Question

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

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

Answer 1

Answer: (D)

$9.66\,m/s^{2}$

Answer 2

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