If acceleration due to gravity at distance $d[ < R]$ from the centre of earth is $\beta$, then its value at distance $d$ above the surface of earth will be [where $R$ is radius of earth]
$\frac{\beta R^2}{(R+d)^3}$
$\frac{\beta R}{2 d}$
$\frac{\beta d}{(R+d)^2}$
$\frac{\beta R^3}{d(R+d)^2}$
Spot the wrong statement :The acceleration due to gravity $‘g’$ decreases if
A pendulum clock is set to give correct time at the sea level. This clock is moved to hill station at an altitude of $2500\, m$ above the sea level. In order to keep correct time of the hill station, the length of the pendulum
Acceleration due to gravity at surface of a planet is equal to that at surface of earth and density is $1.5$ times that of earth. If radius of earth is $R$, radius of planet is .................
The moon's radius is $1/4$ that of the earth and its mass is $1/80$ times that of the earth. If $g$ represents the acceleration due to gravity on the surface of the earth, that on the surface of the moon is