Question

It has been proposed that a spaceship might be propelled in the solar system by radiation pressure, using a large sail made of foil. How large must the surface area of the sail be if the radiation force is to be equal to the Sun's gravitational attraction? Assume that the mass of the ship + sail is 1300 kg, that the sail is perfectly reflecting, and that the sail is oriented perpendicular to the Sun's rays. (With a large sail, the ship is continuously driven away from the Sun.)

Answer 1

A = 8.34 x 10^(5) m²

Step-by-step explanation:

The intensity of the solar radiation is the average solar power per unit area. Thus,

I = P/4A = P/(4(πr²))

Where;

P is average solar power.

r is it's distance from centre of sun

A is area

Now, The rate at which the Sun emits energy has a standard value of 3.90 × 10^(26) W

Thus;

I = [3.90 × 10^(26)]/(4πr²)

I = [3 x 10^(25)]/r²

Now, the formula for radiation pressure is;

P_rad = 2I/c

Where c is speed of light and has a value of 3 x 10^(8) m/s

Thus,

P_rad = 2([3 x 10^(25)]/r²)/(3 x 10^(8))

P_rad = [2.07 x 10^(17)]/r² N/m²

Also, Radiation pressure on ship; P_rad = F/A

Where Force on ship and A is area.

Thus;

F = P_rad x A

So, F = [2.07 x 10^(17)•A]/r²

Now,

F_grav = GMm/r²

Where;

G is gravitational constant with a value = 6.67 x 10^(-11) Nm²/kg²

M is mass of sun with a value of 1.99 x 10^(30)

m is mass of ship and sail = 1300 kg

Thus, plugging in the relevant values to obtain;

F_grav = (6.67×10^(-11) × 1.99 x 10^(30) × 1300)/r²

F_grav = [17.255 x 10^(22)]/r²

Now, equating F to F_grav, we get;

[2.07 x 10^(17)•A]/r² = [17.255 x 10^(22)]/r²

r² will cancel out to give;

2.07 x 10^(17)•A= [17.255 x 10^(22)]

A = [17.255 x 10^(22)]/2.07 x 10^(17)

A = 8.34 x 10^(5) m²