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Q.
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 $m^2$)
Gravitation
Solution:
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 $= 6.94 \times 10^{22}\, m^{2}$