MHT CET · Physics · Gravitation
The mass and radius of the earth and moon are \(M, R\) and \(m, r\) respectively. The distance between their centers is \(d\). The minimum velocity with which a particle of mass \(m_0\) should be projected from the midpoint between them so that it could reach infinity is
- A \(2 \sqrt{\frac{G}{d}(M+m)}\)
- B \(2 \sqrt{\frac{G m}{d}(M+m)}\)
- C \(2 \sqrt{\frac{2 G}{d}(M+m)}\)
- D \(2 \sqrt{\frac{G m(M+m)}{d(R+r)}}\)
Answer & Solution
Correct Answer
(A) \(2 \sqrt{\frac{G}{d}(M+m)}\)
Step-by-step Solution
Detailed explanation
The potential energy at the midpoint is \(\left(-\frac{G m m_0}{d / 2}-\frac{G M m_0}{d / 2}\right)\)
And suppose \(\left(\frac{m_0 V^2}{2}\right)\) kinetic energy is required to just send the particle to infinity.
On considering energy conservation, i.e., initial total energy is equal to the final total energy:
\(\begin{aligned}
& \left(-\frac{G m m_0}{d / 2}-\frac{G M m_0}{d / 2}\right)+\left(\frac{m_0 V^2}{2}\right)=0 \\
& \Rightarrow V^2=\frac{4 G}{d}(M+m) \\
& \Rightarrow V=2 \sqrt{\frac{G(M+m)}{d}}
\end{aligned}\)
And suppose \(\left(\frac{m_0 V^2}{2}\right)\) kinetic energy is required to just send the particle to infinity.
On considering energy conservation, i.e., initial total energy is equal to the final total energy:
\(\begin{aligned}
& \left(-\frac{G m m_0}{d / 2}-\frac{G M m_0}{d / 2}\right)+\left(\frac{m_0 V^2}{2}\right)=0 \\
& \Rightarrow V^2=\frac{4 G}{d}(M+m) \\
& \Rightarrow V=2 \sqrt{\frac{G(M+m)}{d}}
\end{aligned}\)
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