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).

Answer

`6.94xx10^22 m^2`

Explanation

Areal velocity of a planet around the Sun is constant and is given by
`(dA)/(dt)=L/(2m) rArr dA=L/(2m)dt`
Integrating both sides `intdA=L/(2m) intdt rArr A=L/(2m)(T)`
Where L = angular momentum of the planet (earth) about the Sun and m= mass of planet (earth).
Hence, `A=L/(2M)T =1/2 xx4.4xx10^15xx365xx24xx3600 m^2` , Area = `6.94xx10^22 m^2`