Book edition 10th Edition

Author(s) David Halliday

Pages 1328 pages

ISBN 9781118230718

Answers without the blur.

Just sign up for free and you're in.

Short Answer

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 in magnitude to the Sun’s gravitational attraction? Assume that the mass of the ship + sail is 1500 kg, that the sail is perfectly reflecting, and that the sail is oriented perpendicular to the Sun’s rays. See Appendix C for needed data. (With a larger sail, the ship is continuously driven away from the Sun.)

The surface area of the sail, $A = 9.5 \times 10^{5} m^{2}$ .

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Given data

Mass of ship and sail, m=1500kg .

Sail is perfectly reflected.

Step 2: Determining the concept

Here, we need to use the concept of force developed due to the radiation pressure. The force acting on a perfectly reflective surface of the area as a result of incident radiation of intensity I is given by $F \ user 1 = 2 IAc$ , where $c$ is the speed of light.

Formulae are as follows:

For purely reflecting surface, the radiation pressure, $p = \frac{2 I}{c}$ .

Where c is the speed of light, and I is the intensity of radiation.

Attractive force due to Sun on mass m is, $F = \frac{G m M_{s}}{d^{2}}$

Where M_S is the mass of the sun, F is the force, d is the distance.

Step 3: (a) Determining the surface area of the sail

The spaceship is propelled in the solar system by the radiation pressure, and at the same time, it’s attracted by the sun. So write force equations as follows:

$F_{s} = \frac{G m M_{s}}{d^{2}} = p A = f_{r}$

Where A is the surface area of the sail, and p is the radiation pressure, and where F_s, F_r are forces due to sun and radiation.

For purely reflecting surface, radiation pressure, $p = \frac{2 I}{c}$

$\frac{G m M_{s}}{d^{2}} = \frac{2 I A}{c}$

Hence, the Surface area can be calculated as,

$A = \frac{G m M_{s}}{d^{2}} \times \frac{c}{2 I}$

Substitute the values in the above expression, and we get,

$A = \frac{6.67 \times 10^{- 11} \times 1500 \times 1.99 \times 10^{30} \times 3 \times 10^{8}}{{(1.50 \times 10^{11})}^{2} \times 2 \times 1.40 \times 10^{3}} = \frac{6.67 \times 15 \times 1.99 \times 3 \times 10^{29}}{1.50 \times 2 \times 1.40 \times 10^{25}} A = 9.5 \times 10^{5} m^{2}$

Therefore, the surface area of the sail, 9.5x10⁵ m² .

Find Study Materials for

Create Study Materials

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Determining the concept

Step 3: (a) Determining the surface area of the sail

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Select your language

Fundamentals Of Physics

Answers without the blur.

Short Answer

Step by Step Solution

Step 1: Given data

Step 2: Determining the concept

Step 3: (a) Determining the surface area of the sail

Most popular questions for Physics Textbooks

Want to see more solutions like these?

Recommended explanations on Physics Textbooks

Work Energy and Power

Radiation

Astrophysics

Physics of Motion

Scientific Method Physics

Fluids

Torque and Rotational Motion

Nuclear Physics

Fields in Physics

Measurements

Space Physics

Electricity

Physical Quantities and Units

Medical Physics

Electrostatics

Further Mechanics and Thermal Physics

Mechanics and Materials

Engineering Physics

Energy Physics

Force

Particle Model of Matter

Waves Physics

Magnetism

Modern Physics

Atoms and Radioactivity

Dynamics

Electricity and Magnetism

Conservation of Energy and Momentum

Fundamentals of Physics

Kinematics Physics

Circular Motion and Gravitation

Linear Momentum

Rotational Dynamics

Oscillations

Translational Dynamics

Geometrical and Physical Optics

Thermodynamics

Magnetism and Electromagnetic Induction

Electromagnetism

Electric Charge Field and Potential

Turning Points in Physics