Jump to content

Felipe Doria

Members
  • Posts

    8
  • Joined

  • Last visited

Profile Information

  • Favorite Area of Science
    Physics

Felipe Doria's Achievements

Lepton

Lepton (1/13)

0

Reputation

  1. Could someone please help me with the problem set? Imagine that light did not have a constant speed, but behaved in the manner expected from experience. Namely, if the source of the light is rushing toward you, the light will approach you faster; if the source is rushing away from you, the light will approach you slower. This is incorrect, of course, but it's worth investigating the consequences of a non-constant speed of light because the failure to observe those consequences is evidence that the speed of light is constant. With that backdrop, consider a binary star system situated a very large distance L from Earth. Let the angular velocity of the smaller star be ω, as it orbits the larger star in a circle of radius r. What value of ω=v/r will result in the light emitted when the smaller star is traveling directly away from Earth reaching us at the same moment as the light emitted later, when the smaller star's orbit has it moving directly toward earth? (choose one) a) ω=(c/r)*sqrt((πr)/(2L+2πr)) b) ω=(c/r)*((πr)/(2L+πr)) c) ω=(c/r)*sqrt((πr)/(2L+πr)) I think that the time it takes for the light emitted from the smaller star when it is traveling directly away from Earth is equal to the time it takes for the light emitted later plus the time it takes to complete half an orbit. So: t1 = t2 + t3 L/(c-v) = L/(c+v) + pi*r/v How can I get the answer from this? Thank you for your help.
  2. Imagine that you have a multistage rocket in which the first stage blasts off from the ground at v1, the second stage blasts off later with speed v2 relative to the first stage, and the third stage blasts off later still with speed v3 relative to the second stage. A From the perspective of someone stationary on the ground, what is the speed of the second stage? (For ease, assume you are working in units for which c=1.) a) v2 b) v1+v2 c) (v1+v2)/(1+v1v2) d) (v1−v2)/(1−v1v2) B From the perspective of someone stationary on the ground, what is the speed of the third stage? (For ease, assume you are working in units for which c=1). a) (v1−v2−v3+v1v2v3)/(1−v1v2+v1v3+v2v3) b) (v1+v2+v3+v1v2v3)/(1+v1v2+v1v3+v2v3) c) (v1+v2+v3)/(1+v1v2+v1v3+v2v3) d) (v1+v2+v3+v1v2v3)/(1+v1v2v3) C You are on a rocket traveling at a speed of 0.75c away from Earth. You wish to launch a probe that will travel at a speed of 0.9c away from Earth. How fast does the probe need to be going in the rocket's reference frame? a) .15c b) .46c c) .34c d) .25c I've written the answers I think are correct in bold letters, are they correct? I don't quite understand how to apply the formula (v+w) / (1 + (v.w)/c) to 3 variables, could someone please explain B to me? Thank you for your help.
  3. Imagine that light did not have a constant speed, but behaved in the manner expected from experience. Namely, if the source of the light is rushing toward you, the light will approach you faster; if the source is rushing away from you, the light will approach you slower. This is incorrect, of course, but it's worth investigating the consequences of a non-constant speed of light because the failure to observe those consequences is evidence that the speed of light is constant. With that backdrop, consider a binary star system situated a very large distance L from Earth. Let the angular velocity of the smaller star be ω, as it orbits the larger star in a circle of radius r. We want to find the value of ω for which the light emitted by the smaller star, when it's traveling directly away from Earth, arrives at Earth at the same moment as light emitted a little later, when the star is traveling directly toward Earth. Let's work this out in stages: Recall from basic physics that if the angular velocity of the smaller star is ω then its linear speed v is given by v=ωr. In terms of v, what is the time t1 that it will take light emitted by the smaller star to reach Earth, if that light is emitted when the smaller star's orbit has it moving directly away from Earth? (Again, assume—incorrectly—that light behaves as you would expect from everyday experience, with its base speed c being increased or decreased by the motion of the source.) a) t1=L/c b) t1=L/v c) t1=L/(c+v) d) t1=L/(c−v) In terms of v, what is the time t2 that it will take light emitted by the smaller star to reach Earth if that light is emitted when the smaller star's orbit has it moving directly toward Earth? (Again, assume—incorrectly—that light behaves as you would expect from everyday experience, with its base speed c being increased or decreased by the motion of the source.) a) t2=L/c b) t2=L/v c) t2=L/(c+v) d) t2=L/(c−v) In terms of v, what is the time t3 that it takes the smaller star to move from the position relevant to part (A) to the position relevant to part (B)—that is, how long does it take the star to complete one half of a full orbit around the larger star? a) t3=2π/v b) t3=2πr/v c) t3=πr/v d) t3=πr/c What value of ω=v/r will result in the light emitted when the smaller star is traveling directly away from Earth reaching us at the same moment as the light emitted later, when the smaller star's orbit has it moving directly toward earth? a) ω=(c/r)*sqrt((πr)/(2L+2πr)) b) ω=(c/r)*((πr)/(2L+πr)) c) ω=(c/r)*sqrt((πr)/(2L+πr)) The answers I've put written in bold letters(d,c,c) are the ones I think are correct, are they? I didn't put any answer on the last question in bold letters because I don't know how to get the answer, could someone explain it to me please? Thank you for your help!
  4. Thank you for your help! I had just forgotten to add the second 1 and done 2c/1 instead of 2c/2 as dumb as that might sound.
  5. I'm so dumb, I forgot the second 1 in the equation. Thank you!
  6. 1. The problem statement, all variables and given/known data If one light beam is heading toward you from the right, and another is heading toward you from the left, how quickly from the perspective of the right-moving beam is the left-moving beam approaching it? 1) 0 2) c 3) 1.5c 4) 2c 2. Relevant equations (v+w) / (1 + cw/c2) 3. The attempt at a solution I tried plugging in c into the equation for both v and w but it gives me (c + c) / (1 + c2/c2) which equals 2c. It is one of the possible answers but I know nothing can go faster than c so I'm thinking the answer is probably c. Am I right? If so, why? If not, what am I doing wrong? Thank you very much for your help!
×
×
  • Create New...

Important Information

We have placed cookies on your device to help make this website better. You can adjust your cookie settings, otherwise we'll assume you're okay to continue.