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  • The ratio of Earth’s orbital angular momentum (about the Sun) to its mass is  <img alt="

The ratio of Earth’s orbital angular momentum (about the Sun) to its mass is 4.4 \times 10^{15} \mathrm{~m}^2 / \mathrm{s}. The area enclosed by Earth’s orbit approximately ………….. m2

Option: 1

5.94 \times 10^{22}


Option: 2

6.24 \times 10^{22}


Option: 3

6.74 \times 10^{22}


Option: 4

6.94 \times 10^{22}


Answers (1)

best_answer

Areal velocity of a planet round the sun is constant and is given by

\mathrm{\frac{d A}{d t}=\frac{L}{2 m}}

where L = Angular momentum of planet (earth) about sun and m = mass of planet (earth)

Given : \mathrm{\frac{L}{m}=4.4 \times 10^{15} \mathrm{~m}^2 / \mathrm{s}}

 Area enclosed by earth in time T (365 = days) will be

\mathrm{\text { Area }=\frac{d A}{d t} \cdot T=\frac{L}{2 m} \cdot T}

\mathrm{=\frac{1}{2} \times 4.4 \times 10^{15} \times 365 \times 24 \times 3600 \mathrm{~m}^2}

\mathrm{\text { Area } \approx 6.94 \times 10^{22} \mathrm{~m}^2}

Posted by

Rakesh

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