Calculating Distance and Time for Traveling to a Star at 1.0x10^8 m/s

AI Thread Summary
The distance to the star, when traveling at 1.0x10^8 m/s, is calculated to be 37.7 light-years due to Lorentz length contraction. To determine the journey time from the spaceship's perspective, the formula t = distance/speed can be used, with the distance being 37.7 light-years and the speed as ⅓ the speed of light. This results in a journey time of approximately 113.1 years in the spaceship's frame. The calculations can be performed without converting to SI units, as using light-years and years simplifies the process. The discussion emphasizes the importance of understanding relativistic effects in space travel calculations.
msimard8
Messages
58
Reaction score
0
A star is measured to be 40.0 ly from Earth, in the inertial frame in which both star and Earth are at rest.

A) what would you determine this distance to be if you travveled to teh star in a spaceship moving at 1.0x10^8 m/s relative to earth.

answer is 37.7 ly (calculated that, with no problem)

b) How long would you determne the journey to take?

ok this is where i get stumpped

I converted ly to m and got some result.
do u convert this distance to time using v=d/t

if you do that what do you after?

where do i go from here?
thanks
 
Physics news on Phys.org
According to Earth's frame where the star is at rest, it would take the time 120 years since 1.0x10^8 m/s is c/3
The event in Earth's frame is the coordinate (x,ct) = (40, 120) light-years
Use the Lorentz-transformation to obtain the coordinates of that event in the starships rest frame
 
  • Like
Likes LastScattered1090
In addition to what @drmalawi has said, and given the age (15+ years) of the question, can I add some hints for future readers...

37.7 ly for part a) is correct (easily determined using the Lorentz length-contraction formula). This value can then be used to answer part b) directly.

The Earth and star are at rest in some inertial frame. It follows that for an observer on the rocket, the star is seen to approach at speed ⅓c.

The duration of the journey as measured on the rocket’s frame is simply distance/speed, using the values from the rocket's frame. The distance to be covered in the rocket’s frame is 37.7 ly (from part a).

There’ s no need to convert to SI units; work in units with distance in ly, time in years, and c = 1 (ly/y).

v = 1.0x10⁸ m/s = ⅓c (accurate enough for this problem)

t = distance/speed = ##\frac {37.7}{⅓}## years
 
  • Like
Likes LastScattered1090 and malawi_glenn
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top