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This is the question:

NASA is giving serious consideration to the concept of solar sailing. A solar sailcraft uses a large,
low mass sail and the energy-momentum of sunlight for propulsion.
(a) Should the sail be absorbing or reflective?
(b) The total power output of the sun is 3.9x1026 W. How large a sail is nececary to propel a
10,000 kg spacecraft against the gravitational force of the sun? Express your result in square
kilometers.
(c) Explain why your answer to part (b) is independent of the distance from the sun?

You will need to use F=GMm/r^2 as the expression for gravitational force.

How to do it!!!

Btw, the answer to (a) is reflective.

2007-11-03 21:36:37 · 1 answers · asked by Anonymous in Science & Mathematics Physics

1 answers

I would suggest this approach.

The light energy is given by E = h*f, where h = Planck's constant and f = light frequency. The energy impacting the sail Es is the area of the sail divided by the area of the sphere at the distance r from the sun, times the energy output of the sun:

Es = As/(4πr^2) * W*∆t

W = power output of the sun, ∆t is the duration of light impact on the sail

The momentum of the photons reaching the sail is given by p = h*f/c; since h*f = E, p = E/c. Therefore the momentum of the light reaching the sail is

p = Es/c = As/(4πr^2) * W*(∆t/c)

The momentum tranferred to the sail is 2p because of the reflection, and force is change in momentum with time, so the force on the sail is

2p/∆t = As*W/(c*2πr^2); this must equal the gravitational force, so

As*W/(c*2πr^2) = GMm/r^2

The radius (distance from the sun) cancels out leaving

As = 2πcGMm/W

Both the gravitational attraction and the radiation force decrease as 1/r^2, so the distance doesn't matter.

For

c = 3*10^8 m/sec
G = 6.67*10^-11 m^3/(kg*sec^2)
M = 2*10^30 kg
m = 10^4 kg
W = 3.9*10^26 joule/sec

A = 6.45*10^6 m^2 or 6.45 km^2

2007-11-03 21:17:38 · answer #1 · answered by gp4rts 7 · 0 0

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