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First the challenge again, briefly: a spaceship sets off from Earth to a distant star, maintaining a speed of 0.8c. When it reaches the star, it quickly turns around and heads home at 0.2c. What will the average speed of the spaceship be as calculated here on Earth?
Many respondents had the simple case, where the distance between Earth and star remains constant, spot on, with an average speed of 0.32c. It so happens that the distance to the star cancels out and the formula for the average speed is simply:
All velocities are expressed as fractions of c and are hence unit-less. Although v1 and v2 as given are quite relativistic, there are no relativistic effects here - at least not from Earth's point of view. If the star was Alpha Centauri, at 4 ly away, it takes the ship 4/0.8 = 5 earth-years to reach the star and 4/0.2 = 20 earth-years to get back, for an average speed of 8/25 = 0.32c.
However, the challenge question stated a "distant star", so it may not be reasonable to take the distance as constant. Further, since "nearby star" was not used, it seems to rule out Alpha Centauri, Sirius and the like, that travels roughly at the same speed as Earth. It may even be a star outside of the Milk Way, but let's be reasonable and consider stars inside our Galaxy only (I doubt that we will ever fly anything out of our own Galaxy!)
Fortunately, it can be shown that if the relative speeds between Earth and the star are much smaller than the spaceship's return speed (0.2c), the distance again cancels out and it is only the relative speed between Earth and the star that matters. In fact, it is the difference in the outbound and inbound distances that matters, but that can be easily related to the relative velocity between Earth and the star.
As an aside, it's obvious that Earth's movement around the Sun is utterly negligible - it's a 16 light-minutes orbital diameter versus many light-years of travel. What may however matter is our Sun's movement relative to the other star. Let's find out how much it matters.
Essentially, the generalized equation boils down to equation (1) above, but with the star's radial velocity relative to Earth vstar subtracted from the ship's return speed |vreturn|, i.e.,
giving
(I suppose one can write it all in terms of the radial components of the velocity vectors, but it becomes a bit more messy, hence the curious mix of radial velocity vectors and radial speed, a scalar).
Now what will the worst-case relative speed between stars in our Galaxy be? For stars farther away from the galactic center than the Sun, the orbital speeds are more or less the same as Earth's speed, around 217 km/s. For stars closer to the central bulge of our Galaxy, the orbital speeds taper off more or less linearly with radius.
A reasonable order of magnitude guess would be that no two stars in the Galaxy have a relative radial speed exceeding 200 km/s (or 0.00067c). This will be the case with two stars that are roughly 90 degrees apart in the disk, as viewed from the center. To calculate the average speed, we simply use v2 = 0.2 - 0.00067 = 0.1993c in equation (1) above. The answer (va) differs by only 0.25% from the 0.32c, so the error is utterly negligible for all engineering purposes - hence we don't need to know the relative speed accurately.
As a note, in the unlikely case of a star with a more relativistic velocity relative to Earth, the simple subtraction v2 = vship - vstar must be replaced by a relativistic subtraction of velocities:
Finally, although not part of the challenge, it is interesting to analyze the same scenario from the spaceship's point of view, i.e., what will the average velocity (va') of the mission be as measured onboard the spaceship? It's given by:
where
and
These last two are the so-called Lorentz (or gamma) factors for the two velocities. For the Earth frame, gamma1 = gamma2 = 1, so eq. (5) reduces to eq. (1), as one would expect.
For the values v1=0.8c and v2=0.2c as before, this gives the average speed calculated onboard the spaceship as Va' = 0.28c. Some people might be surprised to find that the relativistic calculation of the average speed, as measured by the ship, is less than measured on Earth. This is due to the relatively heavy time dilation and Lorentz contraction during the fast outbound trip, diminishing the contribution of the 0.8c speed.
There is a lot more on time dilation, Lorentz contraction and relativistic addition of speed on the website Relativity 4 Engineers.
Regards, Jorrie
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