A possible means of space flight is to place a perfectly reflecting aluminized sheet into orbit around the Earth and then use the light from the Sun to push this "solar sail." Suppose a sail of area A = 7.20 ✕ 105 m2 and mass m = 6,600 kg is placed in orbit facing the Sun. Ignore all gravitational effects and assume a solar intensity of 1,370 W/m2.(a)What force (in N) is exerted on the sail? (Enter the magnitude.) N(b)What is the sail's acceleration? (Enter the magnitude in µm/s2.) µm/s2(c)Assuming the acceleration calculated in part (b) remains constant, find the time interval (in days) required for the sail to reach the Moon, 3.84 ✕ 108 m away, starting from rest at the Earth. days(d)What If? If the solar sail were initially in Earth orbit at an altitude of 380 km, show that a sail of this mass density could not escape Earth's gravitational pull regardless of size. (Calculate the magnitude of the gravitational field in m/s2.) m/s2(e)What would the mass density (in kg/m2) of the solar sail have to be for the solar sail to attain the same initial acceleration as that in part (b)? kg/m2

(a) The force exerted on the sail can be calculated using the formula for radiation pressure, which is given by P = 2I/c, where I is the intensity of the light and c is the speed of light. The force is then given by F = PA, where A is the area of the sail.

Given that I = 1370 W/m^2, c = 3.00 x 10^8 m/s, and A = 7.20 x 10^5 m^2, we can substitute these values into the formula to get:

P = 2(1370 W/m^2) / (3.00 x 10^8 m/s) = 9.13 x 10^-6 N/m^2

F = (9.13 x 10^-6 N/m^2) * (7.20 x 10^5 m^2) = 6.57 N

(b) The acceleration of the sail can be calculated using Newton's second law, F = ma, where m is the mass of the sail and a is its acceleration. Rearranging for a gives a = F/m.

Given that F = 6.57 N and m = 6600 kg, we can substitute these values into the formula to get:

a = 6.57 N / 6600 kg = 9.95 x 10^-4 m/s^2 = 995 µm/s^2

(c) The time required for the sail to reach the Moon can be calculated using the formula for the distance travelled under constant acceleration, d = 0.5at^2, where d is the distance and t is the time. Rearranging for t gives t = sqrt(2d/a).

Given that d = 3.84 x 10^8 m and a = 9.95 x 10^-4 m/s^2, we can substitute these values into the formula to get:

t = sqrt(2 * 3.84 x 10^8 m / 9.95 x 10^-4 m/s^2) = 8.82 x 10^6 s

Converting this to days gives t = 8.82 x 10^6 s / (60 s/min * 60 min/h * 24 h/day) = 102 days

(d) The gravitational field strength at an altitude of 380 km can be calculated using the formula g = GM/r^2, where G is the gravitational constant, M is the mass of the Earth, and r is the distance from the center of the Earth.

Given that G = 6.67 x 10^-11 N(m/kg)^2, M = 5.97 x 10^24 kg, and r = 6.37 x 10^6 m + 380 x 10^3 m = 6.75 x 10^6 m, we can substitute these values into the formula to get:

g = (6.67 x 10^-11 N(m/kg)^2 * 5.97 x 10^24 kg) / (6.75 x 10^6 m)^2 = 8.87 m/s^2

(e) The mass density of the solar sail required to attain the same initial acceleration as in part (b) can be calculated using the formula ρ = F/(gA), where ρ is the mass density, F is the force, g is the gravitational field strength, and A is the area of the sail.

Given that F = 6.57 N, g = 8.87 m/s^2, and A = 7.20 x 10^5 m^2, we can substitute these values into the formula to get:

ρ = 6.57 N / (8.87 m/s^2 * 7.20 x 10^5 m^2) = 1.05 x 10^-6 kg/m^2