The value of the acceleration due to gravity is g₁ at a height h = $${R \over 2}$$ (R = radius of the earth) from the surface of the earth. It is again equal to g₁ at a depth d below the surface of the earth. The ratio $$\left( {{d \over R}} \right)$$ equals :

A body is moving in a low circular orbit about a planet of mass M and radius R. The radius of the orbit can be taken to be R itself. Then the ratio of the speed of this body in the orbit to the escape velocity from the planet is:

On the x-axis and at a distance x from the origin, the gravitational field due a mass distribution is given by $${{Ax} \over {{{\left( {{x^2} + {a^2}} \right)}^{3/2}}}}$$ in the x-direction. The magnitude of gravitational potential on the x-axis at a distance x, taking its value to be zero at infinity, is:

The mass density of a planet of radius R varies with the distance r from its centre as
$$\rho $$(r) = $${\rho _0}\left( {1 - {{{r^2}} \over {{R^2}}}} \right)$$.
Then the gravitational field is maximum at :