Targeting the Sun: A Hypothetical Look at Laser Power and Sky Position

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To target the Sun with a powerful laser, one must account for the Earth's rotation and the time it takes for light to travel from the Sun to Earth, which is approximately 8 minutes. When aiming, it is suggested to lead the target by aiming slightly ahead, approximately 16 minutes in the Sun's apparent motion across the sky. The discussion also touches on the notion that a laser beam, once fired, travels in a straight line and is not affected by the Earth's movement, making precise calculations essential for a successful hit. Additionally, the effectiveness of such a laser on the Sun is questioned, as it would likely have no significant impact. Ultimately, the conversation explores the physics behind aiming at a celestial body while acknowledging the hypothetical nature of the scenario.
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Say there is a very powerful laser, so powerful it's capable of destroying the Sun. What point in the sky should you aim for in order to hit the sun with a laser beam?

Thanks in advance for your help.
 
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Think of how you lead a duck with a shotgun. Same idea. Just calculate, given your relative motion, where it will be in the number of minutes or hours it will take for your beam to get there. If you happen to be in orbit around the sun, it wouldn't be a good idea to do it at all. (Not that a laser would have any effect anyhow, but I'm staying with your original question.)
 
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You can calculate it for yourself easily enough: 24 hours in a day, 360 degrees in a circle, and the sun is .5 degrees in diameter.
 
Yea, but what about the time it takes to the Sun's light to reach the Earth? What role does that play?
 
If I remember correctly, we are 8 light-minutes away from the sun. The sun makes a full revolution around the Earth in 365 days! (=365*24*60=5,256x10^5 minutes) In 8 minutes, the sun almost doesn't move. So to be on the safe side, aim a little bit off the center of the core and you will hit the core.

I don't know what the effects of the laser will be tought. Maybe you're hoping to transfer the energy of the laser into thermal energy of the core to accelerate the rate of fusion and make it capable of producing iron, which will lead to nova?

I wonder how fast this process can happen. How much energy would be required to make our sun go nova in a lifetime? lol
 
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The whole thing about the laser is just to have a (silly) excuse to wonder how to do this.

I think that the movement that is important here is the rotation of Earth around it's own axis, rather than its movement around the Sun.
 
russ_watters said:
You can calculate it for yourself easily enough: 24 hours in a day, 360 degrees in a circle, and the sun is .5 degrees in diameter.
:bugeye: Russ! Making an assumption! :eek:
The OP never specified where this thing is being fired from. If it has enough power to blow up a star, it sure as hell isn't anywhere near Earth.
 
Sorry, as stupid as this may sound, the laser is asumed to be fired form the Earth.
 
I had forgotten about the Earth's rortation about its axis. Well, 24 hours in a day: determine the tangential speed at the latitude of your canon of doom. you know that the total speed of the laser is c, so you can find the angle with the surface at which the laser will spread. One way to correct for that deviation would be to aim your canon at that same angle but in the opposite direction, so that as a result, the laser leaves perpendiculr to the surface of the earth.
 
  • #10
Something else...
I don't have the data at hand, but a laser does not obey the inverse square law when fired in space (I don't know about in atmosphere). I'll get hold of the info as soon as I can, but it might be a couple of days.
 
  • #11
What do you mean by that? Are you referring to the intensity of the laser?

ASAIK (and i don't know much on this subject!) a laser is a cylindrical plane wave and thus its energy density not fall off with distance, as opposed to a sperical wave, whose amplitude drops-off with 1/r, and thus its energy density as 1/r².
 
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  • #12
I'm assuming the laser doesn't have to be active the entire time it takes for it to reach the sun. Turn the laser on for a second then turn it off. The resulting beam will be a projectile. The Earth's rotation doesn't matter once the projectile is away.
 
  • #13
And yes, the one piece of information I forgot was that 8 minutes. So we have all the info now...
 
  • #14
If you watch the movement of the Sun across the sky, and calculate where it will appear to be in 16 minutes, you should nail it. That's because ti takes about 8 minutes for sunlight to reach Earth. So, if you think o fwatching your laser beem travel to the Sun, you have to realize that it will take about 8 minutes to complete the trip, and will lag 8 minutes behind whatever lead you gave by the time it arrives.

So let's say you're facing South when you fire. As you watch the Sun, it appears to be moving from left to right across the sky. You aim 16 minutes to the right (ahead of it) and shoot. As you watch, you will see the laser taking 8 minutes to reach the Sun. During those 8 minutes, the Sun will continue moving to the right, and your beam will appear to drift to the left (by the same amount). They will arrive at the same location at the same time 8 minutes after you fired.
 
  • #15
And if the weapon didn't happen to be a laser wouldn't you have to compensate for refraction of light from the sun?
 
  • #16
:smile: I'm no physicist, but come on guys...

This is a simple Newtonian situation. The Sun is not moving relative to us. The laser beam is not affected by our motion. If you need to fire a sustained beam, then you'll need to track your target because *your firing platform* is moving/changing orientation. I imagine there is some diffraction in our atmosphere, so the sun is not actually where it appears to be to us, so you need to adjust your aim for that.

But in simple terms, you aim straight at it and fire. That in the minutes it takes for the beam to hit will see your firing platform moved is irrelevant.

So long as you don't hit or "near miss" any substantial mass on the trip, the beam will travel straight and true, and the Sun will be in the same place relative to the beam when the journey is complete.

...and our solar system ceases to exist as we know it!
 
  • #17
First of all, if you hit the sun with a laser it wouldn't do anything. Second, if you want the answer to your question, ask anybody who has studied astronomy.

Also, Lurch is correct... aim ahead

A cool question is: Anything regarding the Aharonov-Bohm effect
 
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