Spacecraft Sail Force Balance

The problem asks for the area of a fully reflecting spacecraft sail required for the gravitational force exerted by the sun to be balanced by the radiation pressure force from the sun's radiation.

We need to consider two main forces acting on the spacecraft:

1. Gravitational pull of the sun.
2. Radiation pressure from the sun.

Gravitational Force on Spacecraft

The gravitational force (\(F_g\)) between the sun (mass \(M\)) and the spacecraft (mass \(m\)) at a distance \(r\) is given by Newton's law of gravitation:

\(F_g = \frac{GmM}{r^2}\)

where \(G\) is the universal gravitational constant.

Radiation Pressure Force on Sail

The sun radiates power \(P\). The intensity (\(I\)) of this radiation at a distance \(r\) from the sun is given by:

\(I = \frac{P}{4\pi r^2}\)

Radiation exerts pressure. For a fully reflecting surface oriented perpendicular to the incident radiation, the radiation pressure (\(p_r\)) is given by:

\(p_r = \frac{2I}{c}\)

where \(c\) is the speed of light.

The force (\(F_r\)) due to radiation pressure on a sail of area \(A\) is:

\(F_r = p_r \times A\)

Substituting the expression for intensity \(I\):

\(F_r = \frac{2}{c} \left(\frac{P}{4\pi r^2}\right) A = \frac{PA}{2\pi c r^2}\)

Balancing the Forces

For the gravitational pull to balance the radiation pressure on the sail, the magnitudes of the two forces must be equal:

\(F_g = F_r\)

\(\frac{GmM}{r^2} = \frac{PA}{2\pi c r^2}\)

Notice that the distance \(r\) from the sun appears on both sides of the equation and cancels out:

\(GmM = \frac{PA}{2\pi c}\)

Now, we can solve for the area \(A\) of the sail:

\(A = \frac{2\pi c GmM}{P}\)

Calculating the Sail Area

We are given the following values:

• Mass of spacecraft, \(m = 1000 \text{ kg} = 10^3 \text{ kg}\)
• Mass of Sun, \(M = 10^{30} \text{ kg}\)
• Sun's radiated power, \(P = 10^{26} \text{ W}\)
• Speed of light, \(c \approx 3 \times 10^8 \text{ m/s}\)
• Universal gravitational constant, \(G \approx 6.674 \times 10^{-11} \text{ N m}^2\text{/kg}^2\)

Let's substitute these values into the formula for \(A\):

\(A = \frac{2\pi \times (3 \times 10^8 \text{ m/s}) \times (6.674 \times 10^{-11} \text{ N m}^2\text{/kg}^2) \times (10^3 \text{ kg}) \times (10^{30} \text{ kg})}{10^{26} \text{ W}}\)

Let's group the numerical coefficients and the powers of 10:

\(A = (2\pi \times 3 \times 6.674) \times \frac{10^8 \times 10^{-11} \times 10^3 \times 10^{30}}{10^{26}} \text{ m}^2\)

\(A = (2\pi \times 3 \times 6.674) \times 10^{8 - 11 + 3 + 30 - 26} \text{ m}^2\)

\(A = (2\pi \times 3 \times 6.674) \times 10^4 \text{ m}^2\)

Calculate the numerical part:

\(2\pi \times 3 \times 6.674 \approx 6.283 \times 3 \times 6.674 \approx 18.85 \times 6.674 \approx 125.77\)

So, the area is approximately:

\(A \approx 125.77 \times 10^4 \text{ m}^2\)

We can rewrite this as:

\(A \approx 1.2577 \times 10^2 \times 10^4 \text{ m}^2\)

\(A \approx 1.2577 \times 10^6 \text{ m}^2\)

Comparing this result to the given options, the value is closest to \(10^6 \text{ m}^2\).