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CBSE Class 11-science Answered

A tunnel is dug along the diameter of the earth. There is a particle of mass m at the centre of the tunnel. Find the minimum velocity given to the particle so that is just reaches to the surface of the earth (R=radius of the earth)
Asked by akkamal03 | 26 May, 2020, 12:11: PM
Expert Answer
Here we are required to find, minimum velocity given to the particle so that is just reaches to the surface of the earth
 
Hence, by applying law of conservation of energy, 
Initial Total energy of particle = Final energy of particle
 
Initially the energy of the particle at the centre of the Earth is given as, 
 
K. E subscript i space plus thin space P. E subscript i space equals space 1 half m v squared space plus space fraction numerator negative G M over denominator R cubed end fraction open square brackets fraction numerator 3 R squared over denominator 2 end fraction minus r squared over 2 close square brackets space equals space fraction numerator negative G M over denominator R cubed end fraction open square brackets fraction numerator 3 R squared over denominator 2 end fraction minus 0 close square brackets space... space left parenthesis S i n c e comma space r space i s space t h e space d i s tan c e space f r o m space t h e space c e n t r e space o f space t h e space E a r t h. space H e r e space r space equals 0 right parenthesis
K. E subscript i space plus thin space P. E subscript i space equals space 1 half m v squared space plus space fraction numerator negative 3 G M over denominator 2 R end fraction space

T o t a l space E n e r g y space a t space s u r f a c e space o f space t h e space E a r t h space i s comma space
K. E subscript f space end subscript plus thin space P. E subscript f space space equals space 0 space plus space fraction numerator negative G M m over denominator R end fraction space

H e n c e comma space
B y space l a w space o f space c o n s e r v a t i o n space o f space e n e r g y comma space
K. E subscript i space plus thin space P. E subscript i space equals space K. E subscript f space end subscript plus thin space P. E subscript f
space 1 half m v squared space plus space fraction numerator negative 3 G M over denominator 2 R end fraction space equals space fraction numerator negative G M m over denominator R end fraction
S o l v i n g space t h e space a b o v e space e q u a t i o n space w e space g e t
v space equals square root of fraction numerator G M m over denominator R end fraction end root space
Answered by Shiwani Sawant | 26 May, 2020, 07:05: PM
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