Question 18: Gravitation

Question

Two spherical planets P and Q have the same uniform density $\rho$, and masses Mp and $\mathrm{MQ}_{\mathrm{Q}}$ and surface areas A and 4 A respectively. Another spherical planet $R$ also has the same uniform density $\rho$, and its mass is $\mathrm{Mp}+\mathrm{MQ}$. The escape velocities from these planets is

Accepted Answer

Let $R_P,R_Q,R_R$ be the radii of $P,Q,R$. Since surface area $S=4\pi R^2$,
For $P$:
$4\pi R_P^2 = A$
$\displaystyle R_P^2=\frac{A}{4\pi}$.
For $Q$:
$4\pi R_Q^2 = 4A$
$\displaystyle R_Q^2=\frac{4A}{4\pi}=\frac{A}{\pi}=4\,\frac{A}{4\pi}$
Ã¢ÂÂ $R_Q=2R_P$.
With uniform density $ho$, mass $M=	frac43\piho\,R^3$, so
$\displaystyle M_Q = \frac{4}{3}\piho\,(2R_P)^3 = 8\,M_P$.
Planet $R$ has $M_R=M_P+M_Q=9M_P$, so
$\displaystyle R_R = \bigl(	frac{3M_R}{4\piho}\bigr)^{1/3}  = (9M_P)^{1/3}\!\bigl(	frac{3}{4\piho}\bigr)^{1/3} = \sqrt[\,3\,]{9}\;R_P\approx2.08\,R_P.$
Escape velocity is
$ V_{m esc}=\sqrt{\frac{2GM}{R}}. $
Ã¢ÂÂ For $P$: $\;V_P=\sqrt{	frac{2GM_P}{R_P}}$.
Ã¢ÂÂ For $Q$:
$ V_Q =\sqrt{\frac{2G\,(8M_P)}{2R_P}} =\sqrt{4\,\frac{2GM_P}{R_P}} =2\,V_P. $
Ã¢ÂÂ For $R$:
$ V_R =\sqrt{\frac{2G\,(9M_P)}{\sqrt[3]{9}\,R_P}} =\sqrt[\,3\,]{9}\;\sqrt{\frac{2GM_P}{R_P}} \approx2.08\,V_P. $
Hence numerically
$ V_R\approx2.08\,V_P,\quad V_Q=2\,V_P,\quad V_P
so the correct ordering is
Option A:  $V_R>V_Q>V_P$.

Answer

$V_R>V_Q>V_P$

Answer

$\mathrm{V}_{\mathrm{R}}<\mathrm{V}_{\mathrm{Q}}<\mathrm{V}_{\mathrm{P}}$

Answer

$V_R=V_Q>V_P$

Answer

$V_R>V_Q=V_P$

Gravitation

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