The distance between Sum and Earth is `1.6xx10^(11)` m and the radius of Earth is `6.4xx10^(6)`m. The ratio of the angular momentum of Earth around the Sum to the angular momentum around its own axis is approximately (Assume Earth as a solid sphere with uniform mass density and rotates around the Sum in a circular path.)

Options

A. `2.0xx10^(2)`
B. `5.1xx10^(8)`
C. `4.3xx10^(6)`
D. `8.7xx10^(12)`

Answer

C

Explanation

Given, the distance between the Sum and Earth, `r_(1)=1.6xx10^(11)m`
Radius of earth `R_(e)=6.4xx10^(6)m` Angular momentum of Earth around the Sum,
`L_(1)=M_(e)v_(1)r_(1)=M_(e).(2pir_(1))/(T_(1)).r_(1)`
`L_(1)=2xx314xx6.0xx10^(24)xx((1.6xx10^(11))^(2))/(365xx24xx60xx60)`
`("":'` Mass of earth, `M_(e)=6.0xx10^(24)kgandT_(1)=365` days)
`L_(1)=3.06xx10^(40)kg.m^(2)//s`
Angular momentum of Earth about its own axis, where, moment `L_(2)=lomega` of inertia of Earth (uniform solid sphere) around its own axis, `I=(2)/(5)M_(e)R_(e)^(2)`.
or `l=(2)/(5)xx6.0xx10^(24)xx(6.4xx10^(6))^(2)`
`=9.8xx10^(37)kg.m^(2)`
The earth's angular velocity, `omega=(2pi)/(1day)`
`=(2pi)/(24xx60xx10)=7.3xx10^(-5)s^(-1)`
`:.L_(2)=9.8xx10^(37)xx7.3xx10^(-5)~~7.2xx10^(33)kg.m^(2)s^(-1)`
From Eqs. (i) and (ii), we get
`(L_(1))/(L_(2))=(3.06xx10^(40))/(7.2xx10^(33))=4.3xx10^(6)`
The ratio of angular momentum of Earth around the Sum to the angular momentum around its own axis is `4.3xx10^(6)` approximately. Hence, the correct option is (c)