Question

What is the decrease in weight of a body of mass $500Kg$ when it is taken into a mine of depth $1000Km$?
(Radius of earth ${\text{R= 6400 km, g = 9.8 m/s}}^2$)

Answer 1

$m=500kg,g=9.8m/s^2$

$R=6500km,d=1000km$

Ref. image

$R_1=(R-d)$

$=(6400-1000)$

$=5400km$

$R_1=5.4\times {10}^{6}m$

We can't use approximation as d is comparable with $R_3$

$M_2=\frac{M}{\frac{4}{3}\pi R^3}\times 4\pi R_1$

$m(\frac{5.4}{6.4}\times \frac{{10}^6}{{10}^6})$

$M_2=06m$

$g=\frac{GM}{R^2}$ ____(1)

$g_1=\frac{G{M}^1}{R_1^2}$ ___(2)

from eq (1) & (2)

$\frac{g_1}{g}=(\frac{M_1}{R_2^2}\times \frac{R^2}{M})$

$g_1=\left(\frac{M_1}{M}\right){\left(\frac{R}{R_1}\right)}^{2}g$

$=\left(\frac{0.6}{M}M\right)\times {\left(\frac{6.4\times {10}^6}{5.4\times {10}^6}\right)}^2\times 9.8$

$=\left(0.6\right)\times 9.8\times {\left(\frac{64}{54}\right)}^2$

$g_1=8.26m/s^2$

Weight on surface of the earth

$w=mg$

$=500\times 9.8$

$w=4900N$

weight of the mass into the mine $\Rightarrow$

$w_1=m{g}_1$

$=500\times 8.26$

$w_1=4130N$

Hence, decrease in weight

$\Delta w=w-w_1$

$=(4900-4130)$

$\Delta w=770N$