Is it snake oil? The amount of power that would be delivered to the ground seems quite minuscule based on that small orbital mirror size.<p>In fact it should be over three orders of magnitude lower than that of normal sunlight on the solar panel, which is roughly 1000 W per square meter.<p>Here are the calculations:<p>---<p>Assumptions:<p>Solar constant: 1366 W/m²<p>Mirror area: 100 m² (10 m x 10 m)<p>Reflectivity of aluminized Mylar: 90%<p>Atmospheric attenuation: 70% of reflected sunlight reaches Earth’s surface<p>Spot diameter on Earth: 500 meters<p>Spot area on Earth: π × (250 m)² ≈ 196,350 m²<p>Calculation:<p>Total incident power = 1366 W/m² × 100 m² = 136,600 W<p>Reflected power (after reflectivity) = 136,600 W × 0.90 = 122,940 W<p>Power reaching Earth’s surface (after atmospheric attenuation) = 122,940 W × 0.70 = 86,058 W<p>Power per square meter actually delivered at Earth’s surface = 86,058 W ÷ 196,350 m² ≈ 0.438 W/m²