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Two identical planets each of mass \mathrm{m} are approaching towards each other due to their mutual gravitational field. If the speed of one planet relative to the other is \mathrm{v}, the K.E. of the system is equal to
 

Option: 1

\mathrm{(1 / 2) m v^2}

 


Option: 2

\mathrm{mv^{2}}
 


Option: 3

\mathrm{(1 / 4) m v^2}

 


Option: 4

None of these


Answers (1)

best_answer

The K.E. of the system of two mass system moving under their mutual interaction can be given as \mathrm{\mathrm{KE}=(1 / 2) \mu v_{\text {res }}^2}

where \mathrm{\mu=\frac{m_1 m_2}{m_1+m_2} \& V_{velocity} =\left(v_1+v_2\right)}

\mathrm{\Rightarrow \mu =\frac{mm}{m+m}=\left ( \frac{m}{2} \right )v^{2}=\frac{mv^{2}}{4}}

Hence option 3 is correct.

 

 

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mansi

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