Question

if a body weighing 40 kg is taken inside the earth to a depth of 143 radius of the earth the weight of the body at that point is

Answer 1

Explaination:

$f = g \times \frac{m1 \times m2}{ {d}^{2} }$

$= 6.67 \times {10}^{ - 11} \times \frac{40 \times 5.97 \times {10}^{24} }{ {143}^{2} }$

$= {10}^{13} \times \frac{6.67 \times 40 \times 5.97}{ {143}^{2} }$

$= 0.077 \times {10}^{13}$

$= 7.7 \times {10}^{11}$