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
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
Updated On: Aug 1, 2022
- $v_Q>v_R > v_P$
- $v_R > v_Q > v_P$
- $v_R/v_P = 3$
- $v_P/v_Q = 1/2$
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The Correct Option is D
Solution and Explanation
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$
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Concepts Used:
Escape Speed
Escape speed is the minimum speed, which is required by the object to escape from the gravitational influence of a plannet. Escape speed for Earth’s surface is 11,186 m/sec.
The formula for escape speed is given below:
ve = (2GM / r)1/2
where ,
ve = Escape Velocity
G = Universal Gravitational Constant
M = Mass of the body to be escaped from
r = Distance from the centre of the mass