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  • The minimum speed of a particle projected from earth's surface so that it will never return is/are  Option: 1 <img alt="\math

The minimum speed of a particle projected from earth's surface so that it will never return is/are

 

Option: 1

\mathrm{\sqrt{\frac{2 G M}{R}}}
 


Option: 2

\mathrm{11.2 \mathrm{~km} / \mathrm{sec}}
 


Option: 3

\mathrm{\sqrt{2 g_0 R}}
 


Option: 4

all of these.


Answers (1)

best_answer

Let the minimum speed of projection of the particle of mass \mathrm{m} be \mathrm{v}

\mathrm{\Rightarrow \mathrm{KE} \, \text{of the particle} \: \: \mathrm{K}_{\mathrm{R}}=\frac{1}{2} \mathrm{mv}^2.}

Gravitational potential energy of the particle.

\mathrm{ =U_R=-\frac{G M m}{R} }

where \mathrm{ \text{M= mass of earth and R= radius of earth.} }
The particle will have to reach the infinity to escape from earth's surface for ever. At infinity for very large distance of separation \mathrm{(r \rightarrow \infty)} gravitation potential at infinity is zero
\mathrm{\Rightarrow \mathrm{U}_{\infty}=0}
Since the particle just reach the infinity, its \mathrm{\mathrm{KE}} at infinity is zero \mathrm{\Rightarrow \mathrm{K}_{\infty}=0}. Conservation of total mechanical energy of the particle between the infinity and earth's surface, we obtain,

\mathrm{ U_R+K_R=U_{\infty}+K_{\infty} }

\mathrm{ \Rightarrow-\frac{G M m}{R}+\frac{1}{2} m v^2=0+0 }

\mathrm{ \Rightarrow v=\sqrt{\frac{2 G M}{R}}=\sqrt{\frac{2 G M}{R^2}} R=\sqrt{2 g_0 R} }

\mathrm{ \Rightarrow v=\left(\sqrt{2 \times 9.8 \times 6.410^6}\right) \mathrm{m} / \mathrm{sec}=11.2 \mathrm{~km} / \mathrm{sec} .}

Since \mathrm{g_0=\frac{G M}{R^2}=V=\sqrt{\frac{2 G M}{R}}}.

Therefore all the choices are correct.

Posted by

SANGALDEEP SINGH

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