Distance traveled, v->c

1. Feb 1, 2006

Pengwuino

I have this equation here and I don't see how I can isolate v, the speed of the ship...

$$\frac{{2(4ly)(9.5x10^{15} \frac{m}{{ly}}}}{v} = \frac{{16y}}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }}$$

y = years
ly = light years

The book doesn't tell you how it got v=0.447c, it just jumped to the conclusion after presenting the equation. It seems like no matter how I try the equation, the v-squared will always cancel and i'll lose my velocity equation. Am I not doing this correctly or is there a more complex method of solving for v?

2. Feb 1, 2006

Cyrus

I will do some cleaning up:

$$a = 2(4ly)(9.5x10^{15})$$

$$b=16y$$

so we write it much cleaner as:

$$\frac{a}{v} = \frac{b}{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }$$

Now I just multiply by the denominator of the RHS and get:

$$\frac{a}{v} [\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }] = b$$

Now I just bring the v inside the square root:

$$a[\sqrt {\frac{1}{v^2} - \frac{{1}}{{c^2 }}}}] = b$$

No v's cancel out.

You can simply it to get,

$$( \frac{1}{v^2} - \frac{1}{c^2}) = \frac{b^2}{a^2}$$

$$\frac{1}{v^2} = \frac{b^2}{a^2} + \frac{1}{c^2}$$

invert both sides and take square root:

$$v = +/- \sqrt{(\frac{b^2}{a^2} + \frac{1}{c^2})^{-1}}$$

Last edited: Feb 1, 2006
3. Feb 1, 2006

quasar987

Just from eying the equation, wouldn't squaring both sides, then multiplying both sides by v²(1-(v/c)²) give you a quadratic equation in v²? the uniqueness of the v would then be due to the fact that one of the root is negative, so not a physically acceptable solution.

4. Feb 1, 2006

Pengwuino

Ok I have another problem where I can't seem to isolate v correctly... A rocket will go to a galaxy 20 light years away and it can only take 40 years round trip according to the rocket passengers. I figured this is how i could write it.

$$\begin{array}{l} T' = \frac{{2L}}{v} = \frac{{T_0 }}{{\sqrt {1 - \frac{{v^2 }}{{c^2 }}} }} \\ \sqrt {1 - \frac{{v^2 }}{{c^2 }}} *2L = T_0 *v \\ \sqrt {4l^2 - \frac{{4l^2 v^2 }}{{c^2 }}} = T_0 v \\ 4l^2 - \frac{{4l^2 v^2 }}{{c^2 }} = T^2 v^2 \\ 4l^2 (1 - \frac{{v^2 }}{{c^2 }}) = T^2 v^2 \\ \frac{1}{{v^2 }} - \frac{1}{{c^2 }} = \frac{{T_0 ^2 }}{{4l^2 }} \\ \end{array}$$

Now im not sure how to isolate v

5. Feb 1, 2006

Cyrus

Add 1/c^2 to both sides, take the inverse, then take the square root, exactly as before.

6. Feb 1, 2006

Pengwuino

How do i do the inverse of two added quantities though?

7. Feb 1, 2006

Cyrus

make them a common fraction and do the addition. Then take the inverse of that new fraction you get.

8. Feb 1, 2006

Pengwuino

How would i do that.... man i can't do basic math!!!

9. Feb 1, 2006

Pengwuino

Do you do this?

$$v^2 = \frac{{4l^2 + c^2 }}{{T_0 }} \]$$

10. Feb 1, 2006

Cyrus

In your latter example you would get:

$$\frac{T^2}{4L^2} + \frac{1}{c^2}$$

$$\frac{c^2T^2}{4L^2c^2}+ \frac{4L^2}{4L^2c^2}$$

$$\frac{c^2T^2+4L^2}{4L^2c^2}$$

The inverse would be:

$$\frac{4L^2c^2}{c^2T^2+4L^2}$$

Last edited: Feb 1, 2006
11. Feb 1, 2006

Pengwuino

Dang that's insane... thats from like 5th grade :( and i forgot!

12. Feb 1, 2006

Pengwuino

Ok and since I know the space ships speed to be 0.894c.... does that mean the time measured from the stationary observers on Earth is ~2.236*40 years using gamma*T proper?

13. Feb 1, 2006

Cyrus

Sorry, im not a physics major. Thats as much help as I can provide.

Last edited: Feb 1, 2006
14. Feb 1, 2006

HallsofIvy

Staff Emeritus
You are great up to this point but remember you want to solve for v- you want to isolate it, not mix it with other things.
$$4l^2= T^2v^2- \frac{4l^2}{c^2}v^2= (T^2- \frac{4l^2}{c^2})v^2$$
$$v^2= \frac{4l^2}{T^2- \frac{4l^2}{c^2}}$$