Question

The density of a solid spherical planet of radius R is given as $\rho = \rho_0 r$, where $\rho_0$ = constant and r is distance measured from centre of planet. The acceleration due to gravity of this planet is half of maximum value at distance x from centre and also at a distance y from the centre. The value of x + y is $R \sqrt{\frac{\alpha}{\beta}}$. Here $\alpha$ and $\beta$ are single digit integer. Find the value of $\alpha - \beta$?

Answer 1

7

Answer 2

The density of the solid spherical planet of radius R is given by $\rho = \rho_0 r$, where $\rho_0$ is a constant and r is the distance from the center.

Step 1: Calculate the mass M(r) enclosed within a sphere of radius r.

Consider a spherical shell of radius $r'$ and thickness $dr'$. The volume of this shell is $dV = 4\pi r'^2 dr'$. The mass of this shell is $dm = \rho(r') dV = (\rho_0 r') (4\pi r'^2 dr') = 4\pi \rho_0 r'^3 dr'$. The mass enclosed within a sphere of radius r ($r \le R$) is $M(r) = \int_0^r dm = \int_0^r 4\pi \rho_0 r'^3 dr'$. $M(r) = 4\pi \rho_0 \left[\frac{r'^4}{4}\right]_0^r = \pi \rho_0 r^4$.

Step 2: Calculate the acceleration due to gravity g(r) at a distance r from the center.

For $r \le R$, the gravitational acceleration is given by $g(r) = \frac{G M(r)}{r^2}$. $g(r) = \frac{G (\pi \rho_0 r^4)}{r^2} = G \pi \rho_0 r^2$.

For $r > R$, the gravitational acceleration is given by $g(r) = \frac{G M_{total}}{r^2}$, where $M_{total}$ is the total mass of the planet. $M_{total} = M(R) = \pi \rho_0 R^4$. So, for $r > R$, $g(r) = \frac{G \pi \rho_0 R^4}{r^2}$.

Step 3: Find the maximum value of g(r).

For $0 \le r \le R$, $g(r) = G \pi \rho_0 r^2$. This function increases with r. The maximum value in this range is at $r=R$, which is $g(R) = G \pi \rho_0 R^2$. For $r > R$, $g(r) = \frac{G \pi \rho_0 R^4}{r^2}$. This function decreases as r increases. The maximum value of g(r) occurs at $r=R$. $g_{max} = g(R) = G \pi \rho_0 R^2$.

Step 4: Find the distances x and y where $g(x) = g(y) = \frac{1}{2} g_{max}$.

$\frac{1}{2} g_{max} = \frac{1}{2} (G \pi \rho_0 R^2)$. We need to find r such that $g(r) = \frac{1}{2} G \pi \rho_0 R^2$.

Case 1: $r \le R$. $g(r) = G \pi \rho_0 r^2$. $G \pi \rho_0 r^2 = \frac{1}{2} G \pi \rho_0 R^2$. $r^2 = \frac{1}{2} R^2$. $r = \sqrt{\frac{1}{2}} R = \frac{R}{\sqrt{2}}$. Since $\frac{1}{\sqrt{2}} < 1$, this distance is within the planet ($r \le R$). Let $x = \frac{R}{\sqrt{2}}$.

Case 2: $r > R$. $g(r) = \frac{G \pi \rho_0 R^4}{r^2}$. $\frac{G \pi \rho_0 R^4}{r^2} = \frac{1}{2} G \pi \rho_0 R^2$. $\frac{R^2}{r^2} = \frac{1}{2}$. $r^2 = 2R^2$. $r = \sqrt{2} R$. Since $\sqrt{2} > 1$, this distance is outside the planet ($r > R$). Let $y = \sqrt{2} R$.

The two distances where the gravity is half of the maximum value are $x = \frac{R}{\sqrt{2}}$ and $y = \sqrt{2} R$.

Step 5: Calculate x + y.

$x + y = \frac{R}{\sqrt{2}} + \sqrt{2} R = R \left(\frac{1}{\sqrt{2}} + \sqrt{2}\right) = R \left(\frac{1 + (\sqrt{2})^2}{\sqrt{2}}\right) = R \left(\frac{1 + 2}{\sqrt{2}}\right) = R \frac{3}{\sqrt{2}}$.

Step 6: Express $x+y$ in the form $R \sqrt{\frac{\alpha}{\beta}}$ and find $\alpha$ and $\beta$.

We have $x + y = R \frac{3}{\sqrt{2}}$. We can write $\frac{3}{\sqrt{2}}$ as $\sqrt{\left(\frac{3}{\sqrt{2}}\right)^2} = \sqrt{\frac{9}{2}}$. So, $x + y = R \sqrt{\frac{9}{2}}$. Comparing this with $R \sqrt{\frac{\alpha}{\beta}}$, we get $\frac{\alpha}{\beta} = \frac{9}{2}$. We are given that $\alpha$ and $\beta$ are single-digit integers. The only pair of single-digit integers $(\alpha, \beta)$ such that $\frac{\alpha}{\beta} = \frac{9}{2}$ is $(\alpha, \beta) = (9, 2)$.

Step 7: Find the value of $\alpha - \beta$.

$\alpha = 9$ and $\beta = 2$. $\alpha - \beta = 9 - 2 = 7$.