# Earth/Moon trip with slow light

I am working on an imaginary world -- call it science fiction if you like -- in which the speed of light is much slower but nothing else is changed. Yes, I know that this is impossible. The speed of light I have chosen is that the distance to the moon is one light year, which means c=12.7 m/sec.

The specific question that interests me is the Apollo moon mission. Since momentum is unchanged, then the astronauts will still be able to go to the moon in four days and return in four days more, as far as their own clock is concerned. The question is, what is the Earth clock doing?

I'm assuming the angular momentum of objects on the surface of the Earth is unchanged from that in our world. So the Earth is rotating at .99962 c and a day is still 24 of our hours. Yes, I know that this is impossible. What interests me is that since the rotation is so fast the clocks on the Earth will be dilated by a factor of 36.45. The velocity of the astronauts relative to the Earth is not constant, so it seems that it would be necessary to integrate the the time dilation of the astronauts relative to the center of the Earth over the entire trip, which is doable but not all that easy. Is there a shortcut for the calculation of the relative time dilations of the two paths: surface of the Earth, and trip to the moon?

The surface of the Moon would be rotating at a mere 0.3c, so once on the moon then the astronauts would lose almost all of their time dilation and have a wait of two or three weeks for the orbiting spacecraft to complete one orbit. As far as the orbiting astronaut is concerned the moon orbit takes two hours.

Anyway, I hope that this crazy problem interests someone who understands these things better than I and can point me in the right direction.

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## Answers and Replies

With c = 12.7 m/s and unchanged gravity, the earth would turn into a black hole, making the trip impossible. Since the speed of sound can't be greater than the speed of light, having rigid objects will also be impossible, unless you change some more constants of nature.

Time Observed = Time / √(1-v2/c2)

So if on earth v = .99962c
And moon v = 0.3c

Tearth = Time / √(1-(.99962c)2/c2)
Tmoon = Time / √(1-(.3c)2/c2)

Doing algebra...

Time = Tearth * √(1-(.99962c)2/c2)

Tmoon = (Tearth * √(1-(.99962c)2/c2)) / √(1-(.3c)2/c2)

Tmoon = Tearth * √((1-(.99962c)2/c2) / (1-(.3c)2/c2))

Tmoon = Tearth * √((1-(.99962c)2) / (1-(.3c)2))

With c = 12.7 m/s and unchanged gravity, the earth would turn into a black hole, making the trip impossible. Since the speed of sound can't be greater than the speed of light, having rigid objects will also be impossible, unless you change some more constants of nature.

Thanks for the feedback. I calculated the Schwarzschild radius of the Earth with such slow light and it turns out to be 8.9 * 10^13 meters, so the entire solar system is well inside this. The moon would also be a black hole, but the solar system would also be inside it's smaller radius. Interstellar travel would require accelerating the Earth and Moon, which seems rather drastic.

The Earth would be rotating very quickly and would be a Kerr black hole. This gives a lot more grist for the mill.

With the Earth rotating so rapidly the distance to the moon would change by a factor of a thousand or so when measured from different locations on the surface.