The value of $g$ at a particular point is $9.8\,m/s^2$. Suppose the Earth suddenly shrinks uniformly to half its present size without losing any mass. The value of $g$ at the same point (assuming that the distance of the point from the centre of the Earth does not shrink) will now be ......... $m/s^2$.

What is the time period of a seconds pendulum on a planet whose mass and radius are twice that of the Earth?

$A$ spherical planet far out in space has a mass $M_0$ and diameter $D_0$. $A$ particle of mass $m$ falling freely near the surface of this planet will experience an acceleration due to gravity which is equal to

The weight of a body on the earth is $400\,N$. Then weight of the body when taken to a depth half of the radius of the earth will be ............ $N$.

Assume there are two identical simple pendulum clocks. Clock-$1$ is placed on the Earth and Clock-$2$ is placed on a space station located at a height $h$ above the Earth's surface. Clock-$1$ and Clock-$2$ operate with time periods of $4\,s$ and $6\,s$ respectively. Then the value of $h$ is $....\,km$ (consider the radius of the Earth $R_E = 6400\,km$ and $g$ on Earth $10\,m/s^2$).

$A$ simple pendulum has a time period $T_1$ when on the earth's surface and $T_2$ when taken to a height $R$ above the earth's surface,where $R$ is the radius of the earth. The value of $T_2/T_1$ is