and ρ is the density. On the other hand the mass of the sphere M r of radius r is π ρ and hence G m M ( . ) If the mass m is situated on the surface of earth, then r = R E and the gravitational force on it is, from Eq. ( .
) M m ( . ) The acceleration experienced by the mass m, which is usually denoted by the symbol g is related to F by Newton’s nd law by relation F = mg . Thus ( . ) Acceleration g is readily measurable.
R E is a known quantity. The measurement of G by Cavendish’s experiment (or otherwise), combined with knowledge of g and R E enables one to estimate M E from Eq. ( . ).
This is the reason why there is a popular statement regarding Cavendish : “Cavendish weighed the earth”. . ACCELERATION DUE TO GRAVITY BELOW AND ABOVE THE SURFACE OF EARTH Consider a point mass m at a height h above the surface of the earth as shown in Fig. .
(a). The radius of the earth is denoted by R E . Since this point is outside the earth, Fig. .
(a) g at a height h above the surface of the earth. its distance from the centre of the earth is ( R E + h ). If F ( h ) denoted the magnitude of the force on the point mass m , we get from Eq. ( .
) : GM m F h ( . ) The acceleration experienced by the point mass is ( )/ F h ≡ and we get F h ( . ) This is clearly less than the value of g on the surface of earth : For , << we can expand the RHS of Eq. ( .
) : ( For << , using binomial expression, R E ( ) ≅ ( . ) Equation ( . ) thus tells us that for small heights h above the value of g decreases by a factor ( / ). Now, consider a point mass m at a depth d below the surface of the earth (Fig.