In free space, two particles of mass m each are initially both at rest at a distance a from each other. They start moving towards each other due to their mutual gravitational attraction. The time after which the distance between them has reduced to a/2 is

Option 1 - <p>( (π+2) / 4√2 ) (a³/Gm)¹/²</p>
Option 2 - <p>( (π-2) / 4√2 ) (a³/Gm)¹/²<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 3 - <p>( (π+2) / 8 ) (a³/Gm)¹/²<br>&lt;!-- [if !supportLineBreakNewLine]--&gt;<br>&lt;!--[endif]--&gt;</p>
Option 4 - <p>( (π-2) / 8 ) (a³/Gm)¹/²</p>
19 Views|Posted 5 months ago
Asked by Shiksha User
1 Answer
V
5 months ago
Correct Option - 3
Detailed Solution:

Let the origin be at the CM of the particles, let the initial positions of the particles be x = a/2 and x = -a/2 and let the instantaneous positions of the particles be x = r and x = -r
Let the instantaneous velocity of each particle be v
Let the time after which the distance between the particles has

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Physics Gravitation 2025

Physics Gravitation 2025

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