Homework Help: Does Enterprise see the shuttle approach?

1. Aug 10, 2009

NikkiNik

1. The problem statement, all variables and given/known data

Deep Space Nine sees Enterprise and a shuttle approach from exactly opposite directions with 0.6 c and 0.4 c, respectively. At what fraction of the speed of light (b) does Enterprise see the shuttle approach?

2. Relevant equations

V2=(v1+ v1,2)/[1 +(v1*v1,2/c^2)]

3. The attempt at a solution

I used -0.6c as v2 and 0.4c as v1 but when I solved it in the end I had 1c= 1v1,2 so I know I did something wrong I just don'e know what

Last edited by a moderator: Jan 7, 2014
2. Aug 10, 2009

diazona

Re: Relativity

The way you've written the formula, you should be using v1,2 = -0.6c and solving for V2. Although I think a better way to write it is
$$u = \frac{v_1 - v_2}{1 + \frac{v_1 v_2}{c^2}}$$

Last edited: Aug 10, 2009
3. Aug 10, 2009

NikkiNik

Re: Relativity

Thanx!

4. Aug 10, 2009

diazona

Re: Relativity

Oh wait, I made a sign error - see my edit. It should be a minus sign on top. (Because, think about it: if both Enterprise and the shuttle were approaching DS9 at the same speed, .5c, you definitely should not be getting 0 for the relative velocity.)

5. Aug 10, 2009

NikkiNik

Re: Relativity

Ok I solved it the way you said and I got 2.63e-1c but that's not right. I even got the same answer when I solved it my way?????

6. Aug 10, 2009

NikkiNik

Re: Relativity

Oh ok

7. Aug 10, 2009

NikkiNik

Re: Relativity

Sorry one more question...if I add the velocities at the top then it would be 1/7.6e-1 and the answer would be 1.32c which I don't think is right. And if I solved it the first way, the answer being 0.263c (positive or negative) is incorrect...what am I doing incorrect?? Sorry I keep replying so quickly

8. Aug 10, 2009

Geezer

Re: Relativity

I think you're plugging in some number incorrectly. When I plug in, I get a relative velocity that's higher than both 0.6c and 0.4c, but that is still less than c.

Think of the expression in the numerator--the v1 - v2 term--as being the classical relative velocity. Classically, the relative speed would be 1c (0.6c + 0.4c), right? Is that what you used?

9. Aug 10, 2009

NikkiNik

Re: Relativity

yes I did (.4c)-(-.6c) =1c

then I didvided by (1 +(-.24c^2/c^2)=0.76

I divided the first answer by the second and got 1.316c ?

10. Aug 10, 2009

Geezer

Re: Relativity

Why are you using -0.6c and +.4c in the denominator? That would be inconsisent with the numerator, where you're allowing the two speeds to have the same algebraic sign...

11. Aug 10, 2009

NikkiNik

Re: Relativity

Ok I see but why would both of the signs be negative and not only one?? Is it because they are going in opposite directions

12. Aug 10, 2009

Geezer

Re: Relativity

Don't think too hard about the signs. Just think about the "relative" speed between the two spaceships. According to either the Enterprise or the Space Shuttle, the other is approaching at 1c classically.

The term in the numerator just gives us the classical relative speeds. Since they are approaching each other, the perceived speed is going to be a speed larger than what the Enterprise perceives for either of them. If you have a negative sign for either of them, the effect will be decreasing the relative speeds, instead of increasing them.