Question

The ratio of Earth's orbital angular momentum (about sun) to its mass is $4.4\times {10}^{15}{m}^2{s}^{-1}$. The area enclosed by the earth's orbit is approximately

• A. $5\times {10}^{22}{m}^2$
• B. $3\times {10}^{22}{m}^2$
• C. $1\times {10}^{22}{m}^2$
• D. $7\times {10}^{22}{m}^2$

Answer 1

The correct option is D $7\times {10}^{22}{m}^2$

Using Keplers law
$\frac{\Delta A}{\Delta t}=\frac{L}{2m}$
Where $\Delta A$ is area swept by the radial vector in time interval $\Delta t$
Now Area of earth's orbit is area swept by radial vector in time period of earth $\left(365days\right)$
Area of earth's orbit $=\frac{L}{2m}\times timeperiod=\frac{4.4\times {10}^{15}}{2}\times 365\times 24\times 60\times 60=7\times {10}^{22}{m}^2$