# Question #484de

Feb 18, 2017

$93.2 N$.

If we rely on the posted correct answer, depth should be $3200 k m$

#### Explanation:

Calculate gravitational force between body and mass of part sphere of earth which lies below 320 km. Force due to outer spherical shell tends to cancel out.

Mass of earth $M = \frac{4}{3} \pi {R}^{3} \rho$
where $R$ is radius of earth $\rho$ is density of earth.
mass of earth sphere below $320 k m$ below earth's surface
${M}_{d} = \frac{4}{3} \pi {\left(R - 3.2 \times {10}^{5}\right)}^{3} \rho$ .....(1)
Acceleration due to gravity at the location of body${g}_{d} = G {M}_{s} / {\left(R - 3.2 \times {10}^{5}\right)}^{2}$
Inserting value from (1) we get
${g}_{d} = G \frac{\frac{4}{3} \pi {\left(R - 3.2 \times {10}^{5}\right)}^{3} \rho}{R - 3.2 \times {10}^{5}} ^ 2$
$\implies {g}_{d} = \frac{4}{3} G \pi \rho \left(R - 3.2 \times {10}^{5}\right)$ ......(2)
We know that acceleration due to gravity at the surface in terms of equation (2)
$g = 9.81 m {s}^{-} 2 = \frac{4}{3} G \pi \rho R$ .....(3)

Dividing (2) by (3) we get
${g}_{d} / 9.81 = \frac{\frac{4}{3} G \pi \rho \left(R - 3.2 \times {10}^{5}\right)}{\frac{4}{3} G \pi \rho R}$
$\implies {g}_{d} = 9.81 \frac{\left(R - 3.2 \times {10}^{5}\right)}{R}$

Taking average radius of earth as $6400 k m$, we get
${g}_{d} = 9.81 \frac{\left(6.4 \times {10}^{6} - 3.2 \times {10}^{5}\right)}{6.4 \times {10}^{6}}$
${g}_{d} = 9.32 m {s}^{-} 1$
Weight of $10 k g$ body $= 10 \times 9.32 = 93.2 N$