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Asked in NEET 2015 · Mutual gravitation and binary systems
No external force acts, so the centre of mass stays put and the displacements satisfy Mx₁=5Mx₂, giving x₁=5x₂.
They touch when their centres are R+2R=3R apart.
So the separation closes by 12R-3R=9R, which means x₁+x₂=9R.
Substituting x₁=5x₂: 6x₂=9R, so x₂=1.5R.
The smaller body therefore covers x₁=5×1.5R=7.5R.
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