Question

The ratio of acceleration due to gravity at a depth $\left(h\right)$ below the surface of earth and at a height $\left(h\right)$ above the surface of earth for $h<<$$R$ ( $R$ = radius of the earth )

• A. is constant
• B. increases linearly with $h$
• C. increases parabolically with $h$
• D. increases parabolically with $h$

Answer 1

The correct option is B increases linearly with $h$

Let, $R$ be the radius of earth and $g$ the acceleration due to gravity on earth’s surface.
Above the surface, $g_{above}^{\prime }=\frac{GM}{(R+h{)}^2}$
Inside the surface, $g_{below}^{\prime }=\frac{GM}{R^2}(1-\frac{h}{R})$

So, $\frac{g_{below}^{\prime }}{g_{above}^{\prime }}=\frac{\frac{GM}{R^2}(1-\frac{h}{R})}{\frac{GM}{(R+h{)}^2}}=\frac{\frac{1}{R^2}(1-\frac{h}{R})}{\frac{1}{(R+h{)}^2}}=\frac{(1-\frac{h}{R})}{\frac{1}{(1+h/R{)}^2}}$

Here, $\frac{h}{R}<<<1$
so, $\frac{g_{below}^{\prime }}{g_{above}^{\prime }}=(1-\frac{h}{R})\times (1+\frac{h}{R}{)}^2=(1-\frac{h}{R})\times (1+\frac{2h}{R}+(\frac{2h}{R}{)}^2)=1+\frac{2h}{R}+(\frac{2h}{R}{)}^2-\frac{h}{R}-2(\frac{h}{R}{)}^2-(\frac{h}{R}{)}^3$
As $\frac{h}{R}<<<1$

so, $\frac{g_{below}^{\prime }}{g_{above}^{\prime }}=1+\frac{h}{R}$
Hence, the ratio $\frac{g_{below}^{\prime }}{g_{above}^{\prime }}$ increases linearly with $h$.