JEE Main 2023 — Physics Mechanics

The formula to calculate the mass $(M)$(M) of any planet of radius $R$R and density $\rho$ρ, considering it to be a sphere, is given by

$$M=\frac{4}{3}\pi {R}^{3}\rho .....................(1)$$M=34​πR3ρ.....................(1)

The escape velocity $({v}_{A})$(vA​) for planet A can be expressed as

$$\begin{matrix}{V}_{A} & = & \sqrt{\frac{2G{M}_{A}}{{R}_{A}}} \\ & = & \sqrt{\frac{2G{\rho }_{A}\frac{4}{3}\pi {{R}_{A}}^{3}}{{R}_{A}}} \\ & = & C\sqrt{{\rho }_{A}.}{R}_{A}............................(2)\end{matrix}$$VA​​===​RA​2GMA​​​RA​2GρA​34​πRA​3​​CρA​.​RA​............................(2)​

where, $G$G is Gravitational constant and $C=\sqrt{\frac{8\pi G}{3}}$C=38πG​​ is a constant.

Similarly, the escape velocity $({v}_{B})$(vB​) of planet B can be expressed as

$${v}_{B}=C\sqrt{{\rho }_{B}}{R}_{B}......................(3)$$vB​=CρB​​RB​......................(3)

Divide equation (2) by equation (3) and solve with the known parameters to obtain the ratio of the densities of the planets.

$$\begin{matrix}\frac{{V}_{A}}{{V}_{B}} & = & \frac{C\sqrt{{\rho }_{A}}{R}_{A}}{C\sqrt{{\rho }_{B}}{R}_{B}} \\ \sqrt{\frac{{\rho }_{A}}{{\rho }_{B}}} & = & \frac{{V}_{A}{R}_{B}}{{V}_{B}{R}_{A}} \\ \frac{{\rho }_{A}}{{\rho }_{B}} & = & {(\frac{{V}_{A}{R}_{B}}{{V}_{B}{R}_{A}})}^{2} \\ & = & {(\frac{1}{2}\times \frac{3}{1})}^{2} \\ & = & \frac{9}{4}............................(4)\end{matrix}$$VB​VA​​ρB​ρA​​​ρB​ρA​​​=====​CρB​​RB​CρA​​RA​​VB​RA​VA​RB​​(VB​RA​VA​RB​​)2(21​×13​)249​............................(4)​

The acceleration due to gravity $({g}_{A})$(gA​) for planet A is given by

$$\begin{matrix}{g}_{A} & = & \frac{G{M}_{A}}{{{R}_{A}}^{2}} \\ & = & \frac{G\frac{4}{3}\pi {{R}_{A}}^{3}\times {\rho }_{A}}{{{R}_{A}}^{2}} \\ & = & \frac{{C}^{2}}{2}{\rho }_{A}{R}_{A}..........................(5)\end{matrix}$$gA​​===​RA​2GMA​​RA​2G34​πRA​3×ρA​​2C2​ρA​RA​..........................(5)​

Similarly, the acceleration due to gravity $({g}_{B})$(gB​) can be expressed as

$${g}_{B}=\frac{{C}^{2}}{2}{\rho }_{B}{R}_{B}...........................(6)$$gB​=2C2​ρB​RB​...........................(6)

Divide equation (5) by equation (6) and solve with the help of known parameters to calculate the required ratio of the accelerations due to gravity in both the planets.

$$\begin{matrix}\frac{{g}_{A}}{{g}_{B}} & = & \frac{\frac{{C}^{2}}{2}{\rho }_{A}{R}_{A}}{\frac{{C}^{2}}{2}{\rho }_{B}{R}_{B}} \\ & = & \frac{{\rho }_{A}}{{\rho }_{B}}\frac{{R}_{A}}{{R}_{B}} \\ & = & \frac{9}{4}\times \frac{1}{3} \\ & = & \frac{3}{4}\end{matrix}$$gB​gA​​​====​2C2​ρB​RB​2C2​ρA​RA​​ρB​ρA​​RB​RA​​49​×31​43​​

$$=\frac{1}{4}\times \frac{{R}_{2}}{{R}_{1}}=\frac{3}{4}$$=41​×R1​R2​​=43​