NEET · Physics · STD 11 - 7. gravitation
Two bodies of mass \(m\) and \(9 m\) are placed at a distance \(R\). The gravitational potential on the line joining the bodies where the gravitational field equals zero, will be ( \(G=\) gravitational constant) :
- A \(-\frac{20 G m}{R}\)
- B \(-\frac{8 G m}{R}\)
- C \(-\frac{12 G m}{R}\)
- D \(-\frac{16 G m}{R}\)
Answer & Solution
Correct Answer
(D) \(-\frac{16 G m}{R}\)
Step-by-step Solution
Detailed explanation
Position of Neutral point (Zero Gravitational Field) \(r_1=\frac{\sqrt{m_1} R}{\sqrt{m_1}+\sqrt{m_2}}=\frac{\sqrt{m} R}{\sqrt{m}+\sqrt{9 m}}=\frac{R}{4}\)
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