0.713c, or 214000 km/s
And initial check without taking into consideration indicates that the
spacecraft is traveling in excess of 305,000 km/s which is faster than the
speed of light. Obviously, this is impossible, so we need to take into
consideration the feature of Lorentz contraction which will cause the earth
based observe to see the spacecraft as being shorter. The formula for
Lorentz contraction is
L = L0(sqrt(1-v^2/c^2))
where
L = observed length
L0 = length when relative velocities is 0
v = speed
c = speed of light
Since the observed time will be L/v = t we can construct the equation:
L/v = t
L = L0(sqrt(1-v^2/c^2))
L0(sqrt(1-v^2/c^2))/v = t
Now solve for v
L0(sqrt(1-v^2/c^2))/v = t
L0(sqrt(1-v^2/c^2)) = tv
sqrt(1-v^2/c^2) = tv/L0
sqrt(1-v^2/c^2)/v = t/L0
(1-v^2/c^2)/v^2 = t^2/L0^2
1/v^2 - 1/c^2 = t^2/L0^2
1/v^2 = t^2/L0^2 + 1/c^2
v^2 = 1/(t^2/L0^2 + 1/c^2)
v = sqrt(1/(t^2/L0^2 + 1/c^2))
Now substitute the known values and calculate v.
v = sqrt(1/((0.82x10^-6 s)^2/(250 m)^2 + 1/(299792458 m/s)^2))
v = sqrt(1/((6.724x10^-13 s^2)/62500 m^2 + 1/8.98755178736818x10^16 m^2/s^2))
v = sqrt(1/(1.07584x10^-17 s^2/m^2 + 1.11265005605362x10^-17 m^2/s^2))
v = sqrt(1/(2.18849005605362E-17 m^2/s^2))
v = sqrt(4.56936049233527x10^16 m^2/s^2)
v = 213760625.3 m/s
So we have a calculated velocity of 213760625.3 m/s. Let's see if that's reasonable.
Due to Lorentz contraction, the observed length will be L0(sqrt(1-v^2/c^2)) meters long. Let's calculate that:
v/c = 213760625.3/299792458
v/c = 0.713028696
L = L0(sqrt(1-v^2/c^2))
L = 250(sqrt(1 - 0.713028696^2))
L = 250(sqrt(1 - 0.508409921))
L = 250(0.701134851)
L = 175.2837127 m
The speed will be L/t, so
175.2837127 m / 0.82x10^-6 s = 213760625.3 m/s
And the result matches the earlier calculated speed. So we have a good cross check on the answer, which when rounded to 3 significant figures is either 0.713c, or 214000 km/s.
