Question

I need PARTS (e), (f), and (g) please NOT (a), (b), (c), and (d)

again, PARTS (e), (f), and (g) please

The intensity of the sun on earth is 1300 W/m2. If this intensity falls on a black surface the size of a football field (what is it's size) (a) what is the radiation pressure on it? (b) what is the force on the football field? (c) Is this force sufficient to levitate a tennis ball (50g) against the earth's gravitation? (d) How large must the black surface be to achieve this? Assume the area of football field to be 110 m x 50 m (you know it certainly to much higher precision than I do) (e) you use a high power constant-wave (CW) laser used in military operations with an output of 100 kW into a circular beam size of diameter 2 cm. What is its radiation pressure? (f) you have a reflective surface of mass 3 g. On earth, can you then shoot this up with this laser? (g) how much laser power would be necessary to levitate the 3 g mass?

Answer 1

Power of the laser is P = 100 kw = 100 times 10^3 w diameter of the beam is d = 2 cm = 2 times 10^-2 m cross - suction of the beam is A = pi (d/2)^2 = pi d^2/4 A = 3.14 times (2 times 0^-2)^2/4 A = 3.14 times 10^-4 m^2 intensity of the laser beam is I = P/A = 100 times 10^3/3.14 times 10^-4 I = 31.847 times 0^7 w/m^2 \r\n now radiation pressure is P_rad = I/c where c be the speed of light P_rad = 31.847 times 10^7/3 times 10^8 P_rad = 1.061 N/m^2 mass of the reflective surface is m = 3g = 3 times 10^-3 kg we know that Pressure = Force/area P_rad = F/A \r\n F = P_rad A this is the force on the reflective surface F = 1.061 times 3.14 times 10^-4 F = 3.331 times^10-4 N wright of the surface is w = mg = 3 times 10^-3 times 9.8 w =