M E – Mass of the Earth and R – Radius of the Earth. Equating the above equation with Newton’s second law, M a GM M Simplifying the above equation we get, a GM Here a A is the acceleration of apple that is equal to ‘g’. Similarly the force experienced by Moon due to Earth is given by Points to Contemplate If Kepler’s third law was “ r T = constant” instead of “ r = constant” what would be the new law of gravitation? Would it still be an inverse square law?
How would the gravitational force change with distance? In this new law of gravitation, will Neptune experience greater gravitational force or lesser gravitational force when compared to the Earth? - - - - Unit Gravitation Mass m R E Earth Mass M R E Earth Figure . Force experienced by a mass on the (i) surface of the Earth (ii) at a distance from the centre of the Earth Now the force experienced by some other object of mass M at a distance r from the center of the Earth is given by, F GM M Using the value of g in equation ( .
), the force F will be, F gM R ( . ) From this it is clear that the force can be calculated simply by knowing the value of g. It is to be noted that in the above calculation G is not required. In the year , Henry Cavendish experimentally determined the value of gravitational constant ‘G’ by using a torsion balance.
He calculated the value of ‘G’ to be equal to . × Nm kg Using modern techniques a more accurate value of G could be measured. The currently accepted value of G is .67259 × Nm kg The above calculation depends on knowing the distance between the Earth and the Moon and the radius of the Earth. The radius of the Earth was measured by Greek librarian Eratosthenes and distance between the Earth and the Moon was measured by