Gravitation

JEE Main level, single correct, medium

123 of 223

Question 123 of 223

The escape velocities of two planets AandBAandB are in the ratio 1:21:2. If the ratio of their radii respectively is 1:31:3, then the ratio of acceleration due to gravity of planet AA to the acceleration of gravity of planet BB will be:

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Answer: (D) 34\dfrac{3}{4}

The formula to calculate the mass (M)\left(M\right) of any planet of radius RR and density ρ\rho, considering it to be a sphere, is given by
M=43πR3ρ.....................(1)M=\dfrac{4}{3}\pi {R}^{3}\rho .....................(1)
The escape velocity (vA)\left({v}_{A}\right) for planet A can be expressed as
VA=2GMARA=2GρA43πRA3RA=CρA.RA............................(2)\begin{matrix}{V}_{A} & = & \sqrt{\dfrac{2G{M}_{A}}{{R}_{A}}} \\ & = & \sqrt{\dfrac{2G{\rho }_{A}\dfrac{4}{3}\pi {{R}_{A}}^{3}}{{R}_{A}}} \\ & = & C\sqrt{{\rho }_{A}.}{R}_{A}............................(2)\end{matrix}
where, GG is Gravitational constant and C=8πG3C=\sqrt{\dfrac{8\pi G}{3}} is a constant.
Similarly, the escape velocity (vB)\left({v}_{B}\right) of planet B can be expressed as
vB=CρBRB......................(3){v}_{B}=C\sqrt{{\rho }_{B}}{R}_{B}......................(3)
Divide equation (2) by equation (3) and solve with the known parameters to obtain the ratio of the densities of the planets.
VAVB=CρARACρBRBρAρB=VARBVBRAρAρB=(VARBVBRA)2=(12×31)2=94............................(4)\begin{matrix}\dfrac{{V}_{A}}{{V}_{B}} & = & \dfrac{C\sqrt{{\rho }_{A}}{R}_{A}}{C\sqrt{{\rho }_{B}}{R}_{B}} \\ \sqrt{\dfrac{{\rho }_{A}}{{\rho }_{B}}} & = & \dfrac{{V}_{A}{R}_{B}}{{V}_{B}{R}_{A}} \\ \dfrac{{\rho }_{A}}{{\rho }_{B}} & = & {\left(\dfrac{{V}_{A}{R}_{B}}{{V}_{B}{R}_{A}}\right)}^{2} \\ & = & {\left(\dfrac{1}{2}\times \dfrac{3}{1}\right)}^{2} \\ & = & \dfrac{9}{4}............................(4)\end{matrix}
The acceleration due to gravity (gA)\left({g}_{A}\right) for planet A is given by
gA=GMARA2=G43πRA3×ρARA2=C22ρARA..........................(5)\begin{matrix}{g}_{A} & = & \dfrac{G{M}_{A}}{{{R}_{A}}^{2}} \\ & = & \dfrac{G\dfrac{4}{3}\pi {{R}_{A}}^{3}\times {\rho }_{A}}{{{R}_{A}}^{2}} \\ & = & \dfrac{{C}^{2}}{2}{\rho }_{A}{R}_{A}..........................(5)\end{matrix}
Similarly, the acceleration due to gravity (gB)\left({g}_{B}\right) can be expressed as
gB=C22ρBRB...........................(6){g}_{B}=\dfrac{{C}^{2}}{2}{\rho }_{B}{R}_{B}...........................(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.
gAgB=C22ρARAC22ρBRB=ρAρBRARB=94×13=34\begin{matrix}\dfrac{{g}_{A}}{{g}_{B}} & = & \dfrac{\dfrac{{C}^{2}}{2}{\rho }_{A}{R}_{A}}{\dfrac{{C}^{2}}{2}{\rho }_{B}{R}_{B}} \\ & = & \dfrac{{\rho }_{A}}{{\rho }_{B}}\dfrac{{R}_{A}}{{R}_{B}} \\ & = & \dfrac{9}{4}\times \dfrac{1}{3} \\ & = & \dfrac{3}{4}\end{matrix}
=14×R2R1=34=\dfrac{1}{4}\times \dfrac{{R}_{2}}{{R}_{1}}=\dfrac{3}{4}

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