91Ó°ÊÓ

• The density of the planet at the center is $15.0×{10}^3{\text{ k²µ/³¾}}^{\text{3}}$.
• The density of the planet at the surface is $2.0×{10}^3{\text{ k²µ/³¾}}^{\text{3}}$.

The law states that every object mainly exerts some gravitational force on a thing that is directly proportional to the mass of the object and inversely proportional to the square of the object's radius.

The product of the gravitational constant and the mass of the planet divided by the square of the radius gives the acceleration due to the gravity of the planet's surface.

From Newton's law of universal gravitation, the acceleration due to gravity can be expressed as-

$g=\frac{GM}{r^2}$ … i)

Here, g is the acceleration due to gravity, G is the gravitational constant, M is the planet's mass, and r is the radius of the earth.

A model is required for describing the density as a function of the radius, which is-

$\begin{array}{l}r=0,\text{ }\mathrm{ÒÏ}\left(0\right)={\mathrm{ÒÏ}}_{center}=15×{10}^3{\text{ k²µ/³¾}}^{\text{3}}\\ r=6.371×{10}^6\text{″¾,}\mathrm{ÒÏ}(6.371×{10}^3\text{m})={\mathrm{ÒÏ}}_{surface}=2×{10}^3{\text{ k²µ/³¾}}^{\text{3}}\end{array}$…i¾±)

Here, the linear model equation is $y=mx+b$ where x and y are the variables, m is the slope, and c is the constant. Hence, y is replaced by $\mathrm{ÒÏ}\left(r\right)$, and b is replaced by ${\mathrm{ÒÏ}}_{center}$. Furthermore, m is replaced by

$\begin{array}{c}m=\frac{y_2−y_1}{x_2−x_1}\\ =\frac{{\mathrm{ÒÏ}}_{surface}−{\mathrm{ÒÏ}}_{center}}{r_{earth}−0}\end{array}$

Hence, putting the values in equation ii), we get a linear model of

$\mathrm{ÒÏ}\left(r\right)={\mathrm{ÒÏ}}_{center}+\frac{{\mathrm{ÒÏ}}_{surface}−{\mathrm{ÒÏ}}_{center}}{r_{earth}}r$

As the goal is to find the acceleration due to gravity, the planet's total mass is necessary to identify. Hence, a finite element and a spherical coordinate must be determined. The differential element’s volume will be

$dv=r^2\sin \mathrm{θ}d\mathrm{ϕ}d\mathrm{θ}dr$…. iii)

The total mass of the planet is expressed as-

$M=∫p\left(r\right)dV$ …. iv)

Replacing the function$dV$in equation iv), we get-

$\begin{array}{c}M={∫}_0^{2\mathrm{Ï€}}d\mathrm{φ}{∫}_0^{\mathrm{Ï€}}\sin \mathrm{θ}d\mathrm{θ}{∫}_0^{r_{earth}}\left(r^2{\mathrm{ÒÏ}}_{center}+\frac{{\mathrm{ÒÏ}}_{surface}−{\mathrm{ÒÏ}}_{center}}{r_{earth}}r\right)\\ M=4\mathrm{Ï€}\left(\frac{r_{earth}^3{\mathrm{ÒÏ}}_{center}}{3}+\frac{r_{earth}^4}{4}\frac{{\mathrm{ÒÏ}}_{surface}−{\mathrm{ÒÏ}}_{center}}{r_{earth}}\right)\end{array}$

Simplifying the expression, we get-

$\begin{array}{l}M=\frac{4\mathrm{Ï€}{\mathrm{ÒÏ}}_{center}{r}_{earth}^3}{3}+\mathrm{Ï€}{r}_{earth}^3({\mathrm{ÒÏ}}_{surface}−{\mathrm{ÒÏ}}_{center})\\ M=\mathrm{Ï€}{r}_{earth}^3\left(\frac{4}{3}{\mathrm{ÒÏ}}_{center}+{\mathrm{ÒÏ}}_{surface}−{\mathrm{ÒÏ}}_{center}\right)\\ M=\mathrm{Ï€}{r}_{earth}^3\left[{\mathrm{ÒÏ}}_{surface}+\frac{1}{3}{\mathrm{ÒÏ}}_{center}\right]\end{array}$

Substituting the values in the above equation, we get-

$\begin{array}{l}M=\mathrm{Ï€}{(6.371×{10}^6\text{″¾})}^3\left(\frac{1}{3}15×{10}^3{\text{ k²µ/³¾}}^{\text{3}}+2×{10}^3{\text{ k²µ/³¾}}^{\text{3}}\right)\\ M=5.68×{10}^{24}\text{ k²µ}\end{array}$… v)

Using the value of equation v) in equation i), we get-

$\begin{array}{l}g=\frac{(6.67×{10}^{−11}{\text{″¾}}^{\text{3}}{\text{/kgs}}^{\text{2}})(5.68×{10}^{24}\text{ k²µ})}{6.371×{10}^6\text{″¾}}\\ g=9.34{\text{″¾/²õ}}^{\text{2}}\end{array}$

Thus, the acceleration due to gravity at the planet's surface is $9.34{\text{″¾/²õ}}^{\text{2}}$.