Question

The ratio of earth?? orbital angular momentum (about the sun) to.its mass is $4.4 \times 10^{15}\, m^2/s$. The area enclosed by earths orbit approximately is (in $m^2$)

• A. $6.94 \times 10^{22}$
• B. $6.94 \times 10^{23}$
• C. $7.94 \times 10^{22}$
• D. $7.94 \times 10^{23}$

Answer 1

Answer: (A)

$6.94 \times 10^{22}$

Answer 2

Let $L =$ Angular momentum of earth about sun. $M =$ Mass of earth $\therefore$ By Keplers law, $\frac{dA}{dt} = \frac{L}{2M}$ or $dA = \frac{L}{2M} dt$ The earth completes its orbital journey in $365$ days. $\therefore$ Area $A = \frac{L}{2M}\times T = \frac{1}{2}\left(\frac{L}{M}\right)T$, where $T = 365$ days. or $A = \frac{\left(4.4 \times 10^{15}\right)\times 365 \times 24\times 60\times 60}{2} m^{2}$ or Area $= 6.94 \times 10^{22}\, m^{2}$ $\therefore$ Area enclosed by earths orbit $= 6.94 \times 10^{22}\, m^{2}$