Question

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

Answer 1

We may recall the formula that depicts the change in gravity-induced acceleration with the depth. You might use two in the place of the depth because we're talking about the point halfway to the center of the planet. By multiplying both sides by the body's mass, you may compare the weight of the body at the surface and depth, and so get the answer.

Complete step-by-step solution:
Here in the problem, given the weight of a body on the Earth's surface, We have to find the weight of the identical body at a depth of halfway below the Earth's center, assuming the density of the Earth is consistent.
Given,
$m = {\text{ }}500{\text{ }}kg$
$g = {\text{ }}9.8m/{s^2}$
$R = {\text{ }}6400{\text{ }}km$
$D{\text{ }} = {\text{ }}1000{\text{ }}km$
${R_1} = \left( {R - d} \right)$
$\Rightarrow {R_1} = \left( {6400 - 1000} \right)$
$\Rightarrow {R_1} = 5400km$
$\Rightarrow {R_1} = 5.4 \times {10^6}m$
${M_2} = \dfrac{M}{{\dfrac{4}{3}\pi {R^3}}} \times 4\pi {R_1}$
$\Rightarrow {M_2} = m\left( {\dfrac{{5.4}}{{6.4}} \times \dfrac{{{{10}^6}}}{{{{10}^6}}}} \right)$
$\Rightarrow {M_2} = 6m$
Now for $g$ we use,
$g = \dfrac{{GM}}{{{R^2}}}$ (eq.1)
So, for${g_1}$ ,
${g_1} = \dfrac{{G{M_1}}}{{{R_1}^2}}$ (eq.2)
From eq. 1 & eq. 2
$\dfrac{{{g_1}}}{g} = \left( {\dfrac{{{M_1}}}{{{R_2}^2}} \times \dfrac{{{R^2}}}{M}} \right)$
$\Rightarrow {g_1} = \left( {\dfrac{{{M_1}}}{M}} \right){\left( {\dfrac{R}{{{R_1}}}} \right)^2}g$
$\Rightarrow {g_1} = \left( {\dfrac{{0.6}}{M}M} \right) \times {\left( {\dfrac{{6.4 \times {{10}^6}}}{{5.4 \times {{10}^6}}}} \right)^2} \times 9.8$
$\Rightarrow {g_1} = \left( {0.6} \right) \times 9.8 \times {\left( {\dfrac{{64}}{{54}}} \right)^2}$
$\Rightarrow {g_1} = 8.26m/{s^2}$
Weight on the surface of the earth,
$w = mg$
$w = 500 \times 9.8$
$w = 4900N$
Weight of the mass into the mine,
{w_1} = m{g_1}$$$${w_1} = 500 \times 8.26
${w_1} = 4130N$
Hence, a decrease in weight
$\Delta w = w - {w_1}$
$\Delta w = \left( {4900 - 4130} \right)$
$\Delta w = 770N$

Note: As you fall a hole to the center of the spherical planet, the value of$\;g$ , and hence your weight, diminishes linearly. You are weightless at the center because the planet's mass pulls evenly in all directions.
The density of the Earth varies, and it is not solid altogether.
Newton's second law of motion, or $F = ma$ (force = mass x acceleration), defines the weight of an object on Earth's surface as the downward force on that object. Other elements, such as the rotation of the Earth, contribute to the total gravitational acceleration and, as a result, impact the weight of the object.