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 earth's orbit is approximately.

• A. $7\times {10}^{22}{m}^2$
• B. $6.02\times {10}^{23}{m}^2$
• C. $7\times {10}^{23}{m}^2$
• D. None of these

Answer 1

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

$\frac{L}{M_e}=4.4\times {10}^{15}$
$\Rightarrow \frac{M_{e}vr}{M_e}=4.4\times {10}^{15}.....\left(1\right)$
$\frac{GMm}{r^2}=\frac{m{v}^2}{r}\Rightarrow GM=v^{2}r$
$\Rightarrow \left(\frac{2\pi r}{365\times 24\times 60\times 60}\right)r=4.4\times {10}^{15}$
$\Rightarrow \pi r^2\approx 7\times {10}^{22}{m}^2$.
Hence $\left(A\right)$ is correct.