Question

 The value of acceleration due to gravity at a depth of 1600km is equal to [Radius of earth=6400km]
\[\begin{align}
  & A.~~~~~~~~9.8m{{s}^{-2}} \\
 & B.~~~~~~~~4.9m{{s}^{-2}} \\
 & C.~~~~~~~~7.35m{{s}^{-2}} \\
 & D.~~~~~~~~19.6m{{s}^{-2}} \\
\end{align}\]

Answer 1

Hint: To find the value of acceleration due to gravity at a depth of 1600km, we can use the formula of acceleration due to gravity at any point inside the earth. We know that the value of acceleration due to gravity at the surface of earth is $9.8m{{s}^{-2}}$.

Complete step by step answer:
 Earth can be considered as a solid sphere with uniform mass distribution. Gravitational field or acceleration due to gravity at any point inside the earth is given by the formula
${{g}_{r}}=\dfrac{GMr}{{{R}^{3}}}$,
Where G is a universal gravitational constant, M is the mass of earth, R is the radius of earth and r is distance of the point from the centre of earth.
Let us consider acceleration due to gravity at earth’s surface to be g. From the formula of acceleration due to gravity at any point inside the earth, we get the value of g to be
$g=\dfrac{GMR}{{{R}^{3}}}=\dfrac{GM}{{{R}^{2}}}$
Also, we know that
$g=9.8m{{s}^{-2}}$
If a point is at depth d, then its distance from center is$R-d$. So acceleration due to gravity at this point is (using formula of acceleration due to gravity at any point inside the earth)
${{g}_{d}}=\dfrac{GMr}{{{R}^{3}}}=\dfrac{GM\left( R-d \right)}{{{R}^{3}}}=\dfrac{GM}{{{R}^{2}}}\left( 1-\dfrac{d}{R} \right)$
${{g}_{d}}=g\left( 1-\dfrac{d}{R} \right)$
We have to find a gravitational field at a depth of 1600 km from the earth’s surface and radius of earth is given to us is 6400 km. So, putting these values in above equation we get
${{g}_{d}}=9.8\left( 1-\dfrac{1600}{6400} \right)=7.35m{{s}^{-2}}$
Hence the correct option is C.
Note: The formula of acceleration due to gravity at a depth d which we obtained was without any approximation. A similar formula is used for height h above the earth’s surface, but that is obtained after approximations. The formula is
${{g}_{h}}=g\left( 1-\dfrac{2h}{R} \right)$ , but it is an approximation and is valid only when$h<