The figure shows not to scale, a cross section through theinterior of Earth. Rather than being uniform throughout, Earth isdivided into three zones: an outercrust,amantle,and an innercore.The dimensions of these zones and the masses contained within them are shown on the figure. Earth has a total mass of$5.98\times {10}^{24}\text{kg}$and a radius of$6370\text{km}$. Ignore rotation and assumethat Earth is spherical. (a) Calculateat the surface. (b) Suppose that a bore hole (theMohole) is driven to the crust–mantle interface at a depth of$25.0\text{km}$; what would be the value ofat the bottom of the hole? (c) Suppose that Earth were a uniform spherewith the same total mass and size. What would be the value ofat a depth of$25.0\text{km}$? (Precise measurements ofare sensitive probes of the interior structure of Earth, although results can be becloud by local variations in mass distribution.)

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

ii) Radius of the earth is$6370\text{\hspace{0.33em}km}$

iii) Mass of the core is$1.93\times {10}^{24}\text{\hspace{0.33em}kg}$

iv) Mass of the crust is $3.94\times {10}^{22}\text{\hspace{0.33em}kg}$

v) Mass of Mantle is$4.01\times {10}^{24}\text{\hspace{0.33em}kg}$

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, $g=\frac{GM}{r^2}$ …(i)

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

$\begin{array}{c}g=\frac{6.67\times {10}^{-11}{\text{\hspace{0.17em}\hspace{0.17em}Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\times 5.98\times {10}^{24}\text{\hspace{0.17em}\hspace{0.17em}kg}}{{\left(6370\times {10}^3\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^2}\\ =9.83{\text{\hspace{0.33em}m/s}}^{\text{2}}\end{array}$

The gravitational acceleration at the Earth’s surface is$9.83{\text{\hspace{0.33em}m/s}}^{\text{2}}$

At bottom of the hole total mass is,

$\begin{array}{c}M=\left(1.93\times {10}^{24}\text{\hspace{0.17em}\hspace{0.17em}kg}+4.01\times {10}^{24}\text{\hspace{0.17em}\hspace{0.17em}kg}\right)\\ =5.94\times {10}^{24}\text{\hspace{0.33em}kg}\end{array}$

And radius$r=6.345\times {10}^6\text{m}$

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

$\begin{array}{c}g=\frac{6.67\times {10}^{-11}{\text{\hspace{0.17em}\hspace{0.17em}Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\times 5.94\times {10}^{24}\text{\hspace{0.17em}\hspace{0.17em}kg}}{{\left(6.345\times {10}^6\right)}^2}\\ =9.84{\text{\hspace{0.33em}m/s}}^{\text{2}}\end{array}$

The gravitational acceleration at the bottom of the hole is$9.84{\text{\hspace{0.33em}m/s}}^{\text{2}}$

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

$R=6.345\times {10}^6\text{\hspace{0.17em}\hspace{0.17em}m}$

$\begin{array}{c}M=\left(\frac{R^3}{R_e^3}\right)M_e\\ =\left(\frac{{\left(6.345\times {10}^6\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^3}{{\left(6.370\times {10}^6\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^3}\right)\cdot \left(5.98\times {10}^{24}\text{\hspace{0.17em}\hspace{0.17em}kg}\right)\\ =5.91\times {10}^{24}\text{\hspace{0.33em}kg}\end{array}$

So, acceleration is given as:

$\begin{array}{c}g=\frac{GM}{r^2}\\ =\frac{6.67\times {10}^{-11}{\text{\hspace{0.17em}\hspace{0.17em}Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\times 5.91\times {10}^{24}\text{\hspace{0.17em}\hspace{0.17em}kg}}{{\left(6.345\times {10}^6\text{\hspace{0.17em}\hspace{0.17em}m}\right)}^2}\\ =9.79{\text{\hspace{0.33em}m/s}}^{\text{2}}\end{array}$

The gravitational acceleration at the bottom of the hole is$9.79{\text{\hspace{0.33em}m/s}}^{\text{2}}$