The ratio of Earth's orbital angular momentum (about the Sun) to its mass is 4.4 × 10¹⁵ m²s^–1. The area enclosed by the earth's orbit is approximately.

7 × 10²² m²

Since Areal velocity = $\frac{\text{Area Swept}}{\text{Time for one Revolution of Earth about the sun}}$

Further Areal velocity = $\frac{L}{2 M}$

⇒ Area Swept = $\left(\frac{L}{2 M}\right)\left(\begin{array}{c}\text { Time for one } \\ \text { Revolution of Earth } \\ \text { about the sun }\end{array}\right)$

Area Swept = $\frac{1}{2}\left(4.4 \times 10^{15}\right)(365 \times 24 \times 60 \times 60)$

⇒ Area Swept = $7 \times 10^{22} ~m^2$