A broken-down spaceship (star Wreck?) flies across an international soccer field (Regulation length = 100 m) at a speed of 0.55c.
A.) How long does the spaceship pilot measure the field to be?
B.) From the spaceship pilot’s perspective, how long does it take to make this dash?
C.) How long does this spaceship’s flight take, according to ground abservers?
To answer these questions, we need to consider the concept of time dilation and the relativistic effects at near-light speeds. The formula used to calculate time dilation is:
t' = t / √(1 - (v^2/c^2))
Where:
t' = the measured time for the observer (spaceship pilot)
t = the reference time (ground observer's perspective)
v = velocity of the spaceship
c = speed of light
A.) How long does the spaceship pilot measure the field to be?
To calculate the measured length of the field for the spaceship pilot, we can rearrange the time dilation formula:
t' = t / √(1 - (v^2/c^2))
In this case, the field's length (100m) will be measured by the spaceship pilot. The velocity of the spaceship (v) is given as 0.55c (where c is the speed of light). Let's calculate the measured length:
t' = 100 m / √(1 - (0.55c)^2/c^2)
t' = 100 m / √(1 - 0.55^2)
t' = 100 m / √(1 - 0.3025)
t' = 100 m / √(0.6975)
t' = 100 m / 0.8363
t' ≈ 119.76 m
Therefore, the spaceship pilot would measure the field's length to be approximately 119.76 meters.
B.) From the spaceship pilot’s perspective, how long does it take to make this dash?
To calculate the time it takes for the spaceship pilot to travel across the field, we can use the time dilation formula as follows:
t' = t / √(1 - (v^2/c^2))
In this case, we are given the reference time (t) as the length of the field (100m). The velocity of the spaceship (v) is given as 0.55c. Let's calculate the measured time:
t' = 100 m / √(1 - (0.55c)^2/c^2)
t' = 100 m / √(1 - 0.55^2)
t' = 100 m / √(1 - 0.3025)
t' = 100 m / √(0.6975)
t' = 100 m / 0.8363
t' ≈ 119.76 m / c
Since the speed of light (c) is approximately 3 x 10^8 m/s, we can calculate:
t' ≈ (119.76 m / 3 x 10^8 m/s)
t' ≈ 3.992 x 10^(-7) seconds
Therefore, from the spaceship pilot's perspective, it would take approximately 3.992 x 10^(-7) seconds to make this dash.
C.) How long does this spaceship’s flight take, according to ground observers?
To find how long the spaceship flight takes according to ground observers, we use the reference time (t) and the velocity of the spaceship (v) in the time dilation formula:
t' = t / √(1 - (v^2/c^2))
In this case, the reference time (t) would be the time it takes for the spaceship to cross the field. The velocity of the spaceship (v) is given as 0.55c. Let's calculate the measured time:
t' = 100 m / √(1 - (0.55c)^2/c^2)
t' = 100 m / √(1 - 0.55^2)
t' = 100 m / √(1 - 0.3025)
t' = 100 m / √(0.6975)
t' = 100 m / 0.8363
t' ≈ 119.76 seconds
Therefore, according to ground observers, the spaceship's flight would take approximately 119.76 seconds.