The ratio of the earth's orbital angular momentum (about the Sun) to its mass is `4.4 xx 10^(15) m^(2) s^(-1)`. The area enclosed by the earth's orbit is approximately-___________m^(2).

Options

A. `1xx10^(22) m^(2)`
B. `3xx10^(22) m^(2)`
C. `5xx10^(22)m^(2)`
D. `7xx10^(22) m^(2)`

Answer

D

Explanation

Areal velocity =Area swept`//` time for one revolution of earth about sun So, Area=(Area velocity)(time periode)
`=L/(2m)xx365xx86400`