Two planets have the same density but different radii. The acceleration due to gravity would be ........

The variation of acceleration due to gravity $g$ with distance $d$ from the centre of the earth is best represented by ($R =$ Earth's radius)

$A$ simple pendulum is oscillating with frequency $F$ on the surface of the earth. It is taken to a depth $R/3$ below the surface of the earth ($R =$ radius of earth). The frequency of oscillation at depth $R/3$ is

The ratio of the accelerations due to gravity at heights $1280 \ km$ and $3200 \ km$ above the surface of the earth is (Radius of the earth $= 6400 \ km$)

Earth is assumed to be a sphere of radius $R$. If $g_{\phi}$ is the value of effective acceleration due to gravity at latitude $30^{\circ}$ and $g$ is the value at the equator,then the value of $|g - g_{\phi}|$ is ($\omega$ is the angular velocity of rotation of the Earth,$\cos 30^{\circ} = \frac{\sqrt{3}}{2}$).

An object is taken to a height above the surface of the Earth at a distance of $\frac{5}{4} R$ from the center of the Earth,where the radius of the Earth is $R = 6400 \, km$. The percentage decrease in the weight of the object will be $....\%$.