A lighthouse beacon alerts ships to the danger of a rocky coastline.
Part A
According to the lighthouse keeper, with what speed does the light leave the lighthouse?
v/c=
A boat is approaching the coastline at speed 0.5. According to the captain, with what speed is the light from the beacon approaching her boat?
v/c=
I understand both answer depends on c,but what are they in terms of v/c=?
the answer is "1" for both questions. i just did it on mastering physics.
To answer these questions in terms of v/c, let's consider the concept of relativistic velocity addition.
Part A:
According to special relativity, the speed of light in a vacuum is denoted by "c". Since the question asks for the speed at which the light leaves the lighthouse, we can say that the speed of light leaving the lighthouse is equal to the speed of light in a vacuum, which is "c".
Therefore, v/c = c/c = 1.
Part B:
The speed of light relative to an observer moving at a velocity "v" can be calculated using the relativistic velocity addition formula:
v' = (v + u) / (1 + (v*u/c^2))
where:
v' = relative speed of light from the beacon approaching the boat
v = velocity of the boat relative to the lighthouse
u = speed of light leaving the lighthouse (which is equal to "c")
c = speed of light in a vacuum
Since the boat is approaching the coastline at a speed of 0.5 (v = 0.5), and we know that u = c = 1, we can substitute these values into the equation:
v' = (0.5 + 1) / (1 + (0.5 * 1 / 1))
Simplifying,
v' = 1.5 / (1 + 0.5)
v' = 1.5 / 1.5
Therefore, v' = 1.
So, according to the captain, the speed of the light from the beacon approaching her boat is equal to 1 times the speed of light (c), which means it's moving at the same speed as light in a vacuum.
To determine the speed of the light leaving the lighthouse and approaching the boat in terms of v/c (where v represents the speed of the boat and c represents the speed of light in a vacuum, approximately 299,792,458 meters per second), we can use the concept of relative velocity.
Part A:
The speed of the light leaving the lighthouse, in terms of v/c, can be calculated by using the formula: v'/c = (v - v0)/(1 - v * v0 / c^2), where v' represents the velocity of the light and v0 is the velocity of the lighthouse (which should be 0 in this case as the lighthouse keeper is stationary). However, since v0 = 0, the formula simplifies to: v'/c = v/c.
Therefore, the speed of the light leaving the lighthouse, in terms of v/c, is equal to v/c.
For the boat approaching the coastline at speed 0.5, we need to calculate the speed of the light from the beacon approaching the boat, in terms of v/c. We can again use the same formula: v'/c = (v - v0)/(1 - v * v0 / c^2), where v' represents the velocity of the light and v0 is the velocity of the boat (0.5 in this case).
Therefore, the speed of the light from the beacon approaching the boat, in terms of v/c, is given by: v'/c = (0.5 - 0)/(1 - (0.5 * 0) / c^2) = 0.5 / (1 - 0) = 0.5.
In both cases, the speed of the light leaving the lighthouse and approaching the boat is equal to v/c (in other words, it is equal to the ratio of the boat's speed to the speed of light in a vacuum).