The value of the acceleration due to gravity is $g_{1}$ at a height $h = \frac{R}{2}$ ($R$ = radius of the earth) from the surface of the earth. It is again equal to $g_{1}$ at a depth $d$ below the surface of the earth. The ratio $\left(\frac{d}{R}\right)$ equals

Consider Earth to be a sphere of radius $R_e$ rotating about its own axis with angular speed $\omega$. If $g_{E}$ and $g_{P}$ are the accelerations due to gravity at the equator and the poles respectively,then $(g_{P}-g_{E})$ is given by $\left[\cos (0^{\circ})=\sin (\frac{\pi}{2})=1, \sin (0^{\circ})=\cos (\frac{\pi}{2})=0\right]$

For a sphere of mass $M$ and radius $R$,if the gravitational forces at distances $r_1$ and $r_2$ from the center are $F_1$ and $F_2$ respectively,then:

The acceleration due to gravity at height $h$ above the earth if $h \ll R$ (radius of earth) is given by

Obtain the general equation of gravitational force at a distance $r$ from the centre of the Earth and derive the equation for acceleration due to gravity on the surface of the Earth.

If both the mass and the radius of the earth decrease by $1\%$,the value of the acceleration due to gravity will