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

Answer

A

Explanation

According to Kepler's second law, the areal velocity of the radius vector joining a planet of mass (m) to the sun is constant i.e. `(dA)/(dt)` = constant `= (L)/(2m)`.
where L is the angular momentum of the planet.
In this case, the areal velocity of the earth
`= ("Area swept")/("Time (T) of one revolution of the earth about the sun")`
`because (dA)/(dt)=(L)/(2m)`
`therefore dA = (L xx dt)/(2m)=(1)/(2)xx((L)/(m))xx dt [dt = 24 h]`
` = (1)/(2)xx4.4xx10^(15)xx24` hour
`= 2.2xx10^(15)xx24xx60xx60xx365`
`div 7xx10^(22)m^(2)`