How Large Should a Solar Sail Be to Counteract the Sun's Gravitational Pull?

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SUMMARY

The discussion focuses on calculating the necessary surface area of a solar sail to counteract the Sun's gravitational pull. Given a mass of 1780.4 kg for the ship and sail, and using the radiation pressure formula, the required surface area is determined to be approximately 1.14 × 10^6 m². This calculation incorporates the Sun's energy emission rate of 3.90 × 10^26 W and the gravitational constant of 6.67 × 10^-11 N m²/kg². The result indicates that the sail must be just over one square kilometer in area to achieve equilibrium with the Sun's gravitational force.

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ft92
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Homework Statement


t has been proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil.

Assume that the mass of the ship + sail is 1780.4 kg, that the sail is perfectly reflecting, and that the sail is oriented perpendicular to the Sun's rays. How large must the surface area of the sail be if the radiation force is to be equal in magnitude to the Sun's gravitational attraction?

The rate at which the Sun emits energy is 3.90 × 1026 W. The Sun's mass is 1.99 × 1030 kg. The gravitational constant is 6.67 × 10-11 N m2/kg2.

Homework Equations


Preflect= 2Ef cos2ϑ /c

The Attempt at a Solution



Ef= (3.9×1026 W)/A = (3.9 ×1026)/ 4πR2 = 3.1×1025/R2

let surface area be Af :

force on ship= Preflect Af= (2×3.1×1025×cos20)(Af)/ 3×108×R2

F= 2.07×1017 (Af)/R2

Fgravitational at R= GMm/R2 = (6.67×10-11×1.99×1030×1780.4)/R2

let F=Fgrav =(6.67×10-11×1.99×1030×1780.4)/(2.07×1017)= 1.14 ×106m2 do you agree with my result?
 
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The result looks reasonable. Just over a kilometer square.
 
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