So two ships leave earth in opposite directions

[pigasuspig]

This has probably been answered before, but I couldn't find it online.

So two spaceships leave Earth in opposite directions. Suppose they are both traveling away from Earth at .4c.

From one of the ships, how fast would the other appear to be moving away?

And then, same question, except both ships are going .8c away from Earth. If it would be impossible for the other ship to be moving 1.6c away, how fast would it be moving away (ignoring Earth)?

 

[Pengwuino]

Use the relativistic formulas.

(U+V)/sqr(1-(v^2)/(c^2))

 

[pigasuspig]

unfortunately, I'm not a physics student - I'm just curious. How does the formula work?

 

[Janus]

Use the relativistic formulas.

(U+V)/sqr(1-(v^2)/(c^2))

That should be:

$$\frac{U+V}{1+\frac{UV}{c^2}}$$

To answer the OP's question, U is the velocity of one ship with respect to Earth and V is the velocity of the other ship. The answer is the velocity of the ship with respect to each other as measured from either ship.

For this particular example this gives:
$$\frac{0.4c+0.4c}{1+\frac{(0.4c)(0.4c)}{c^2}}= .69c$$

 

[Pengwuino]

oops!

I was using the original equation and trying to remember what the 2 velocity one was and forgot the bottom portion.

 

[εllipse]

To expand upon what Janus has said, you will notice that any time you fill in two velocities which are less than the speed of light, you will be returned a velocity less than the speed of light. Even if you use $$U = V = .99999c$$.

Also note that if you fill in c for one of the velocities, you will be returned c, no matter what the other velocity is:

$$\frac{U + c}{1 + \frac{Uc}{c^2}} = \frac{U + c}{\frac{c}{c} + \frac{U}{c}} = \frac{U + c}{\frac{c + U}{c}} = c(\frac{U + c}{c + U}) = c$$

So everyone always measures the speed of light to be c.

 

[pigasuspig]

Thank you all for your help - it's great to know such sites as these exist.