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.

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