Mirroring sun with parabolic dish

[DaveC426913]

A long while ago I had this idea for a plot of a story. Now I'm not so sure my logic is right.

If I were to place a (perfectly) mirrored parabolic dish out near Pluto, point it at the sun and put my (tiny point-like) spaceship at the focus, how big would the parabolic dish have to be to give me an amount of sunlight equivalent to a spot in Earth's orbit?

I'd originally assumed that, as long as the apparent size of the disc of the dish were equivalent to the apparent size of the disc of the sun (i.e. ~0.5 degrees), that would be enough.

Say you placed your mirror 93 miles away from your spaceship and made the mirror 0.8 miles in diameter. It would appear the same size as the sun from Earth (93 million miles distance at .8 million miles diameter). Because the mirror's focus is set to 93 miles, when you looked at the mirror, you'd see a disc as bright as the sun and as large as the sun.

Is my logic correct?

 

[sysreset]

I think you forgot one factor, which is the angular density of radiation form the sun. At a distance 30AU a mirror will only collect 1/900 the radiation as the same mirror at 1AU. So I think your scenario would work for a spaceship placed, say, in the shadow of an asteroid 1AU from the sun, but not 30AU. unless you increased the apparent size of the dish by a factor of 900.

 

[DaveC426913]

I think you forgot one factor, which is the angular density of radiation form the sun. At a distance 30AU a mirror will only collect 1/900 the radiation as the same mirror at 1AU. So I think your scenario would work for a spaceship placed, say, in the shadow of an asteroid 1AU from the sun, but not 30AU. unless you increased the apparent size of the dish by a factor of 900.

Yes, this is the conclusion I'm coming to. Pluto, being 40AUs more distant would need a mirror that is 40^2 larger in area, or 32 miles in diameter, to capture the same cone of radiation as a spot near Earth.

Huh, I don't even have to calulate the apparent size with trig. The Plutonian sun would have to appear correspondingly larger in area. ie. 0.5x40 = 20 degrees in diameter.

Drat. My cool plot device won't work very well then.


The hero in his tiny underdog spaceship is totally outgunned by a larger, faster ship. They are out in the boondocks of the solar system out by Pluto. He heads for a giant communications dish, tens of kilometers in diameter, which is pointed to the inner solar system. At this distance, it is effectively pointed at the sun (which is why it's coated flat black).

He vapourizes a hunk of mercury, sending the expanding cloud toward the dish. It coats the dish, making a jerry-rigged visible light reflector.

He then parks inside the focus of the dish, luring the enemy ship to come to rest at the focus. As the enemy ship enters the focus, a small area of its hull suddenly finds itself very close to the Sun, which acts like a cutting torch. The enemy captain barely has time to feel foolish before he and his atmo vent into space.

 

[sysreset]

I hope you haven't been testing this mercury splatter technique too much!

 

[tony873004]

Here's my attempt:

You would collect sunlight over pi*0.4^2 = 0.502654824574367 square miles of mirror surface. This sunlight would be 1/40^2 = 0.000625 times as strong as sunlight on Earth. So it would be the equivalent of collecting light with a mirror with a surface area of 0.000625*0.502654824574367 = 3.14159265358979E-04 square miles. This would be a mirror with a diameter of 2*sqr(3.14159265358979E-04/pi) = 0.02 miles 0.02*5280 = 105.6 feet. You would burn a pinpoint hole through your spaceship if it were at the "perfect" focus of this mirror. But if you moved your spacecraft towards or away from the focus so that the beam of light was spread 105.6 feet wide, that should be the equivalent of 1 AU sunlight.

 

[Gokul43201]

This would be a mirror with a diameter of 2*sqr(3.14159265358979E-04/pi) = 0.02 miles 0.02*5280 = 105.6 feet.

Your progression from 15 significant figures down to 1 and then up to 4, is quite strange!

 

[tony873004]

Your progression from 15 significant figures down to 1 and then up to 4, is quite strange!

regarding:

2*sqr(3.14159265358979E-04/pi) = 0.02 miles 0.02*5280 = 105.6 feet.

I've always been taught not to round the intermediate steps. Leave all digits until the final answer and then round the final answer, which is what I did here.

The 0.02 is actually not 1 significant figure. It's also 15. It's just that my word processor's built-in calculator fails to display the trailing zeros. Look at where 0.02 came from: 2*sqr(3.14159265358979E-04/pi). The first 2 is simply to get a diameter out of a formula that would otherwise give radius. So it's not 1 significant digit . It's infinite: 2.00000000000000000000000000000... The rest of the equation is pi*10^-4 to 15 significant digits, divided by pi, which the word processor's built-in calculator recognizes to 15 digits. This gives 0.000100000000000. If I had asked the word processor to display this intermediate step, it would have displayed 0.0001. Taking the square root of this does not change the significant digits, and multiplying it by a pure 2 doesn't either. That's why 0.02 is actually 0.020000000000000 but the calculator doesn't display the trailing 0's. 5280 is simply feet in a mile. I'm not sure if a mile is defined as exactly 5280 feet, making it a pure number as well, which would justify a 15 digit answer of 105.600000000000 feet although it would be displayed as 105.6, or if 5280 is feet in a mile rounded off to 4 digits, which would justify only 105.6 feet. In either case, it would be displayed the same.

But since earlier in the problem, Pluto's solar distance was expressed as 40AU, which is only 1 significant digit, I think my only significant digit mistake is not giving an answer of 100 feet.