Can you solve this

Dear Student,
The above problem can be solved by determining the gravitational potential at the initial point (r = R/2 from the centre of the earth) and the final point (r = 2R from the centre of the earth.
Assuming the earth to be a sphere of uniform mass density.
Gravitational potential at any point above the surface, i.e. , r > R is
g left parenthesis r right parenthesis space equals space minus fraction numerator G M over denominator r squared end fraction V left parenthesis r right parenthesis space equals space integral g left parenthesis r right parenthesis d r equals negative fraction numerator G M over denominator r end fraction w h e r e comma G space i s space t h e space g r a v i t a t i o n a l space c o n s tan t space a n d space M space i s space t h e space m a s s space o f space t h e space e a r t h

For an interior point (r < R), the gravitational potential can be determined as:
g left parenthesis r right parenthesis space equals space fraction numerator G M left parenthesis r right parenthesis over denominator r squared end fraction w h e r e comma G space i s space t h e space g r a v i t a t i o n a l space c o n s tan t space a n d space M left parenthesis r right parenthesis space i s space t h e space m a s s space o f space t h e space s p h e r e space o f space r a d i u s space r M left parenthesis r right parenthesis space equals space fraction numerator M over denominator begin display style bevelled fraction numerator 4 pi R cubed over denominator 3 end fraction end style end fraction cross times 4 over 3 πr cubed equals Mr cubed over straight R cubed Thus comma straight V left parenthesis straight r right parenthesis space equals space integral straight g left parenthesis straight r right parenthesis dr equals integral fraction numerator GMr squared over denominator straight r squared straight R cubed end fraction dR equals fraction numerator GMr squared over denominator 2 straight R cubed end fraction plus straight C For space straight r space equals space straight R comma space straight V left parenthesis straight R right parenthesis space equals space minus GM over straight R Thus comma open square brackets straight V left parenthesis straight r right parenthesis close square brackets subscript at space straight r equals straight R end subscript equals fraction numerator GMR squared over denominator 2 straight R cubed end fraction plus straight C space equals space minus GM over straight R therefore straight C space equals space minus fraction numerator 3 GM over denominator 2 straight R end fraction So comma at space straight r less than straight R straight V left parenthesis straight r right parenthesis space equals space open parentheses fraction numerator GMr squared over denominator 2 straight R cubed end fraction minus fraction numerator 3 GM over denominator 2 straight R end fraction close parentheses straight V left parenthesis straight r right parenthesis space equals fraction numerator GM over denominator 2 straight R end fraction open parentheses straight r squared over straight R squared minus 3 close parentheses
Change in V(r) from r = R/2 to r = 2R:
V left parenthesis 2 R right parenthesis minus V left parenthesis R divided by 2 right parenthesis equals negative fraction numerator G M over denominator 2 R end fraction minus fraction numerator G M over denominator 2 R end fraction open parentheses fraction numerator R squared over denominator 4 R squared end fraction minus 3 close parentheses equals negative fraction numerator G M over denominator 2 R end fraction plus fraction numerator 11 G M over denominator 8 R end fraction equals fraction numerator 7 G M over denominator 8 R end fraction equals 7 over 8 g R A l s o comma V left parenthesis 2 R right parenthesis minus V left parenthesis R divided by 2 right parenthesis equals fraction numerator W o r k space d o n e over denominator M end fraction W o r k space d o n e equals M cross times 7 over 8 g R equals 7 over 8 M g R
Ans (c)

Regards

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