Question

(A) Find the height from the earth’s surface where g will be 25% of its value on the surface of earth.
($$R=6400km$$)
(B) Find the percentage increase in the value of g at a depth h from the surface of earth.

Answer 1

Hint: Use the formula for acceleration due to gravity at height h from the surface of the earth. Substitute $${\displaystyle \frac{g}{4}}$$ for acceleration due to gravity at height h and solve the equation for h. For the second part, take the ratio of difference in the acceleration due to gravity at the surface of the earth and acceleration due to gravity at given depth to the acceleration due to gravity at the surface of the earth to determine the percentage decrease in the acceleration due to gravity.

Formula used:
$$g_h=g{\left({\displaystyle \frac{R}{R+h}}\right)}^2$$
Here, $$g_h$$ is the acceleration due to gravity at height h from the surface of the earth, g is the acceleration due to gravity at the surface of the earth, and R is the radius of the earth.
$$g_d=g\left(1-{\displaystyle \frac{h}{R}}\right)$$
Here, $$g_d$$ is the acceleration due to gravity at depth h below the earth’s surface.

Complete step by step answer:
(A) We know the formula for variation of acceleration due to gravity at with altitude h.
$$g_h=g{\left({\displaystyle \frac{R}{R+h}}\right)}^2$$
Here, $$g_h$$ is the acceleration due to gravity at height h from the surface of the earth, g is the acceleration due to gravity at the surface of the earth, and R is the radius of the earth.
We have given the acceleration due to gravity at height h is 25%. Therefore, $$g_h=25\mathrm{\%}g={\displaystyle \frac{g}{4}}$$.
Substitute $${\displaystyle \frac{g}{4}}$$ for $$g_h$$ in the above equation.
$${\displaystyle \frac{g}{4}}=g{\left({\displaystyle \frac{R}{R+h}}\right)}^2$$
$$\Rightarrow {\displaystyle \frac{1}{4}}={\left({\displaystyle \frac{R}{R+h}}\right)}^2$$
Take the square root of the above equation, we get,
 $${\displaystyle \frac{R}{R+h}}={\displaystyle \frac{1}{2}}$$
$$\Rightarrow 2R=R+h$$
$$\Rightarrow h=R$$
$$\therefore h=6400km$$
Therefore, at height 6400 km, the acceleration due to gravity is 25% of the acceleration due to gravity at the surface of the earth.
(B) We know the formula for the variation of the acceleration due to gravity with depth below the earth’s surface.
$$g_d=g\left(1-{\displaystyle \frac{h}{R}}\right)$$
Here, $$g_d$$ is the acceleration due to gravity at depth h below the earth’s surface.
Rearrange the above equation as follows,
$$g_d=g-g{\displaystyle \frac{h}{R}}$$
$$\Rightarrow g-g_d=g{\displaystyle \frac{h}{R}}$$
$$\therefore {\displaystyle \frac{g-g_d}{g}}={\displaystyle \frac{h}{R}}$$
In the above equation, $${\displaystyle \frac{g-g_d}{g}}$$ is the decrease in the acceleration due to gravity with respect to the acceleration due to gravity at the surface of the earth.
Since in the section (A), we have calculated $$h=R$$, the above equation becomes,
$${\displaystyle \frac{g-g_d}{g}}={\displaystyle \frac{R}{R}}$$
$${\displaystyle \frac{g-g_d}{g}}=1$$
The percentage decrease in the acceleration due to gravity is,
$$\left({\displaystyle \frac{g-g_d}{g}}\right)\mathrm{\%}=1\times 100$$
$$\left({\displaystyle \frac{g-g_d}{g}}\right)\mathrm{\%}=100\mathrm{\%}$$
Thus, the acceleration due to gravity at height equal to radius of the earth, the decrease in the acceleration due to gravity is 100% that is $$0m/s^2$$.

Note:
In the section (B). it is incorrectly mentioned an increase in the acceleration due to gravity.
The acceleration due to gravity does not increase with the depth rather it decreases with the depth.