# Laser moving across the moon

1. Sep 9, 2013

### leroyjenkens

1. The problem statement, all variables and given/known data
So basically I'm given the distance to the moon (384,000,000 meters), and I need to find out how many degrees per second I'd need to move the laser for the spot touching the moon to move faster than the speed of light.

2. Relevant equations
$w=\frac{v}{r}$

3. The attempt at a solution
I have two ways of doing this and they both give me different answers.
The easy way I thought of would be to make a right triangle with the x being the distance to the moon, and the y being the distance light travels in a second. Then finding the angle of such a triangle gave me $\frac{299,792,458}{384,000,000}=tanθ$, and then I took the inverse tangent to find the angle, which is 38°. So the answer would be 38° per second I'd have to move the laser for the end of the beam to move faster than the speed of light. Well, I'd have to move it any amount greater than that to break the light barrier.

The other way I used angular velocity, which is $w=\frac{v}{r}=\frac{299,792,458}{384,000,000}=0.78$ radians per second, and then I multiply that by 57.3 to turn it into degrees, which gives me 44.8° per second. One of these is wrong. Possibly both. Any suggestions?

Thanks.

2. Sep 9, 2013

### Jufro

Your first method you are solving for θ.

This is just an angle (degrees, radians). What you want is radians / second.

So, from just the unit standpoint, the second way returns the correct answer

3. Sep 9, 2013

### leroyjenkens

The first way is solving for θ, but it's solving for the angle where the opposite side is the distance light travels in a second. So wouldn't that be degrees per second?

The problem asks for degrees per second, that's why I converted to degrees.

4. Sep 9, 2013

### Enigman

tanθ doesn't have any units and therefore tan inverse would give you just angle.
A better approach would be through differentiation of $x/3.84*10^7=tanθ$ and Differentiating wrt time

5. Sep 9, 2013

### Jufro

Tanθ has no units. So, dividing as you did does not make sense.

Again, I am not sure about the equation from the second method, but the first method does not work out.

**Edit**
What Enigman said

6. Sep 9, 2013

### vela

Staff Emeritus
It turns out your first method is less accurate. You could do a similar calculation based on the same idea, however. The radius of the moon is 1740 km, so half the angle the moon subtends is $\theta=\tan^{-1} \frac{1740}{384000}$. The time it would take the spot to move one moon radius is $t = 1740/300000\text{ s}$. You should find that $\theta/t = 43^\circ$ per second, which agrees with what you got with your second method.

So why do you suppose your first method didn't yield an accurate result?

7. Sep 9, 2013

### D H

Staff Emeritus
No, it's not. You dropped the units. Don't do that. With units intact, here's what you attempted to solve for by your first method:
$$\frac{299,792,458\,m/s}{384,000,000\,m} \approx 0.7807\,s^{-1} = \tan\theta$$
The left hand side has units of inverse time but the right hand side is dimensionless. That's inconsistent dimensionally, so it's wrong. Period.

Don't discard units. They provide a nice sanity check to see if you are on the wrong track.

8. Sep 9, 2013

### vela

Staff Emeritus
He didn't drop the units. (Well, he dropped the units, but his mistake isn't the one you think he made.) He said that y was the distance light travels in one second. It's the same time he used to divide the angle he calculated to find the angular speed.

Last edited: Sep 9, 2013
9. Sep 9, 2013

### D H

Staff Emeritus
Good point. So, leroyjenkins, what is your answer to vela's question in post #6?

10. Sep 9, 2013

### leroyjenkens

Thanks for the responses, guys.
Well, I was initially skeptical if I could create a right triangle for this problem. I'm guessing the reason I got the wrong answer is that I can't do that? I guess maybe because the hypotenuse is going to be longer than the adjacent side, which I'm not supposed to have. I'm not exactly sure why though.
I was thinking about this problem on the way home from school. I was going to use it to check my original answer, which was done using the angular speed formula.

11. Sep 9, 2013

### Enigman

12. Sep 9, 2013

### D H

Staff Emeritus
Suppose that you have to cross some railroad tracks on the way home, and today you got stuck behind the crossing guard. To pass the time you decide to keep your eye on the locomotive as long as you can. How fast do you need to turn your head when the locomotive crosses in front of you? How about when the train has almost finished crossing the road and the locomotive is far away?

13. Sep 9, 2013

### leroyjenkens

Thanks. I'm gonna read through that when I get home from work tomorrow. This is the kinda stuff that got me into physics. Unfortunately, there's a few physics classes I have to take that I'm not interested in.
I'd have to turn my head the speed of the locomotive when it's directly in front of me. When it's far off to my left or right, I don't need to turn my head very fast at all.
So correct me if I'm wrong, but my right triangle answer is wrong because I need to create an arc, not a flat side of a triangle, which means I can't use a right triangle. The arc will make the spot of light travel the same speed the entire duration of movement, while the right triangle will have the spot of light traveling at a decreasing speed as it climbs to the top of the opposite side of that triangle.

14. Sep 9, 2013

### D H

Staff Emeritus
You can use a triangle, not just the big honkin' triangle that you used. For example, if you had used a time of one microsecond instead of one second you would have been fine. For small values of θ, tan θ ≈ θ.