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  • The ratio of Earth's orbital angular momentum (about the Sun) to its mass is <img alt="4.4 \times10^{15} \mathrm{~m}^2 \mathrm{~s}^{-1}" src="https://entrancecorner.oncodecogs.com/gif.latex?4.4

The ratio of Earth's orbital angular momentum (about the Sun) to its mass is 4.4 \times10^{15} \mathrm{~m}^2 \mathrm{~s}^{-1}. The area enclosed by the earth's orbit is approximately.
 

Option: 1

1 \times 10^{22} \mathrm{~m}^2

 


Option: 2

3 \times 10^{22} \mathrm{~m}^2
 


Option: 3

5 \times 10^{22} \mathrm{~m}^2
 


Option: 4

7 \times 10^{22} \mathrm{~m}^2


Answers (1)

best_answer

\mathrm{{Since\: Areal \: velocity =\frac{\text { Area Swept }}{\text { Time for one Revolution of\: Earth about the sun }} \text{}}}


\mathrm{ \text { Further Areal velocity }=\frac{\mathrm{L}}{2 \mathrm{M}} }

\mathrm{ \Rightarrow \text { Area Swept }=\left(\frac{\mathrm{L}}{2 \mathrm{M}}\right)\left(\begin{array}{c} \text { Time for one } \\ \text { Revolution of Earth } \\ \text { about the sun } \end{array}\right) }

\mathrm{ \text { Area Swept }=\frac{1}{2}\left(4.4 \times 10^{15}\right)(365 \times 24 \times 60 \times 60) }

\mathrm{ \Rightarrow \text { Area Swept }=7 \times 10^{22} \mathrm{~m}^2 }

Hence option 4 is correct.



 

Posted by

Deependra Verma

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