How long does it take for light to reach Earth in a moving rocket ship?

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SUMMARY

The discussion focuses on calculating the time it takes for light to reach Earth from a rocket ship moving at 0.50c. The key equation to use is t = d / (c - v), where d is the distance to Earth (6.00 x 10^12 m), c is the speed of light (3 x 10^8 m/s), and v is the speed of the rocket (0.50c). Participants emphasize the importance of understanding relative motion and dismiss the need for time dilation in this specific scenario, confirming that the calculation can be simplified using basic kinematic equations.

PREREQUISITES
  • Understanding of special relativity concepts, particularly relative motion.
  • Familiarity with the speed of light (c = 3 x 10^8 m/s).
  • Basic knowledge of kinematic equations, specifically v = d/t.
  • Awareness of Lorentz transformations and time dilation, even if not directly applicable here.
NEXT STEPS
  • Research the application of the equation t = d / (c - v) in relativistic contexts.
  • Study the principles of special relativity, focusing on relative motion and its implications.
  • Explore Lorentz transformations and their relevance in different frames of reference.
  • Review kinematic equations and their applications in physics problems involving light and speed.
USEFUL FOR

Students studying physics, particularly those tackling problems in special relativity, as well as educators looking for examples of relative motion and light speed calculations.

justinh8
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Physics Relativity Question?

Homework Statement


A rocket ship is moving directly toward Earth with a constant speed of 0.50 c relative to Earth. It turns on its headlight, and the light beam travels toward Earth with speed c relative to the ship. According to the rocketship astronauts, Earth is 6.00 x 10^12 m away when the headlight is turned on. In the frame of the astronauts, how long does it take for the light to reach Earth? (Hint: in the frame of the astronauts, Earth is approaching the ship with a speed of 0.5c)


Homework Equations


c = 3x10^8 m/s


The Attempt at a Solution


I just got 20000 seconds but i know its wrong
 
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Can you show your working?
 


He divided 6x1012 by 3x108. I think even justinh8 knows that doesn't qualify as "an attempt at a solution".

justinh8, you haven't been following along in class. Read the appropriate chapter in your textbook on length contraction and Lorentzian Transform.
 


You need to take into account time dilation.
 


There was no lesson for this, it was just a do you remember from grade 11 physics but i do not recall any lorentzian transformations in grade 11. Plus we haven't gotten our textbooks yet so i have no clue. However, i think i know one equation that might be relevant. Its ct^2 = vt^2 + ct^2. Its from grade 11 but not sure if it would work...
 
Last edited:


justinh8 said:
There was no lesson for this, it was just a do you remember from grade 11 physics but i do not recall any lorentzian transformations in grade 11. Plus we haven't gotten our textbooks yet so i have no clue. However, i think i know one equation that might be relevant. Its ct^2 = vt^2 + ct^2. Its from grade 11 but not sure if it would work...

Hmm that's odd then. Well the equation to use is t=\frac{t'}{\sqrt{1-\frac{v^2}{c^2}}}, \text{Where } t'= \text{dilated time and } t = \text{stationary time}
 


Bread18 said:
You need to take into account time dilation.
I'm pretty sure he doesn't need to be concerned about time dilation.

Everything is given and asked in the frame of the rocket ship.
 


SammyS said:
I'm pretty sure he doesn't need to be concerned about time dilation.

Everything is given and asked in the frame of the rocket ship.

I think you're right, my mistake.
 


In which case, I think all he has to take into account is relative motions.
 
  • #10


I just talked to my teacher today and he said there is nothing to do with time dilation, he just said think simple so i think it might be using just the v= d/t and just rearrange it to find time??
 

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