Question

Two spherical planets P and Q have the same uniform density p, masses $M_P$ and $M_Q$, and surface areas A and 4A, respectively. A spherical planet R also has uniform density p and its mass is $(M_P + M_Q)$. The escape velocities from the planets P, Q and R, are $v_P$, $v_Q$ and $v_R$, respectively. Then

• A. $v_Q>v_R > v_P$
• B. $v_R > v_Q > v_P$
• C. $v_R/v_P = 3$
• D. $v_P/v_Q = 1/2$

Answer 1

Answer: (D)

$v_P/v_Q = 1/2$

Answer 2

Surface area of Q is four times. Therefore, radius of Q is two times. Volume is eight times. Therefore, mass of Q is also eight times. So, let $\, \, \, \, \, \, \, \, \, M_P = M \, and \, R_P = r$ Then, $\, \, \, \, \, \, \, \, \, M_Q = 8 \, M \, and \, R_Q = 2r$ Now, mass of R is $(M_P + M_Q)$ or 9 M. Therefore, radius of R is $(9 )^{1/3r}$. Now, escape velocity from the surface of a planet is given by $\, \, \, \, \, \, \, \, v=\sqrt{\frac{2GM}{r}}$ (r = radius of that planet) $\, \, \, \, \, \, \, \, v_P=\sqrt{\frac{2GM}{r}}$ $\, \, \, \, \, \, \, \, v_Q=\sqrt{\frac{2G(8M)}{2r}}$ $\, \, \, \, \, \, \, \, v_R=\sqrt{\frac{2G(9M)}{(9)^{1/3}r}}$ From here we can see that, $\frac{v_p}{v_Q}= \frac{1}{2}$ and $\, \, \, \, \, \, \, \, \, \, \, \, \, v_R > v_Q > v_P$