# Faster than speed of light spaceship.

Hi all,

I new to this physics thing and have a couple of ideas where i know im wrong but would love if someone could explain why to me.

My main mind boggler(for myself):
If spaceship one takes off from one side of a planet at say 90% the speed of light and
spaceship two takes off from the other side of the planet in the exact opposite direction also at 90% the speed of light.

What would the observations be from a person on spaceship one in respect to spaceship two?
Would spaceship two be traveling faster than the speed of light in respect to spaceship one?

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Hi all,

I new to this physics thing and have a couple of ideas where i know im wrong but would love if someone could explain why to me.

My main mind boggler(for myself):
If spaceship one takes off from one side of a planet at say 90% the speed of light and
spaceship two takes off from the other side of the planet in the exact opposite direction also at 90% the speed of light.

What would the observations be from a person on spaceship one in respect to spaceship two?
Would spaceship two be traveling faster than the speed of light in respect to spaceship one?
You've specified those speeds from an earth based perspective. So that means each spaceship would see the earth move towards them at 90%, and the other spaceship would only be a little bit more, for example 95%.

"if a ship is moving relative to the shore at velocity v, and a fly is moving with velocity u as measured on the ship, calculating the velocity of the fly as measured on the shore is what is meant by the addition of the velocities v and u"

So in this case v = the earth velocity and u = the other ship velocity.

The result gives you the other ship velocity in relation to the first ship.

tiny-tim
Homework Helper
Welcome to PF!

Hi mooneyg3! Welcome to PF!

To add to what Aaron_Shaw has said …

So, if you put u = (1 - a)c and v = (1 - b)c, then that's c(2 - a - b)/(2 - a - b + ab), which is obviously less than c.

Thanks guys! :) Im still just dealing with the theory of it all and its mind boggling! but i like that. Guess I should look at the math!

Just one more question on this....

Say a spaceship were to take off from earth at say ~99% the speed of light and witnessed time dialation so as one day on the spaceship was two days on earth (I think this is what happens).

Once the spaceship reaches a constant speed it can be considered stopped(I may be wrong, probably am, and speed may not be the right word but I think it is)

Anyway... a pod takes off from the spaceship at a slightly greater speed than it is traveling away from the earth. because this pod is traveling at near the speed of light relative to the spaceship it witnesses time dialation so 1 day on the pod is 2 days on the spaceship.

Now this pod is traveling at a slow speed relative to the earth.

So my question is once this pod reaches earth will it observe time moving very slow because 1 day on pod = 2 days on spaceship and two days on spaceship = 4days on earth?

JesseM
Thanks guys! :) Im still just dealing with the theory of it all and its mind boggling! but i like that. Guess I should look at the math!

Just one more question on this....

Say a spaceship were to take off from earth at say ~99% the speed of light and witnessed time dialation so as one day on the spaceship was two days on earth (I think this is what happens).
Keep in mind that as long as both observers are moving inertially (constant direction and speed), time dilation is completely relative. If the ship's clock is slowed down by a factor of 2 in the rest frame of the Earth, then the Earth's clock is also slowed down by a factor of 2 in the rest frame of the ship; there's no frame-independent truth about whose clock is "really" running slow (speed is relative too--in the ship's frame the Earth is the one moving at a large fraction of light speed).

Also, to have a time dilation factor of 2, the speed would need to be about 0.866c. The time dilation equation says that a clock moving at speed v will have its ticks stretched by a factor of $$\frac{1}{\sqrt{1 - v^2/c^2}}$$, so to get a factor of 2, you need $$2 = \frac{1}{\sqrt{1 - v^2/c^2}}$$, then you can square both sides and then take 1 over each side to get the equation (1 - (v/c)^2) = 1/4, so (v/c)^2 = 3/4, so v/c = sqrt(3)/2 = about 0.866.
mooneyg3 said:
Anyway... a pod takes off from the spaceship at a slightly greater speed than it is traveling away from the earth. because this pod is traveling at near the speed of light relative to the spaceship it witnesses time dialation so 1 day on the pod is 2 days on the spaceship.

Now this pod is traveling at a slow speed relative to the earth.

So my question is once this pod reaches earth will it observe time moving very slow because 1 day on pod = 2 days on spaceship and two days on spaceship = 4days on earth?
All that matter's is the pod's speed relative to the Earth. If it's moving very slowly relative to the Earth (because its direction relative to the ship was the opposite of the ship's direction relative to the Earth), then the pod's clock and the Earth's clock will be ticking at nearly the same rate.

Thanks! :)