Question

The ratio of Earth's orbital angular momentum (about the 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. $1\times {10}^{22}{m}^2$
• B. $3\times {10}^{22}{m}^2$
• C. $5\times {10}^{22}{m}^2$
• D. $7\times {10}^{22}{m}^2$

Answer 1

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

Area enclosed by a body in an orbit is given by: $\frac{dA}{dt}=\frac{L}{2m}$
Total orbital area enclosed by the earth is the area covered in one revolution i.e 365days.
$\therefore A=\frac{L}{2m}365\times 24\times 3600=\frac{4.4\times {10}^{15}}{2}365\times 86400=6.938\times {10}^{22}$ m${}^2$