Energy and gravitation problem

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NASA's exploration of solar sailing involves using light and particle momentum from the sun to propel spacecraft with large sails. To determine the minimum thickness of a sail that balances gravitational attraction, calculations must consider the sail's density and the solar radiation intensity at Earth's orbit. The gravitational force is derived from the mass of the sail and the distance from the sun, while the radiation pressure from sunlight provides the necessary counterforce. The area of the sail plays a crucial role, as both mass and force are proportional to it, allowing for simplifications in the equations. Understanding these dynamics is essential for effective solar sail design and operation.
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Homework Statement


NASA is considering solar sailing: using the momentum of light and of massive particles emitted from the sun to help push a spacecraft equipped with large, diaphanous sails. Assume that the density of the material from which the sails are made is about 1000kg.m^{-3}.

a) What is the minimum thickness of a sail such that the use of the momentum from the sun's light is enough to balance the gravitational attraction (of the sail alone) towards the sun? At the Earth's orbit (150 million km), the "solar constant" or intensity of solar radiation is: 1.4kW.m^{-2}.

m_{sun} = 1.99E30, G = 6.67E-11

b) How does the answer depend on distance from the sun?


Homework Equations



Force = Gmm/r^2
Volume = Area x Thickness
Density = Mass/Volume

The Attempt at a Solution



I'm not entirely sure how to start this. I guess I would need to find out the Area somehow, as well as the mass of the sail, that could help determine the gravitational attraction.
 
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Consider the area 1 m^2.

ehild
 
You might also want to investigate "radiation pressure".
 
ehild said:
Consider the area 1 m^2.

ehild

Where did you get that from?
 
The given data are the density of the sail material and the intensity of the light. So both mass and force are directly proportional to the area of the sail. The area will cancel from the equation of equilibrium, just like it cancels in the equation of planetary motion: it does not matter if a fly orbits around a planet or a big space station, they move with the same velocity at the same orbit. If you do not like 1m2 for the area, just call it A.

ehild
 
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