A spaceship is moving at a speed of 0.6 c (60% the speed of light) relative to an observer on Earth. How does time pass on the spaceship relative to time on Earth?
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Jawaban:
In the theory of special relativity proposed by Albert Einstein, there is a phenomenon known as time dilation. Time dilation suggests that as an object approaches the speed of light, time passes more slowly for that object compared to an observer at rest. This effect becomes more significant as the object's speed gets closer to the speed of light (c).
In your scenario, the spaceship is moving at a speed of 0.6 c (60% the speed of light). According to time dilation, time on the spaceship will pass more slowly than time on Earth. The extent of this time dilation can be calculated using the Lorentz factor:
γ = 1 / √(1 - v² / c²)
Where:
- γ (gamma) is the Lorentz factor.
- v is the velocity of the spaceship (0.6 c in this case).
- c is the speed of light.
Substituting the values into the equation:
γ = 1 / √(1 - (0.6 c)² / c²)
γ = 1 / √(1 - 0.36) [Simplifying (0.6 c)² / c² to 0.36]
γ = 1 / √0.64 [Subtracting 1 from 0.36]
γ = 1 / 0.8 [Taking the square root of 0.64]
γ = 1.25
So, the Lorentz factor (γ) is 1.25. This means that time on the spaceship passes 1.25 times more slowly than time on Earth, or in other words, for every second that passes on Earth, only 0.8 seconds pass on the spaceship.
This effect becomes more pronounced as the spaceship's speed approaches the speed of light. At such high speeds, time dilation can have a significant impact on the experience of time for astronauts on the spaceship compared to people on Earth.
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Jawaban:
According to the theory of relativity, time passes slower for objects that are moving at high speeds relative to an observer. This effect is known as time dilation.
In this case, the spaceship is moving at a speed of 0.6 c relative to an observer on Earth. According to time dilation, time will pass more slowly on the spaceship compared to time on Earth. The amount of time dilation can be calculated using the formula:
t' = t / sqrt(1 - v^2/c^2)
where:
t' = time on the spaceship (as measured by someone on Earth)
t = time on Earth (as measured by someone on Earth)
v = velocity of the spaceship relative to Earth
c = speed of light
Plugging in the given values, we get:
t' = t / sqrt(1 - (0.6c)^2/c^2)
t' = t / sqrt(1 - 0.36)
t' = t / sqrt(0.64)
t' = t / 0.8
This means that time on the spaceship will pass at a rate of 0.8 times the rate of time on Earth. In other words, for every 1 second that passes on Earth, only 0.8 seconds will pass on the spaceship.