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NASA is giving serious consideration to the concept of \(solar\) \(sailing\). A solar sailcraft uses a large, low-mass sail and the energy and momentum of sunlight for propulsion. (a) Should the sail be absorbing or reflective? Why? (b) The total power output of the sun is 3.9 \(\times\) 10\(^{26}\) W. How large a sail is necessary to propel a 10,000-kg spacecraft against the gravitational force of the sun? Express your result in square kilometers. (c) Explain why your answer to part (b) is independent of the distance from the sun.

Short Answer

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(a) The sail should be reflective for momentum effect. (b) Sail area: 50 km². (c) Force dependence cancels distance effect.

Step by step solution

01

Reflective or Absorbing Sail

A solar sail should be reflective because it can effectively use the momentum of sunlight. When sunlight hits a reflective surface, it bounces off and imparts more momentum to the sail, compared to an absorbing surface where energy might be lost in heat.
02

Calculate Force from Solar Radiation

The force exerted by sunlight on a perfectly reflective surface is given by the formula: \[ F = 2\frac{P}{c} \]where \( P = 3.9 \times 10^{26} \text{ W} \) is the power output of the sun, and \( c = 3 \times 10^8 \text{ m/s} \) is the speed of light. Calculating this gives:\[ F = 2 \times \frac{3.9 \times 10^{26}}{3 \times 10^8} = 2.6 \times 10^{18} \text{ N/m}^2 \]
03

Calculate Required Sail Area

The gravitational force on the spacecraft due to the sun is given by:\[ F_g = \frac{G M m}{r^2} \]where \( G = 6.674 \times 10^{-11} \text{ Nm}^2/\text{kg}^2 \), \( M = 1.989 \times 10^{30} \text{ kg} \) (mass of sun), \( m = 10,000 \text{ kg} \) (mass of spacecraft). Since we need the light force equal to the gravitational force:\[ F = F_g = m g_s \]Then solving for the area \( A \) gives:\[ A = \frac{m g_s c}{2P} = \frac{10,000 \times 9.8 \times 3 \times 10^8}{7.8 \times 10^17} = 5 \times 10^7 \text{ m}^2 \] or \( 50 \text{ km}^2 \).
04

Independence from Distance

The answer is independent of distance from the sun because the power per unit area received from the sun, when multiplied by the area and distance from sun, cancels out the effect of distance in the calculation of physics of light pressure. The inverse square law of gravitation and radiation pressure balances, making the required sail area independent of distance.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solar Radiation Pressure
The concept of solar radiation pressure is linked to the force exerted by sunlight on objects it encounters. Sunlight is not just light; it carries energy and momentum, which can exert pressure on surfaces it hits. This pressure is used in solar sailing as propulsion for spacecraft.

When sunlight hits a surface, the radiation pressure can be calculated using the formula: \[ F = \frac{P}{c} \]where:
  • \( P \) is the power with which sunlight hits the surface (in watts), and
  • \( c \) is the speed of light (approximately \( 3 \times 10^8 \) meters per second).
A reflective surface doubles this force since light bounces back, effectively transferring more momentum. This principle helps propel solar sails through space using just sunlight.
Gravitational Force
In solar sailing, it's crucial to understand the gravitational force that acts on the spacecraft. This force is due to the sun's massive gravitational pull. It's calculated using Newton's law of universal gravitation: \[ F_g = \frac{G M m}{r^2} \]where:
  • \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \text{Nm}^2/\text{kg}^2) \),
  • \( M \) is the mass of the sun \((1.989 \times 10^{30} \text{ kg})\),
  • \( m \) is the mass of the spacecraft, and
  • \( r \) is the distance between the center of the sun and the spacecraft.
In solar sailing, the solar sail must balance this gravitational force using solar radiation pressure to effectively propel the spacecraft without fuel.
Reflective Surfaces
Reflective surfaces are key to solar sailing. A reflective sail is designed to bounce sunlight off its surface, increasing the momentum transfer compared to absorbing surfaces.

An absorbing surface would primarily turn the light energy into heat, which is not useful for propulsion. Whereas, a reflective surface sends the photons away, which means:
  • The momentum imparted by the reflected photons to the sail is doubled, enhancing the thrust.
  • This maximizes the efficiency of the sail, making solar sailing a feasible propulsion method.
This choice of surface optimizes the use of solar radiation pressure, essential for maintaining momentum and speed in space.
Momentum Transfer
Momentum transfer is a critical concept in understanding how solar sails work. When light hits a surface, it carries momentum even though it doesn't have mass. This momentum can be transferred to the object it strikes.
  • For a reflective surface, the process is more efficient because the light photons rebound, essentially "pushing" the sail twice with the same energy.
  • This continuous exchange of momentum from photons to the sail generates a constant force that propels the spacecraft forward.
Understanding momentum transfer helps explain why such lightweight sails are sufficient to move even large spacecraft despite the small force exerted by sunlight alone.

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Most popular questions from this chapter

A cylindrical conductor with a circular cross section has a radius \(a\) and a resistivity \(\rho\) and carries a constant current \(I\). (a) What are the magnitude and direction of the electricfield vector \(\vec{E}\) at a point just inside the wire at a distance \(a\) from the axis? (b) What are the magnitude and direction of the magneticfield vector \(\vec{B}\) at the same point? (c) What are the magnitude and direction of the Poynting vector \(\vec{S}\) at the same point? (The direction of \(\vec{S}\) is the direction in which electromagnetic energy flows into or out of the conductor.) (d) Use the result in part (c) to find the rate of flow of energy into the volume occupied by a length \(l\) of the conductor. (\(Hint\): Integrate \(\vec{S}\) over the surface of this volume.) Compare your result to the rate of generation of thermal energy in the same volume. Discuss why the energy dissipated in a current-carrying conductor, due to its resistance, can be thought of as entering through the cylindrical sides of the conductor.

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