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A tunnel is dug along the diameter of the earth (Radius R & mass M). There is a particle of mass 'm' at the centre of the tunnel. The minimum velocity given to the particle so that it just reaches to the surface of the earth is :

Option: 1

\sqrt{\frac{G\; M}{R}}      


Option: 2

\sqrt{\frac{G\; M}{2\; R}}


Option: 3

\sqrt{\frac{2\; G\; M}{ R}}


Option: 4

it will reach with the help of negligible velocity.


Answers (1)

best_answer

As we have learnt in

 

Gravitational Potential Energy at centre of earth relative to infinity -

U=m\left ( -\frac{3}{2}\frac{GM}{R} \right )

m\rightarrow mass of body

M\rightarrow Mass of earth

- wherein

{Ucentre=mVcentre}\\ {Vcentre\rightarrow Potential\: at\: centre}

 

 

Gravitational Potential energy at a point -

W=-\frac{GMm}{r}

U=-\frac{GMm}{r}

U\rightarrow gravitational potential energy 

M\rightarrow Mass of source body

m\rightarrow mass of test body

r\rightarrow distance between two

- wherein

Always negative in the gravitational field because Force is attractive in nature.

 

 

Let the minimum speed imparted to the particle of mass m so that it just reaches surface of earth is v.

Applying conservation of energy

\frac{1}{2}\; mv^{2}+\left [ -\frac{3}{2} \; \frac{G\; M}{R}m\right ]=-\frac{G\; M}{R}m+0

Solving we get  V=\sqrt{\frac{G\: M}{R}}

Posted by

Deependra Verma

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