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A solar sail consists of a very light, thin plain surface which can absorb all the photons incident on it. Consider such a sail is placed at a very large distance fron the sun, noormal to the incident rays. What is the mass per unit area of the sail such that it will neither be blown away or pulled in by the sun.
Given:
mean distance of the earth from the sun (semi major axis) = 1.5 X 10 Km
c=3 X 108 m/s
1 earth year = X 107 s.
Express your answer in mg/m2 and round it off to the nearest integer.

Amit Saxena , 11 Years ago
Grade upto college level
anser 1 Answers
Askiitians Tutor Team

Last Activity: 4 Hours ago

To determine the mass per unit area of a solar sail that will maintain a balance between the gravitational pull of the Sun and the pressure exerted by sunlight, we need to analyze the forces acting on the sail. The gravitational force pulling the sail towards the Sun must equal the radiation pressure pushing it away. Let's break this down step by step.

Understanding the Forces

1. **Gravitational Force**: The gravitational force acting on the sail can be calculated using Newton's law of gravitation. The formula is:

F_gravity = G * (M_sun * m_sail) / r^2

Where:

  • G is the gravitational constant, approximately 6.674 × 10-11 N(m/kg)2.
  • M_sun is the mass of the Sun, about 1.989 × 1030 kg.
  • m_sail is the mass of the sail, which we can express as the product of the mass per unit area (σ) and the area (A).
  • r is the distance from the Sun to the sail, which is approximately 1.5 × 1011 m (the mean distance of the Earth from the Sun).

2. **Radiation Pressure**: The pressure exerted by sunlight on the sail can be calculated using the formula:

P_radiation = I / c

Where:

  • I is the intensity of sunlight at the distance of the Earth, approximately 1361 W/m2.
  • c is the speed of light, about 3 × 108 m/s.

The force due to radiation pressure can then be calculated as:

F_radiation = P_radiation * A

Setting the Forces Equal

For the sail to remain stationary, these two forces must be equal:

F_gravity = F_radiation

Substituting the expressions we derived:

G * (M_sun * σ * A) / r^2 = (I / c) * A

We can cancel the area (A) from both sides, as long as A is not zero:

G * (M_sun * σ) / r^2 = I / c

Solving for Mass per Unit Area

Now, we can rearrange this equation to solve for σ:

σ = (I / c) * (r^2 / (G * M_sun))

Substituting the known values:

σ = (1361 W/m2 / (3 × 108 m/s)) * ((1.5 × 1011 m)2 / (6.674 × 10-11 N(m/kg)2 * 1.989 × 1030 kg))

Calculating Each Component

1. Calculate the radiation pressure:

P_radiation = 1361 / (3 × 108) ≈ 4.537 × 10-6 N/m2

2. Calculate the gravitational force component:

G * M_sun ≈ 6.674 × 10-11 * 1.989 × 1030 ≈ 1.327 × 1020 m3/s2

3. Now plug these values into the equation for σ:

σ = (4.537 × 10-6) * ((1.5 × 1011)2 / (1.327 × 1020))

4. Calculate the area term:

(1.5 × 1011)2 = 2.25 × 1022 m2

5. Now substitute back:

σ = (4.537 × 10-6) * (2.25 × 1022 / 1.327 × 1020)

σ ≈ (4.537 × 10-6) * (16.95) ≈ 7.686 × 10-5 kg/m2

Final Conversion to mg/m²

To express this in mg/m², we convert kilograms to milligrams (1 kg = 1,000,000 mg):

σ ≈ 7.686 × 10-5 kg/m2 * 1,000,000 mg/kg ≈ 76.86 mg/m2

Rounding this to the nearest integer gives us:

Mass per unit area of the sail ≈ 77 mg/m2

This means that for the solar sail to maintain its position without being pulled in or blown away by the Sun, it should have a mass per unit area of approximately 77 mg/m².

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