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MHT CET 2026 (14 May, Shift 1) · Gravitational Potential & Energy

Gravitational Potential & Energy question from MHT CET 2026 (14 May, Shift 1)

Q4MHT CET 2026 · 14 MAY · SHIFT 1Medium

Gravitational Potential & Energy

Two particles of masses mm and MM are placed a distance dd apart. The gravitational potential at the position where the gravitational field due to them is zero is VV. Then

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Answer: (d)

  1. Null point at xx from mm: xd−x=mM⇒x=m dm+M\frac{x}{d-x} = \sqrt{\frac{m}{M}} \Rightarrow x = \frac{\sqrt{m}\,d}{\sqrt{m}+\sqrt{M}}, d−x=M dm+Md - x = \frac{\sqrt{M}\,d}{\sqrt{m}+\sqrt{M}}
  2. V=−Gmx−GMd−x=−G(m+M)d(m+M)V = -\frac{Gm}{x} - \frac{GM}{d-x} = -\frac{G(\sqrt{m}+\sqrt{M})}{d}\left(\sqrt{m}+\sqrt{M}\right)
  3. V=−Gd(m+M)2V = -\frac{G}{d}(\sqrt{m}+\sqrt{M})^2
  4. So the answer is (d).
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