# Solving for v using algebra

1. Aug 31, 2013

### leroyjenkens

No matter how many math classes I take, algebra is always the most difficult part.
1. The problem statement, all variables and given/known data

Solve for v.

$$\frac{8.1356x10^{16}}{v}=\frac{16}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

c is the speed of light.

3. The attempt at a solution

I tried this several times and keep getting the wrong answer.

I inverse both sides, multiply both sides by 16, divide 16 by 8.1356x1016, square both sides, multiply both sides by c2, add v2 to both sides, which doesn't change the left side, because the coefficient on the v2 on the left is so large, then I divide both sides by 3.86777x10-32, and then square root both sides to get v, which is 5x1015, which is velocity that's almost twice the speed of light. So obviously that's incorrect.
I feel one of my steps is illegal.

Thanks.

2. Aug 31, 2013

### Integral

Staff Emeritus
You are telling us what you wanted to do. We need to see what you did.

PF has a very nice mathematics interface. Please use it to show us what you are doing.

3. Aug 31, 2013

### leroyjenkens

Here is original equation $$\frac{8.1356x10^{16}}{v}=\frac{16}{\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

Inverse it $$\frac{v}{8.1356x10^{16}}=\frac{\sqrt{1-\frac{v^{2}}{c^{2}}}}{16}$$

Multiply both sides by 16 $$\frac{16v}{8.1356x10^{16}}={\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

Divide 16 by 8.1356x1016 $$1.966665x10^{-16}v={\sqrt{1-\frac{v^{2}}{c^{2}}}}$$

Square both sides and multiply both sides by c2 $$3.48x10^{-15}v^{2}={{c^{2}-{v^{2}}}}$$

And here's where I realize I made a mistake. I thought it was a big number on the left, but it was actually a small number. However, I still get a wrong answer.

$$v^{2}=c^{2}$$ I add v2 to both sides, which gives me practically 1v2. So the speed is actually the speed of light. That's incorrect. The correct answer is 1.42x108, as it says in the book. It's close to the speed of light, but not close enough for me to just get v=c.

Thanks.

4. Aug 31, 2013

### DeltaFunction

Something's not right here. The equation doesn't balance for the book answer of 1.42x10^8

5. Sep 1, 2013

### vela

Staff Emeritus
When the answer is so close to $c$, you need to solve for the difference between v and c, otherwise you run into numerical complications.

You're ending up with something of the form $\varepsilon v^2 = c^2 - v^2$ where $\varepsilon << 1$. If you solve for $v/c$, you'll get
$$\frac{v}{c} = \sqrt{\frac{1}{1+\varepsilon}}.$$ You want to expand the righthand side in a Taylor series about $\varepsilon=0$. In other words, use the approximations
\begin{align*}
\frac{1}{1+x} &\cong 1-x \\
(1+x)^{1/2} &\cong 1+\frac{1}{2}x
\end{align*}

EDIT: Actually, based on your book's answer, you shouldn't have to do this. There's something wrong with your original equation, as DeltaFunction noted.

6. Sep 1, 2013

### Integral

Staff Emeritus
Where did the initial equation come from?

7. Sep 1, 2013

### Ray Vickson

Sometimes it is easier to use symbols instead of numbers, then substitute in the numbers at the end. So, write your equation as
$$\frac{a}{v} = \frac{b}{\sqrt{1 - \frac{v^2}{c^2}}}$$.
Square both sides and multiply through by both denominators, to get
$$a^2 \left( 1 - \frac{v^2}{c^2}\right) = b^2 v^2.$$
This is a simple linear equation in the variable $v^2$.

8. Sep 1, 2013

### D H

Staff Emeritus
You probably made an earlier mistake as well. A physics textbook is not going to ask you to solve $8.1356\times10^{16}/v = 16/\sqrt{1-(v/c)^2}$. That is your equation, not the text's.

What is the problem as it is stated in your text?

9. Sep 2, 2013

### leroyjenkens

It's an example problem for time dilation. The equation is written exactly like that in the book, and they solve for v. Must be a mistake in the book. I'll just show it to the professor Wednesday and see what he says.
Thanks for the responses.

10. Sep 2, 2013

### Ray Vickson

If you solve the problem exactly as I suggested in my previous post, namely
$$\frac{a}{v} = \frac{b}{\sqrt{1 -\frac{v^2}{c^2}}} \Longrightarrow v = \frac{ac}{\sqrt{a^2 + b^2 c^2 }}$$
and then substitute $a = 8.1356 \times 10^8, b = 16, c = 2.9979 \times 10^8,$
you should get $v = 2.997899999 \times 10^8 \doteq 2.9979 \times 10^8.$ Presumably, $v = c$ except for roundoff errors.

11. Sep 2, 2013

### leroyjenkens

Yeah I solved it that way, thanks. And I did get v = c.

The problem has a ship traveling 4.3 lightyears to a planet, and it only has 16 years worth of food for the crew. It's asking how fast the ship would have to travel to make it there in 16 years. So the speed won't be as fast as the speed of light.
We have two systems, one is stationary (the one observing the ship) and the other is moving with the ship. Both systems experience a different time duration of flight.
Using the time dilation equation, we have...

$$T=\frac{2L}{v}=\frac{T'_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

Where T is the time the observers measure in the stationary reference frame.
L is the distance to the star.
$T'_0$ is the time the people measure in the moving reference frame.
Converting light years into meters:

$$\frac{2(4.0678x10^{16})}{v}=\frac{T'_0}{\sqrt{1-\frac{v^2}{c^2}}}$$

The book got the answer for v as $v=0.473c=1.42x10^8\frac{m}{s}$
And T (stationary reference frame) = 18.2 years.

I didn't think I'd need to post the whole thing, but maybe I made a mistake.
I'm not getting their v answer. Using their v answer, I get the T answer they got, though.

12. Sep 2, 2013

### vela

Staff Emeritus
Recheck your units. You'll see they don't work out the way you think they do.

13. Sep 3, 2013

### DeltaFunction

All is clear now :) you have made a mistake. When in doubt, put everything in SI

14. Sep 4, 2013

### Ray Vickson

This equation---exactly as written---is difficult to solve without access to at least double (but preferably higher) precision numerics. Taking c = 2.9979e8, your equation becomes
$$\frac{A}{\beta} = \frac{1}{\sqrt{1-\beta^2}}$$
where
$$A = \frac{8.1356e16}{16 \times 2.9979e8} = 1.696103939e7$$
and $\beta = v/c$.
Since $0 \leq \beta < 1,$ $A/\beta > 1.696103939e7,$ so $\beta$ must be very near 1 to make the right-hand-side large.

Using 10-digit accuracy the solution is $\beta = 1.000000000$. Using 15-digit accuracy gives $\beta = .999999999999998$. Using 30-digits accuracy we have $\beta = .999999999999998261938743055454$.