If the earth is a point mass of `6 xx 10^(24) kg` revolving around the sun at a distance of `1.5 xx 10^(8) `km and in time, `T = 3.14 xx 10^(7)s`, then the angular momentum of the earth around the sun is

Options

A. `1.2 xx 10^(18) "kg m"^(2)//s`
B. `1.8 xx 10^(29) "kg m"^(2)//s`
C. `1.5 xx 10^(37) "kg m"^(2)//s`
D. `2.7 xx 10^(40) "kg m"^(2)//s`

Answer

D

Explanation

Angular momentum `L = mvr = m omega r^(2) = m xx (2pi)/(T)xxr^(2)`
`=(2xx3.14xx6xx10^(24)xx(1.5xx10^(11))^(2))/(3.14xx10^(7))=2.7xx10^(40)" kg-m"^(2)" s"^(-1)`