Two Relativity Problems on relative length

In summary, when travelling at 0.8c, the second ship would appear to be 60m long to the observer, while the first ship would appear to be 1m long to the second ship. Similarly, a sphere with a rest length of 3m travelling at 1.5*10^8 m/s would appear to have a height of 3m and a width of 2.6m to an observer on Earth. This is due to the phenomenon of length contraction, where objects appear shorter in the direction of motion when moving at high speeds. The equation used to calculate this effect is l = (1 - v^2/c^2)^1/2, where l is the moving length,
  • #1
Kevin Hsu
1
0

Homework Statement


Travelling at 0.8c in your 100m long spaceship you pass a second, identical ship t rest. How long does the second ship appear to you, and how long does our ship appear to it?


Homework Equations


l = (1 - v2/c2)^1/lo


The Attempt at a Solution


l = (1 - v2/c2)^1/lo
l = [1 - (0.8c)2/(3.00*108)2]^1/100m
l = 0.36^1/100
l = 1m

I cannot find any mistakes in my calculations, but the answer I calculated differs from the one found in the answer key, 60m. If anyone can please point out the error for me, it'd be greatly appreciated.

Homework Statement


A sphere of diameter 3m travels past the Earth at 1.5*10^8 m/s, what height and width does it have to an observer on earth?


Homework Equations


l = (1 - v2/c2)^1/lo


The Attempt at a Solution


I plugged into the equation lo = 3m and v = 1.5*10^8 m/s.
The equation then yielded 0.9m to be the length. Again, it differs from the
answer key, height = 3m and width = 2.6m.

What I don't understand completely is that, does the sphere stretch into an eclipse when it is accelerated to a fraction of the speed of light, so that the height and the width would be different? How then, would one calculate the height and the width separately?


Thanks in advance for any help/pointers!
 
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  • #2
gamma = 5/3
so, presumably the moving length is 3/5 the rest length ... or 60m.
... when you computed gamma, it looks like you divided by an extra (3x10^8)^2 ... you don't need to do this explicitly because (0.8c)^2/(c)^2 = (0.8)^2.
 

FAQ: Two Relativity Problems on relative length

1. What is the concept of relative length in the theory of relativity?

Relative length is the measurement of an object's length as observed by an observer in relative motion. According to the theory of relativity, an object's length can appear different to different observers depending on their relative motion.

2. How does the concept of relative length affect our understanding of space and time?

The concept of relative length is a fundamental aspect of the theory of relativity and it plays a crucial role in our understanding of space and time. It helps to explain how space and time are not absolute but rather are dependent on the observer's frame of reference.

3. Can you provide an example of a real-life situation where relative length is observed?

One example of relative length in action is the phenomenon of length contraction, where an object's length appears shorter when it is moving at high speeds relative to the observer. This can be observed in particle accelerators, where subatomic particles are accelerated to near-light speeds and their length is measured.

4. What is the difference between relative length and absolute length?

Absolute length refers to the physical length of an object as measured in its own frame of reference, while relative length refers to the length of an object as observed by an observer in a different frame of reference. The two lengths can appear different due to the effects of relativity.

5. How does the concept of relative length relate to the concept of time dilation?

Time dilation is another aspect of the theory of relativity that is closely related to relative length. As an object's velocity increases, its relative length decreases and its relative time also slows down. This means that an object in motion experiences time at a slower rate compared to an object at rest, as observed by an external observer.

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