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Two identical hypothetical isolated planets \mathrm{A \& B} each of mass \mathrm{M} have the coordinates as shown. The minimum speed of projection of a body from \mathrm{A \: to \: B} is

 

Option: 1

\mathrm{\sqrt\frac{GM}{a}}


Option: 2

\mathrm{\sqrt\frac{2GM}{3a}}


Option: 3

\mathrm{\sqrt\frac{GM}{3a}}


Option: 4

None of these


Answers (1)

best_answer

When the body just passes the point \mathrm{P} have the net gravitational field force/intensity is zero, it will be attached towards the planet \mathrm{B}. That means the velocity of projection in order to escape from \mathrm{A} is that required to just reach the point \mathrm{P}.

The potential difference between M \& $P$ is

\mathrm{ V=\left|V_p-V_M\right|=\left\{-\frac{G M}{a}-\frac{G M}{2 a}\right\} \mid }

\mathrm{ -\left\{-\frac{G M}{(3 / 2) a}-\frac{G M}{(3 / 2) a}\right\} \Rightarrow V=\left|-\frac{3 G M}{2 a}+\frac{4 G M}{3 a}\right|=\frac{G M}{6 a} }

\mathrm{ \therefore \quad\left|\Delta U_{g r}\right|=\left|K E_{M \rightarrow P}\right| \Rightarrow m V=\frac{1}{2} m^2 }

\mathrm{ \Rightarrow V=\sqrt{2 V}=\sqrt{\frac{G M}{3 a}}}

Hence option 2 is correct.





 

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