AP EAMCET · PHYSICS · Gravitation
Two bodies of masses \(m_1\) and \(m_2\) initially at rest at infinite distance apart move towards each other under gravitational force of attraction. Their relative velocity of approach when they are separated by a distance \(r\) is ( \(\mathrm{G}=\) universal gravitational constant.)
- A \(\left[\frac{2 G\left(m_1-m_2\right)}{r}\right]^{1 / 2}\)
- B \(\left[\frac{2 G\left(m_1+m_2\right)}{r}\right]^{1 / 2}\)
- C \(\left[\frac{r}{2 G\left(m_1 m_2\right)}\right]^{1 / 2}\)
- D \(\left[\frac{r}{2 G} m_1 m_2\right]^{1 / 2}\)
Answer & Solution
Correct Answer
(B) \(\left[\frac{2 G\left(m_1+m_2\right)}{r}\right]^{1 / 2}\)
Step-by-step Solution
Detailed explanation
Initially when the two masses are at an infinite distance from each other, their gravitational potential energy is zero. When they are at a distance \(r\) from each, the gravitational \(P E\) is \(\mathrm{PE}=\frac{-G m_1 m_2}{r^2}\) The minus sign-indicates that there is a…
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