Question

the depth at which the value of g becomes 25 percentage of that at the surface of the earth .( radius of the earth = 6400km).

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Answer 1

Answer:

Explanation:

g=GM/Re² =>at any depth h below the earth g becomes g'=GM'/(Re-h)².

Here M’=M[(4/3)π(Re-h)³/{(4/3)πRe³]

=M×{(Re-h)/Re}³.  

So g’/g=[GM{(Re-h)³/Re³}/(Re-h)²]/[GM/Re²]

={(Re-h)/Re}.So g'/g={1-(h/Re)}  

As the question indicates g'/g=0.25

=>[1-(h/Re)]=1/4

=>h/Re=1-(1/4)=3/4

=>h=(3/4)Re=(3/4)×6400=4800km.

ANS. 4800 km.

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Answer 2

Answer:

The value of g at a depth is given by:

gd=g(1−Rd)

Here, gd=4g

⇒4g=g(1−Rd)

⇒Rd=43

⇒d=0.75×6400=4800 km