Q.What is the decrease in weight of a body of mass 600 kg when it is taken in a mine of depth 5000 m? [Radius of earth = 6400 km, g = 9.8 m/s²]

Acceleration due to gravity decreases linearly with depth below Earth's surface: $g_d=g\left(1-\dfrac{d}{R}\right)$; the decrease in weight is $mg\dfrac{d}{R}$.

Inside the Earth (treating it as a uniform sphere), the acceleration due to gravity at depth $d$ below the surface is

$g_d=g\left(1-\frac{d}{R}\right)$

where $R$ is the radius of the Earth.

The decrease in $g$ is

$\Delta g=g-g_d=g\frac{d}{R}$

So the decrease in weight of a body of mass $m$ is

$\Delta W=m\,\Delta g=mg\frac{d}{R}$

Substituting $m=600$ kg, $g=9.8\ \text{m/s}^2$, $d=5000$ m, $R=6400\ \text{km}=6.4\times10^6$ m: …