A spherical dust grain of radius 0.50 µm and density 1000 kg/m³ is at the Earth's distance from the…

Step 1. The force of light. $\pi a^2 = 7.854 \times 10^{-13}\ \text{m}^2$. So $F_L = \dfrac{I\pi a^2}{c} = \dfrac{1361 \times 7.854 \times 10^{-13}}{3.00 \times 10^{8}}\ \text{N} = 3.56 \times 10^{-18}\ \text{N}$.

Step 2. The mass and the pull of gravity. $m = \rho\tfrac{4}{3}\pi a^3 = 5.236 \times 10^{-16}\ \text{kg}$. In SI units, $GM_S = 1.327 \times 10^{20}$ and $r^2 = 2.238 \times 10^{22}$. Then $F_G = \dfrac{GM_Sm}{r^2} = \dfrac{1.327 \times 10^{20} \times 5.236 \times 10^{-16}}{2.238 \times 10^{22}}\ \text{N} = 3.11 \times 10^{-18}\ \text{N}$.

Step 3. The ratio. $\dfrac{F_L}{F_G} = 1.15$. The light wins for this grain.

Step 4. Equal forces. Setting $F_L = F_G$ gives $a = \dfrac{3Ir^2}{4cGM_S\rho}$, which is $0.574\ \mu\text{m}$.