One possible means of achieving space flight is to place a perfectly reflecting

Question

One possible means of achieving space flight is to place a perfectly reflecting aluminized sheet into Earth 's orbit and to use the light from the Sun to push this solar sail. Suppose such a sail, of area 5.20 104 m2 and mass 6500 kg, is placed in orbit facing the Sun. Ignore all gravitational effects, and assume a solar intensity of 1340 W/m2. (Hint: The radiation pressure by a reflected wave is given by 2 (average power per area)/c.)
(a) What force is exerted on the sail?
F = __?__ N

(b) What is the sail's acceleration?
a = __?__ m/s2

(c) How long does it take for this sail to reach the Moon, 3.84 108 m away?
t = __?__ days

Explanation / Answer

      Given:       The area of the sail, A = 5.20x104 m2       The mass of the sail, m = 6500 kg       The solar intensity, I = 1340 W/m2       The speed of the light, c = 3x108 m/s       The speed of the light, c = 3x108 m/s ---------------------------------------------------------------------- Solution: a)       The radiation pressure is                      P = 2I/c                         = (2)(1340 W/m2)/(3x108 m/s)                         = 893.33x10-8 N/m2       The force F realated with the The radiation pressure P is                         = (2)(1340 W/m2)/(3x108 m/s)                         = 893.33x10-8 N/m2       The force F realated with the The radiation pressure P is                      F = PA                          = (893.33x10-8 N/m2)(5.20x104 m2)                          = 0.46453 N ----------------------------------------------------------------------- b)       According to Newton's second law of motion, the force       can be deffined as                        F = ma
      Therefore, the acceleration of the sail's is                         a = F/m                            = (0.46453 N)/(6500 kg)                            = 7.1467x10-5 m/s2 ------------------------------------------------------------------------ c)       The distance, s = 3.84x108 m       Using the kinematic relation,                            s = ut + (1/2)at2                                s = 0 + (1/2)at2         Therefore, the time taken by the sail to reach the moon is                            t = 2s/a                               = (2)(3.84x108 m)/(7.1467x10-5 m/s2)                               = 3278150.188 s                               = (3278150.188 s)(1 day/86400 s)                               = 37.94 days                               = 38.00 days (nearly)                            = (0.46453 N)/(6500 kg)                            = 7.1467x10-5 m/s2 ------------------------------------------------------------------------ c)       The distance, s = 3.84x108 m       Using the kinematic relation,                            s = ut + (1/2)at2                                s = 0 + (1/2)at2         Therefore, the time taken by the sail to reach the moon is                            t = 2s/a                               = (2)(3.84x108 m)/(7.1467x10-5 m/s2)                               = 3278150.188 s                               = (3278150.188 s)(1 day/86400 s)                               = 37.94 days                               = 38.00 days (nearly)                               = (2)(3.84x108 m)/(7.1467x10-5 m/s2)                               = 3278150.188 s                               = (3278150.188 s)(1 day/86400 s)                               = 37.94 days                               = 38.00 days (nearly)                               = 38.00 days (nearly)

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