Question

The ratio of earth’s 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)

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 $=6.94\times 10^{22}{m}^2$