91Ó°ÊÓ

i) Mass of the earth is$5.98×{10}^{24}\text{kg}$

ii) Radius of the earth is$6370\text{ k³¾}$

iii) Mass of the core is$1.93×{10}^{24}\text{ k²µ}$

iv) Mass of the crust is $3.94×{10}^{22}\text{ k²µ}$

v) Mass of Mantle is$4.01×{10}^{24}\text{ k²µ}$

Using the formula for gravitational acceleration in which acceleration depends on mass of the earth and square of distance from surface of the earth, we can find the value of acceleration at different points.

Formula:

Gravitational acceleration, $g=\frac{GM}{r^2}$ …(¾±)

Gravitational acceleration on the surface of Earth’s surface using equation (i), we get

$\begin{array}{c}g=\frac{6.67×{10}^{−11}{\text{ â¶Ä‰N³¾}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{ }×5.98×{10}^{24}\text{ â¶Ä‰k²µ}}{{\left(6370×{10}^3\text{ â¶Ä³¾}\right)}^2}\\ =9.83{\text{ m/²õ}}^{\text{2}}\end{array}$

The gravitational acceleration at the Earth’s surface is$9.83{\text{ m/²õ}}^{\text{2}}$

At bottom of the hole total mass is,

$\begin{array}{c}M=\left(1.93×{10}^{24}\text{ â¶Ä‰k²µ}+4.01×{10}^{24}\text{ â¶Ä‰k²µ}\right)\\ =5.94×{10}^{24}\text{ k²µ}\end{array}$

And radius$r=6.345×{10}^6\text{m}$

Using equation (i), the acceleration at the bottom of the hole is:

$\begin{array}{c}g=\frac{6.67×{10}^{−11}{\text{ â¶Ä‰N³¾}}^{\text{2}}{\text{/kg}}^{\text{2}}×5.94×{10}^{24}\text{ â¶Ä‰k²µ}}{{\left(6.345×{10}^6\right)}^2}\\ =9.84{\text{ m/²õ}}^{\text{2}}\end{array}$

The gravitational acceleration at the bottom of the hole is$9.84{\text{ m/²õ}}^{\text{2}}$

Mass is uniformly distributed so mass at 25 km, therefore we have radius

$R=6.345×{10}^6\text{ â¶Ä³¾}$

$\begin{array}{c}M=\left(\frac{R^3}{R_e^3}\right)M_e\\ =\left(\frac{{\left(6.345×{10}^6\text{ â¶Ä³¾}\right)}^3}{{\left(6.370×{10}^6\text{ â¶Ä³¾}\right)}^3}\right)â‹\dots \left(5.98×{10}^{24}\text{ â¶Ä‰k²µ}\right)\\ =5.91×{10}^{24}\text{ k²µ}\end{array}$

So, acceleration is given as:

$\begin{array}{c}g=\frac{GM}{r^2}\\ =\frac{6.67×{10}^{−11}{\text{ â¶Ä‰N³¾}}^{\text{2}}{\text{/kg}}^{\text{2}}×5.91×{10}^{24}\text{ â¶Ä‰k²µ}}{{\left(6.345×{10}^6\text{ â¶Ä³¾}\right)}^2}\\ =9.79{\text{ m/²õ}}^{\text{2}}\end{array}$

The gravitational acceleration at the bottom of the hole is$9.79{\text{ m/²õ}}^{\text{2}}$