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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)`)

Text Solution

Verified by Experts

The correct Answer is:
`6.94xx10^22 m^2`
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`
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