$A$ body weighs $45 \ N$ on the surface of the earth. The gravitational force on a body due to the earth at a height equal to half the radius of the earth will be: (in $N$)

At what depth below the surface of the earth,will the acceleration due to gravity $g$ be half of its value at $1600 \, km$ above the surface of the earth?

Assertion $(A)$: $A$ particle of mass $m$ dropped into a hole made along the diameter of the Earth from one end to the other possesses simple harmonic motion.
Reason $(R)$: Gravitational force between any two particles is inversely proportional to the square of the distance between them.

Write the equation of gravitational acceleration which is used for any height $h$ from the surface of the Earth.

If the radius of the earth contracts by $2\%$ and its mass remains the same,then the weight of a body at the earth's surface:

The magnitudes of the gravitational field at distances $r_1$ and $r_2$ from the center of a uniform sphere of radius $R$ and mass $M$ are $F_1$ and $F_2$ respectively. Then: