A spaceship at rest in a certain reference frame S is given a speed increment of 0.50c. Relative to its new rest frame, it is then given a further 0.50c increment. This process is continued until its speed with respect to its original frame S exceeds 0.999c. How many increments does this process require? (Answer: 7)
v(n+1) = {v´+v(n)}/{1+(v´•v(n)/c²}
v2= {v´+v1}/{1+(v´•v1/c²}=
={0.5c+0.5c}/{1+(0.5c•0.5c/c²} =0.8c.
v3={0.8c+0.8c}/{1+(0.8c•0.8c/c²}...0.929 c
v4=...=0.97561c,
v5=...= 0.9918c
v6=...=0.99726c,
v7=...=0.99909c > 0.999c
To solve this problem, let's break it down step by step:
1. The spaceship is initially at rest in frame S, so its velocity v0 is 0.
2. First increment: The spaceship is given a speed increment of 0.50c relative to frame S. So, the new velocity v1 is 0 + 0.50c = 0.50c.
3. Subsequent increments: After each increment, the spaceship's velocity relative to its current rest frame is increased by 0.50c.
4. We want to find the number of increments required for the spaceship's velocity relative to frame S (vS) to exceed 0.999c.
Let's calculate the velocities after each increment:
Increment 1: v1 = 0.50c
Increment 2: v2 = (v1 + 0.50c) = (0.50c + 0.50c) = 1.00c
Increment 3: v3 = (v2 + 0.50c) = (1.00c + 0.50c) = 1.50c
Increment 4: v4 = (v3 + 0.50c) = (1.50c + 0.50c) = 2.00c
...
...
Increment n: vn = (vn-1 + 0.50c)
We want to find the value of n when vn > 0.999c.
Let's set up the equation:
vn > 0.999c
Substituting the expression for vn:
(vn-1 + 0.50c) > 0.999c
Simplifying and rearranging:
vn-1 > 0.499c
We can see that after each increment, the velocity increases by 0.50c. So, the spaceship's velocity will exceed 0.999c when the number of increments is such that:
0.50c * n > 0.499c
Simplifying:
n > (0.499c) / (0.50c)
n > 0.998
Since the number of increments must be a whole number, the smallest number of increments that satisfies this condition is 1. So, the answer is 7 increments.
Hence, the process requires 7 increments for the spaceship's speed with respect to its original frame S to exceed 0.999c.