Question

Assuming the earth to be a uniform sphere of radius $6400$km and density $5.5$ g/c.c, find the value of g on its surface. $G=6.66\times {10}^{-11}N{m}^{2}k{g}^{-2}$.

• A. $3.82m{s}^{-2}$.
• B. $9.82m{s}^{-2}$.
• C. $19.82m{s}^{-2}$.
• D. $2m{s}^{-2}$.

Answer 1

Answer: (B)

$9.82m{s}^{-2}$.

Here, $R=6400\times {10}^{3}m=6.4\times {10}^{6}m$
$\rho =5.5g/c.c.=5.5\times {10}^3$ $kg/m^3$
Now, $g=\frac{GM}{R^2}=\frac{G}{R^2}\times \frac{4}{3}\pi R^3\times \rho$
$=\frac{4}{3}\pi GR\rho$
$=\frac{4}{3}\times \frac{22}{7}\times 6.66\times {10}^{-11}\times 6.4\times {10}^6\times 5.5\times {10}^3$
$=9.82m{s}^{-2}$.