Question

Assuming earth to be sphere of uniform density what is the value of acceleration due to gravity at a point $100km$ below the earth surface (Given $R=6380\times 10^{3}m$)

• A. $3.10m/s$
• B. $5.06m/s^2$
• C. $7.64m/s^2$
• D. $9.66m/s^2$

Answer 1

Answer: (D)

Here : Depth $=100km=100\times 10^{3}m$
Radius of earth $R=6380\times 10^{3}m$
Acceleration due to gravity below the earth's surface is given by
$g,=g\left(1-\frac{d}{R}\right)$
$g^{\prime }=9.8\left[1-\frac{100\times 10^3}{6380\times 10^3}\right]$
$=9.8\left[1-\frac{1}{63.8}\right]$
$=9.8\times \frac{62.8}{63.8}=9.66m/s^2$