A strange question about the sun

[weeman]

This isn't a majorly debatable question, just something that I was randomly pondering:

In a hypothetical situation, if we could stand on the surface of the sun how far would the horizon be? On Earth, if you're in a relatively flat region (i.e. Kansas, Oklahoma, etc.) you can basically see the same distance in every direction. So, with the sun being as large as it is, and not having tall mountains or deep valleys, how far could you see to the horizon? Do astronomers have calculations for horizons depending on the size of the celestial body? Do astronomers even waste time trying to figure it out?

Not sure why I was pondering this, but it just struck me as interesting.

 

[kg]

Great question. Here is the formula (acording to Wikepedia) for finding the distance to the horizon.
http://en.wikipedia.org/wiki/Horizon#Ap ... e_formulas
I'm only a professional farrier so if it doesn't involve smacking something with a hammer I'm not qualified. Maybe someone with a calculator could....
Besides the obvious, (the sun being so bright all you would probably see was the back of your own skull as you were being vaporized) the sun might be massive enough for general relativity to play a role in how far away the vilsible horizon was.

 

[bdewoody]

It's just a simple math problem. If you set your eye at the surface the horizon is zero. At six feet height it would be approximately 31 miles.

 

[R1]

So if the Sun's surface was flat and solid, how many miles would the horizon appear to be
to a six foot tall demon standing on the Sun ?

(I think that's what weeman asks.)

 

[InnyBinny]

So if the Sun's surface was flat and solid, how many miles would the horizon appear to be
to a six foot tall demon standing on the Sun ?

(I think that's what weeman asks.)

Pretty much what bdewoody said. The answer varies a lot depending on what height you are and how many d.p.'s you use.

[sun radius = 6.96*10^8m, 6ft = 1.8288m]

The equation of the sun's circle is:

x^2 + y^2 = (6.96*10^8)^2

And your line of sight is at a tangent to that circle, with y intercect of the sun's radius plus your height:

y = mx + 6.96*10^8 + 1.8288

The gradient of the line is at a tangnet to the circle, so the derivative of the circle:

2x + 2ydy/dx = 0 -> dy/dx = -x/y

[m and dy/dx are both gradient]

Sub in the gradient function:

y = -x^2/y + (6.96*10^8 + 1.8288)
->
x^2 + y^2 = (6.96*10^8 + 1.8288)y

but x^2 + y^2 = (6.96*10^8)^2, so:

(6.96*10^8 + 1.8288)y = (6.96*10^8)^2
y = 695999998.171...

Angle is the cos^-1 of A/H:

o = cos^-1(1391999998.171/1392000000)
= 0.0000724926 rad

arc length = r*o = 696000000*0.0000724926

= 50454.8 m

50 km or 31 miles.

Hope I haven't done anything wrong. :?

EDIT: Changed sun's value from diameter to radius. Woops.

 

[bdewoody]

It's just a simple right triangle. I think I learned the math in the 9th grade.

 

[InnyBinny]

It's just a simple right triangle. I think I learned the math in the 9th grade.

Really? How? Hmm, that would make it a lot quicker. :lol:

The formula derived from what I did above is in the Wikipedia link someone else gave

http://en.wikipedia.org/wiki/Horizon#Ap ... e_formulas

arclength = R cos^-1 (R/[R + h])

 

[bdewoody]

1st leg of triangle is a radial from the center of the sphere to the surface = R

2nd leg is R + 6' the hypotenuse

the right angle is from the horizon point to the eye of the viewer that distance is X

X^2 + R^2 = (R+6)^2

plug in the radius of the sun in feet and solve for X

 

[bdewoody]

So if the Sun's surface was flat and solid, how many miles would the horizon appear to be
to a six foot tall demon standing on the Sun ?

(I think that's what weeman asks.)

the sun's surface if it had one would not be flat it would approximately be the surface of a sphere. If it were flat the horizon would be infinity.

 

[InnyBinny]

1st leg of triangle is a radial from the center of the sphere to the surface = R

2nd leg is R + 6' the hypotenuse

the right angle is from the horizon point to the eye of the viewer that distance is X

X^2 + R^2 = (R+6)^2

plug in the radius of the sun in feet and solve for X

Oh, we were figuring out two very slightly different things...the arclength of the circle from where one's standing vs the linear distance from eyes to horizon.

the sun's surface if it had one would not be flat it would approximately be the surface of a sphere. If it were flat the horizon would be infinity.

I think he means if the sun was perfect, not bumpy...

 

[bdewoody]

1st leg of triangle is a radial from the center of the sphere to the surface = R

2nd leg is R + 6' the hypotenuse

the right angle is from the horizon point to the eye of the viewer that distance is X

X^2 + R^2 = (R+6)^2

plug in the radius of the sun in feet and solve for X

Oh, we were figuring out two very slightly different things...the arclength of the circle from where one's standing vs the linear distance from eyes to horizon.

the sun's surface if it had one would not be flat it would approximately be the surface of a sphere. If it were flat the horizon would be infinity.

I think he means if the sun was perfect, not bumpy...

My way was much simpler and for the need at hand accurate enough.