It has been proposed to propel spacecraft through the Solar System with a large sail that is struck by photons from the Sun.

(a). Which would be more effective, a black sail that absorbs photons or a shiny sail that reflects photons back toward the Sun? Explain briefly

(b). Suppose thatphotons hit a shiny sail per second, perpendicular to the sail. Each photon has energy. What is the force on the sail? Explain briefly

Short Answer

Expert verified

(a). A shiny sail that reflects back the photons towards the sun will be more effective.

(b). The force (F)on the sail is2NEc.

Step by step solution

01

The concept of rate of change of momentum

The applied force equals the rate of change of momentum.

Step1:Explanation for why shining sail is more effective sail

(a)

A shining sail that reflects photons back toward the Sun would be more effective; the shiny sail absorbs the energy emitted by the Sun and reflects the photons back to the Sun's surface. In this instance, the change in photon momentum will be higher, and the net force will be bigger (the net force can be expressed as the rate of change of momentum).

02

Calculation for net force

(b)

If N photons per second strike a gleaming sail perpendicular to the sail, and each photon has energy E, the energy of each photon (a massless radiation, hence no rest mass) is:

E=pcp=Ec

On shiny sail photon is absorbed and re-emitted in the opposite direction, so the change in its momentum:

Δp=pf-pi=-p-p=-2p=-2Ec

The magnitude of change in momentum for N photons/second is

ΔpN/s=2NEc

The net force for N photons/second is the rate of change of momentum:

F=ΔpΔt=2NEc

Thus, the force ()on the sail is2NEc .

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Under what conditions is the momentum of a system constant? Can the x component of momentum be constant even if they component is changing? In what circumstances? Give an example of such behavior.

In a reference frame where a ∆+ particle is at rest it decays into a proton and a high-energy photon (a “gamma ray”): +P++γ+. The mass of the ∆+ particle is 1232 MeV/c2and the mass of the proton is 938 MeV/c2(1 MeV = 106eV). Calculate the energy of the gamma ray and the speed of the proton.

In a nuclear fission reactor, each fission of a uranium nucleus is accompanied by the emission of one or more high-speed neutrons, which travel through the surrounding material. If one of theneutrons is captured in another uranium nucleus, it can trigger fission, which produces more fast neutrons, which could make possible a chain reaction

Two asteroids in outer space collide and stick together. The mass of each asteroid, and the velocity of each asteroid before the impact, are known. To find the momentum of the stuck-together asteroids after the impact, what approach would be useful? (1) Use the Energy Principle. (2) Use the Momentum Principle. (3) It depends on whether or not the speed of the asteroids was near the speed of light. (4) Use the relationship among velocity, displacement, and time. (5) It depends on whether the collision was elastic or inelastic.

A particle of mass m, moving at speed v=45c, collides with an identical particle that is at rest. The two particles react to produce a new particle of mass M and nothing else. (a) What is the speed V of the composite particle? (b) What is its mass M?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free