Question

The ratio of the 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 $\underset{–}{}{m}^2$.

• A. $6.94\times {10}^{22}{m}^2$
• B. $4.24\times {10}^{22}{m}^2$
• C. $4.92\times {10}^{22}{m}^2$
• D. $5.54\times {10}^{22}{m}^2$

Answer 1

The correct option is A $6.94\times {10}^{22}{m}^2$

Areal velocity of a planet around the Sun is constant and is given by
$\frac{dA}{dt}=\frac{L}{2m}\Rightarrow dA=\frac{L}{2m}dt$
Integrating both sides $\int dA=\frac{L}{2m}\int dt\Rightarrow A=\frac{L}{2m}\cdot T$
where $L=$ angular momentum of the planet (earth) about the Sun and $m=$ mass of planet (earth).
Hence, $A=\frac{L}{2m}T=\frac{1}{2}\times 4.4\times {10}^{15}\times 365\times 24\times 3600m^2$
$Area=6.94\times {10}^{22}{m}^2$