known quantities r and T, is ( . ) Here T is the time period of revolution of the planet.
Substituting this value of v in equation ( . ) we get, a ( . ) Substituting the value of ‘a’ from ( . ) in Newton’s second law, F ma , where ‘m’ is the mass of the planet.
F mr ( . ) From Kepler’s third law, k constant ( . ) - - - - Unit Gravitation F GM M Here R m - distance of the Moon from the Earth, M m – Mass of the Moon The acceleration experienced by the Moon is given by a GM The ratio between the apple’s acceleration to Moon’s acceleration is given by a a . From the Hipparchrus measurement, the distance to the Moon is times that of Earth radius.
R m = 60R. a A / a m = The apple’s acceleration is times the acceleration of the Moon. The same result was obtained by Newton using his gravitational formula. The apple’s acceleration is measured easily and it is .
m s − . Moon orbits the Earth once in . days and by using the centripetal acceleration formula, (Refer unit ). a a 00272 which is exactly what he got through his law of gravitation.
EXAMPLE . Moon and an apple are accelerated by the same gravitational force due to Earth. Compare the acceleration of the two. The gravitational force experienced by the apple due to Earth F GM M Here M A – Mass of the apple,