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The ratio of earth's orbital angular mom...

The ratio of earth's orbital angular momentum (about the sun ) to its mass is `4.4xx10^(15) m^(2)//s`. The area enclosed by earth's orbit approximately is (in `m^(2)`)

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The correct Answer is:
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Areal velocity of a planet around the sun is constant and is given by
`(dt)/(dt)=L/(2m)impliesdA=L/(2m)dt`
Integrating both sides `int dA=L/(2m)intdtimpliesA=L/(2m)/T`
where `L=` angular momentum of the planet (earth abut the sun and `m=` mass of planet (earth).
Hence `A=L/(2m)T=1/2xx4.4xx10^(15)xx365xx24xx3600m^(2)`
Area `=6.94xx10^(22)m^(2)`
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