Two spherical planets P and Q have the same u | vedclass

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(A) The escape velocity $V_e$ is given by $V_e = \sqrt{\frac{2GM}{R}}$. Since $M = \rho \cdot \frac{4}{3}\pi R^3$,we have $V_e = \sqrt{\frac{2G}{R} \c

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$(B, D)$

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(A) The escape velocity $V_e$ is given by $V_e = \sqrt{\frac{2GM}{R}}$. Since $M = \rho \cdot \frac{4}{3}\pi R^3$,we have $V_e = \sqrt{\frac{2G}{R} \cdot \rho \cdot \frac{4}{3}\pi R^3} = \sqrt{\frac{8\pi G \rho}{3}} R$. Thus,$V_e \propto R$.Given surface area $A_P = A = 4\pi R_P^2$ and $A_Q = 4A = 4\pi R_Q^2$. This implies $R_Q^2 = 4R_P^2$,so $R_Q = 2R_P$.For planet $R$,$M_R = M_P + M_Q$. Since density $\rho$ is uniform,$\frac{4}{3}\pi R_R^3 \rho = \frac{4}{3}\pi R_P^3 \rho + \frac{4}{3}\pi R_Q^3 \rho$. Thus,$R_R^3 = R_P^3 + R_Q^3 = R_P^3 + (2R_P)^3 = R_P^3 + 8R_P^3 = 9R_P^3$. So,$R_R = 9^{1/3} R_P$. Comparing radii: $R_R = 9^{1/3} R_P \approx 2.08 R_P$,$R_Q = 2 R_P$,and $R_P = R_P$. Therefore,$R_R > R_Q > R_P$,which implies $V_R > V_Q > V_P$. This confirms statement $(B)$ is correct.For statement $(D)$,$\frac{V_P}{V_Q} = \frac{R_P}{R_Q} = \frac{R_P}{2R_P} = \frac{1}{2}$. This confirms statement $(D)$ is correct.

Column-$I$	Column-$II$
$(1)$ Value of escape velocity on the surface of the Earth	$(a)$ $2.38 \, km \, s^{-1}$
$(2)$ Value of escape velocity on the surface of the Moon	$(b)$ $7.92 \, km \, s^{-1}$
	$(c)$ $11.2 \, km \, s^{-1}$

Similar Questions

The ratio of the escape velocity of a planet to the escape velocity of the Earth will be: Given: Mass of the planet is $16$ times the mass of the Earth and the radius of the planet is $4$ times the radius of the Earth.

The escape velocity of an object on a planet whose $g$ value is $9$ times that of Earth and whose radius is $4$ times that of Earth in $km/s$ is:

Match Column-$I$ with Column-$II$.

Column-$I$ Column-$II$

$(1)$ Value of escape velocity on the surface of the Earth $(a)$ $2.38 \, km \, s^{-1}$

$(2)$ Value of escape velocity on the surface of the Moon $(b)$ $7.92 \, km \, s^{-1}$

$(c)$ $11.2 \, km \, s^{-1}$

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Two planets $A$ and $B$ have the same material density. If the radius of $A$ is twice that of $B,$ then the ratio of the escape velocity $\frac{V_A}{V_B}$ is

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