Question

What the decrease in weight ofa body of mass 600 kg when it is taken in a mine of depth 5000 m ? [Radius of earth a 6400 km, g = 9.8 $m/s^2$]

Answer 1

Given that $m=600$ kg, $d=5000m$,
$R=6400km$ $=6.4\times {10}^{6}m$
Weight of the body on the surface of the Earth $=600\times 9.8=5880N$
At depth $d$, gravitation acceleration is
$g_d=g[1-\frac{d}{R}]$
$\therefore g_d=g[1-\frac{5}{6400}]=9.8\times 0.999$
$\therefore g_d=9.7902m/s^2$
Weight on surface $=mg$
$=600\times 9.8$
$\therefore$ Weigh on surface$=5880N$
Weight of the body at depth$=m{g}_d$
$=600\times 9.7902$
$=5874N$
$\therefore$ Decrease in weight $mg-m{g}_d$
$=5880N-5874N$
$\therefore$ Decrease in weight$=6N$