Question

If the earth is a point mass of $6\times {10}^{24}kg$ revolving around the sun at a distance of $1.5\times {10}^{8}km$ and in time of $T=3.14\times {10}^7$ seconds, then the angular momentum of the earth around the sun is :-

• A. $1.2\times {10}^{18}kg-m^2/s$
• B. $1.8\times {10}^{20}kg-m^2/s$
• C. $1.5\times {10}^{37}kg-m^2/s$
• D. $2.7\times {10}^{40}kg-m^2/s$

Answer 1

The correct option is D $2.7\times {10}^{40}kg-m^2/s$

Given,

$m=6\times {10}^{24}kg$

$r=1.5\times {10}^{8}km$

$T=3.14\times {10}^{7}sec$

angular velocity, $\omega =\frac{2\pi }{T}$

$\omega =\frac{2\times 3.14}{3.14\times {10}^7}=2\times {10}^{-7}rad/s$

Angular momentum,

$L=m\omega r^2$

$L=6\times {10}^{24}\times 2\times {10}^{-7}\times (1.5\times {10}^8\times {10}^3{)}^2$

$L=2.7\times {10}^{40}kg{m}^2/s$

The correct option is D.