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

AI Thread Summary
To counteract the Sun's gravitational pull using a solar sail, the surface area must be calculated based on the balance between radiation pressure and gravitational force. Given the mass of the ship and sail at 1780.4 kg, the required surface area is derived from the radiation force equation, resulting in a value of approximately 1.14 × 10^6 m². This calculation considers the Sun's energy emission rate and gravitational constant. The proposed sail size is just over one square kilometer, which seems reasonable for practical applications. The discussion confirms the feasibility of using a large, perfectly reflecting sail for propulsion in the solar system.
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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