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  • The ratio of earth’s orbital angular momentum (about the sun) to its mass is <img alt="\mathrm{\text { is } 4.4 \times 10^{15} \mathrm{~m}^2 / \mathrm{s}}" src="https://entrancecorner.oncodec

The ratio of earth’s orbital angular momentum (about the sun) to its mass is \mathrm{\text { is } 4.4 \times 10^{15} \mathrm{~m}^2 / \mathrm{s}}. The area enclosed by earth’s orbit is approximately

Option: 1

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


Option: 2

6.02 \times 10^{23} \mathrm{~m}^2


Option: 3

7 \times 10^{23} \mathrm{~m}^2


Option: 4

none of these


Answers (1)

best_answer

\mathrm{\begin{aligned} \frac{L}{M_e}=4.4 \times 10^{15} & \Rightarrow \frac{M_e v r}{M_e}=4.4 \times 10^{15} \\ & \Rightarrow\left(\frac{2 \pi r}{365 \times 24 \times 60 \times 60}\right) r=4.4 \times 10^{15} \\ & \Rightarrow \pi r^2=6.938 \times 10^{22} \mathrm{~m}^2 . \end{aligned}}

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Pankaj Sanodiya

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