Answer
Verified
114.9k+ 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.
Recently Updated Pages
Uniform Acceleration - Definition, Equation, Examples, and FAQs
JEE Colleges - Detailed Description of Top JEE Colleges
JEE Main 2023 January 25 Shift 1 Question Paper with Answer Key
Geostationary Satellites and Geosynchronous Satellites for JEE
Complex Numbers - Important Concepts and Tips for JEE
JEE Main 2023 (February 1st Shift 2) Maths Question Paper with Answer Key
Trending doubts
Learn About Angle Of Deviation In Prism: JEE Main Physics 2025
JEE Main Login 2045: Step-by-Step Instructions and Details
JEE Main Exam Marking Scheme: Detailed Breakdown of Marks and Negative Marking
Collision - Important Concepts and Tips for JEE
Ideal and Non-Ideal Solutions Raoult's Law - JEE
Free Radical Substitution Mechanism of Alkanes for JEE Main 2025
Other Pages
Current Loop as Magnetic Dipole and Its Derivation for JEE
A particle performs SHM of amplitude A along a straight class 11 physics JEE_Main
JEE Main 2023 April 6 Shift 1 Question Paper with Answer Keys & Solutions
JEE Main 2023 January 30 Shift 2 Question Paper with Answer Keys & Solutions
Inductive Effect and Acidic Strength - Types, Relation and Applications for JEE
NCERT Solutions for Class 11 Physics Chapter 4 Laws of Motion