# Moon's orbit period as seen from a spaceship traveling at 0.8c

I am studying special relativity and I am trying to figure out the following small problem which occurred to me:

An observer, the pilot of a spaceship flying to or from earth at v = 0.8c, is observing earth from his spaceship.

He observes the moon orbiting the earth once every 45 days, 1.667 times slower than the period seen from earth.

He also observes the earth is 1.667 times more massive than it seems to a person on earth.

However, doesn't that mean the moon should orbit earth faster than normal, not slower?

What is the solution to this problem?

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Please have a look at our homework policy, in which you will find that whether or not this is actual homework, since you are asking for a specific solution to a specific problem, it is to be classified as homework. Since I think you are asking a kind of conceptual question, I am not voting to close, though. –  ACuriousMind Jul 23 '14 at 20:30
Thanks, indeed, this is a conceptual question that occurred to me while trying to understand special relativity. I am studying physics on my own. –  nir Jul 23 '14 at 20:42
I recommend that you consider the distinction between observer in relativity and 'looking out the window'. They're not the same concept: en.wikipedia.org/wiki/Observer_%28special_relativity%29 This isn't a trivial consideration - what one sees looking out the window is much different from what one observes according to an inertial reference frame. –  Alfred Centauri Jul 23 '14 at 20:54
Thanks. If I understand you correctly, I meant an observer. I edited the question in an attempt to reflect this. –  nir Jul 23 '14 at 21:08
I think it may be pretty important which way the ship is flying - is it on a path in the plane of the Earth-Moon orbit? Perpendicular to it? On some more general trajectory? Though maybe it doesn't matter - having trouble sorting it out in my head on the fly... –  Kyle Jul 24 '14 at 0:49

If I understand your question correctly, you find an apparent paradox pursuing the consequences of plugging the relativistic correction factors ($\gamma$) in both sides of Kepler's third law. On one side, one should measure the increase of the period because of time dilation, but on the other side, one should find a decrease in the period because the relativistic mass correction of the earth moving respect to you. You ignore the consequences of a possible Lorentz contraction of the orbit's size, but that is not important since none of Kepler's laws is consistent with special relativity or with the adequate corrections of gravitational fields when measured from moving observers.
For you, moving at $0.8c$, the moon completes an orbit in 45 days indeed. It is like a giant clock ticking slower just because of Lorentz time dilation. The enormous gain of earth's kinetic energy respect to you does indeed modify its gravitational field, but in a very different way compared to just increasing the mass of the earth by a factor of $\gamma$. In fact, the gravitational field changes in just the adequate way for you to observe the moon's orbit modified according to the predictions of special relativity. For example, if you move in the plane of the orbit towards the earth, you will see the orbit of the moon (originally a circle, say) becomes an ellipse but with earth in the center, not in any of the foci. This is only to illustrate that gravitational fields, from moving observers, do not behave as a normal Newtonian field with an increased mass.