
Planet A has mass M and radius R. Planet B has half the mass and half the radius of planet A. If the escape velocities from the planets A and B are \[{v_A}\] and ${v_B}$ respectively, then $\dfrac{{{v_A}}}{{{v_B}}} = \dfrac{n}{4}$ ,The value of n is
(A) $3$
(B) $2$
(C) $4$
(D) $5$
Answer
161.4k+ views
Hint: In order to solve this question, we will first find the escape velocity for both the planets A and B using the general formula of escape velocity and then we will find the ratio of the escape velocity of planets A and B and then compare with the given form to solve for the value of n.
Formula used:
The escape velocity for a planet is given by:
$v = \sqrt {\dfrac{{2GM}}{R}} $
Where, G is the gravitational constant, M is the mass of the Planet and R is the radius of the planet.
Complete answer:
We have given that, mass of the planet A is M and its radius is R then its escape velocity is calculated as ${v_A} = \sqrt {\dfrac{{2GM}}{R}} \to (i)$ and for planet B, let its mass is M’ then according to the question $M' = \dfrac{M}{2}$ and let its Radius is R’ then $R' = \dfrac{R}{2}$ so its escape velocity is given by:
${v_B} = \sqrt {\dfrac{{2G'M'}}{{R'}}} $
$
{v_B} = \sqrt {\dfrac{{2G\dfrac{M}{2}}}{{\dfrac{R}{2}}}} \\
{v_B} = \sqrt {\dfrac{{2GM}}{R}} \to (ii) \\
$
On dividing equations (i) and (ii) we get,
$\dfrac{{{v_A}}}{{{v_B}}} = 1$
Now, comparing this value with given form $\dfrac{{{v_A}}}{{{v_B}}} = \dfrac{n}{4}$ we get,
$
\dfrac{n}{4} = 1 \\
\Rightarrow n = 4 \\
$
Hence, the correct option is (C) $4$.
Note: It should be remembered that escape velocity is the minimum velocity needed for an object to escape the gravitational field effect of any planet, star or any celestial object, and escape velocity is independent of the mass of the object, it only depends upon Radius and mass of the planet.
Formula used:
The escape velocity for a planet is given by:
$v = \sqrt {\dfrac{{2GM}}{R}} $
Where, G is the gravitational constant, M is the mass of the Planet and R is the radius of the planet.
Complete answer:
We have given that, mass of the planet A is M and its radius is R then its escape velocity is calculated as ${v_A} = \sqrt {\dfrac{{2GM}}{R}} \to (i)$ and for planet B, let its mass is M’ then according to the question $M' = \dfrac{M}{2}$ and let its Radius is R’ then $R' = \dfrac{R}{2}$ so its escape velocity is given by:
${v_B} = \sqrt {\dfrac{{2G'M'}}{{R'}}} $
$
{v_B} = \sqrt {\dfrac{{2G\dfrac{M}{2}}}{{\dfrac{R}{2}}}} \\
{v_B} = \sqrt {\dfrac{{2GM}}{R}} \to (ii) \\
$
On dividing equations (i) and (ii) we get,
$\dfrac{{{v_A}}}{{{v_B}}} = 1$
Now, comparing this value with given form $\dfrac{{{v_A}}}{{{v_B}}} = \dfrac{n}{4}$ we get,
$
\dfrac{n}{4} = 1 \\
\Rightarrow n = 4 \\
$
Hence, the correct option is (C) $4$.
Note: It should be remembered that escape velocity is the minimum velocity needed for an object to escape the gravitational field effect of any planet, star or any celestial object, and escape velocity is independent of the mass of the object, it only depends upon Radius and mass of the planet.
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