mass m also exerts equal and opposite force on m . So the force of attraction ( F ) experienced by m due to m is in the direction of positive ‘y’ axis ie., r j = . F j =− F j The direction of the force is shown in the figure, - - - - Unit Gravitation hollow sphere also as another point mass. Essentially the entire mass of the hollow sphere appears to be concentrated at the center of the hollow sphere.
It is shown in the Figure . (a). There is also another interesting result. Consider a hollow sphere of mass M.
If we place another object of mass ‘m’ inside this hollow sphere as in Figure . (b), the force experienced by this mass ‘m’ will be zero. This calculation will be dealt with in higher classes. M O M O M Hollow sphere of mass O No force on m Figure .
A mass placed in a hollow sphere. The triumph of the law of gravitation is that it concludes that the mango that is falling down and the Moon orbiting the Earth are due to the same gravitational force. Newton’s inverse square Law: Newton considered the orbits of the planets as circular. For circular orbit of radius r, the centripetal acceleration towards the center is a ( .
) The torque experienced by the Earth due to the gravitational force of the Sun is given by τ = × = × − = F GM M S Since r r r r r ( ) = , So τ = dL dt . It implies that angular momentum L is a constant vector. The angular momentum of the Earth about the Sun is constant throughout the motion. It is true for all the planets.
In fact, this constancy of angular momentum leads to the Kepler’s second law. The expression