Question

find the ratio of the acceleration due to gravity at a height R/2 from the the surface of the earth to the value at a depth R/2 from the surface of the Earth (R- radius of the earth)

Answer 1

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As we know that :

$\large{\boxed{\boxed{\sf{g \: = \: G \dfrac{M}{R^2}}}}}----(1)$

Where,

• g is gravitational force
• G is gravitational constant
• M is mass of earth
• R is radius of earth

Now, if the radius of earth decreases or shrinks to 50% of its original radius then we can take Radius as R/2

Substitute value of R' which is R/2 in (1) we get,

$\large{\boxed{\boxed{\sf{g' \: = \: G \dfrac{M}{R' ^2}}}}}$

$\implies {\sf{g' \: = \: G \dfrac{M}{\dfrac{R}{2}}^2}}$

$\implies {\sf{g' \: = \:G\dfrac{M}{\dfrac{R^2}{4}}}}$

$\implies {\sf{g' \: = \: \dfrac{g}{4}}}$

$\implies {\sf{\dfrac{g'}{g} \: = \: \dfrac{1}{4}}}$

➠ Ratio is 1:4