Question

Two spherical planets P and Q have the same uniform density $\rho$, masses ${M_P}$ and ${M_Q}$, and surface area A and 4A, respectively. A spherical planet R also has uniform density $\rho$ and its mass is $\left( {{M_P} + {M_Q}} \right)$. The escape velocities from the planets P, Q and R, are ${V_P}$, ${V_Q}$ and ${V_R}$, respectively. Then
This question has multiple correct options.
${\text{A}}{\text{. }}\dfrac{{{V_P}}}{{{V_Q}}} = \dfrac{1}{2} \\\ {\text{B}}{\text{. }}{V_R} > {V_Q} > {V_P} \\\ {\text{C}}{\text{. }}\dfrac{{{V_R}}}{{{V_P}}} = 3 \\\ {\text{D}}{\text{. None of these}} \\\$

Answer 1

We know the expression for the escape velocity for a planet of certain mass and radius. From the given information, we need to find out the relations between the masses and also the radius of the three given planets. Then inserting them into the expression for escape velocity, we can compare equations to check which of the given options are correct.
Formula used:
The expression for the escape velocity of a planet is given as
$V = \sqrt {\dfrac{{2GM}}{R}}$

Complete answer:
First consider the spherical planet P. Its mass is given as ${M_P}$ and its density is $\rho$. Its surface area is given as A. If ${R_P}$ is the radius of planet P then we can write the surface area as
${A_P} = 4\pi R_P^2 \\\ A = 4\pi R_P^2 \\\ {R_P} = \sqrt {\dfrac{A}{{4\pi }}} = r(let) \\\$
Therefore, the volume of the planet P is given as
${v_P} = \dfrac{4}{3}\pi R_P^3 = \dfrac{4}{3}\pi {\left( {\dfrac{A}{{4\pi }}} \right)^{\dfrac{3}{2}}}$
Therefore, mass can be written as
${M_P} = \rho {v_P} = \rho \dfrac{4}{3}\pi {\left( {\dfrac{A}{{4\pi }}} \right)^{\dfrac{3}{2}}} = m(let)$
Similarly, for the spherical planet Q, we are given its mass as ${M_Q}$ and its density is $\rho$. Its surface area is given as 4A. If ${R_Q}$ is the radius of planet P then we can write the surface area as
${A_Q} = 4\pi R_Q^2 \\\ 4A = 4\pi R_Q^2 \\\ {R_Q} = \sqrt {\dfrac{{4A}}{{4\pi }}} = 2r \\\$
Therefore, the volume of the planet Q is given as
${v_Q} = \dfrac{4}{3}\pi R_Q^3 = \dfrac{4}{3}\pi {\left( {\dfrac{{4A}}{{4\pi }}} \right)^{\dfrac{3}{2}}}$
Therefore, mass can be written as
${M_Q} = \rho {v_P} = \rho \dfrac{4}{3}\pi {\left( {\dfrac{{4A}}{{4\pi }}} \right)^{\dfrac{3}{2}}} = 8m$
For planet R, we are given its mass to be ${M_R} = \left( {{M_P} + {M_Q}} \right) = m + 8m = 9m$
${M_R} = \left( {{M_P} + {M_Q}} \right) \\\ \rho {V_R} = \rho {V_P} + \rho {V_Q} \\\ R_R^3 = R_P^3 + R_Q^3 = {r^3} + {\left( {2r} \right)^3} = 9{r^3} \\\ {R_R} = {9^{\dfrac{1}{3}}}r \\\$
Now finally we can write the expressions for the escape velocities for the planets in the following way.
${V_P} = \sqrt {\dfrac{{2G{M_P}}}{{{R_P}}}} = \sqrt {\dfrac{{2Gm}}{r}} \\\ {V_Q} = \sqrt {\dfrac{{2G{M_Q}}}{{{R_Q}}}} = \sqrt {\dfrac{{2G \times 8m}}{{2r}}} = 2{V_P} \\\ {V_R} = \sqrt {\dfrac{{2G{M_R}}}{{{R_R}}}} = \sqrt {\dfrac{{2G \times 9m}}{{{9^{\dfrac{1}{3}}}r}}} = {9^{\dfrac{1}{3}}}{V_P} = 2.08{V_P} \\\$
Therefore, we can write the following expressions based on the above values.
$\dfrac{{{V_P}}}{{{V_Q}}} = \dfrac{1}{2} \\\ {V_R} > {V_Q} > {V_P} \\\ \dfrac{{{V_R}}}{{{V_P}}} = 2.08 \\\$

Therefore, we can say that only options A and B are correct.

Note:
It should be noted that with increase in the mass of a planet, the escape velocity of the planet increases while with increase in radius of the planet, its escape velocity decreases. Though the density of the given planets is the same, the increase in mass dominates over the increase in radius and escape velocity increases from P to R.