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Science Forum Index » Astro - Amateur Forum » Distance to horizon on Moon and Earth
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| canopus56 |
 Posted: Sun Feb 18, 2007 2:48 am |
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What is the visible distance from the horizon on the Moon and Earth
and for any given planet of radius R, how can this be computed? -
Canopus56 |
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| Brian Tung |
| Posted: Sun Feb 18, 2007 2:48 am |
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canopus56 wrote:
Quote: What is the visible distance from the horizon on the Moon and Earth
and for any given planet of radius R, how can this be computed? -
For any airless, spherical world of radius R, an observer at altitude h
(h << R) can see to a distance of sqrt(2Rh). Thus, for instance, the
Moon has a radius of 3,500 km, more or less. An observer at an altitude
of 10 m (that is, 0.01 km) can see to a horizon distance of
sqrt(2*3,500*0.01) = sqrt(70)
or about 8.3 km. On the Earth, atmospheric refraction extends this
distance, and per a suggestion from Andrew Young, the formula is closer
to sqrt(2.3Rh).
--
Brian Tung <brian@isi.edu>
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.html |
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| Brian Tung |
| Posted: Sun Feb 18, 2007 12:39 pm |
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Bill wrote:
Quote: Actually, this is an approximation. It works well when h is
small compared to R but the exact formula is:
Dist = sqrt(2Rh + h*h)
I did state h << R as a condition for the validity of the formula.
Incidentally, although the formula you cite is more accurate for an
airless world, the Earth isn't airless. The recommendation from Andrew
Young is to pretend that the Earth's radius is about 15 percent larger
than it actually is. This correction probably goes away once you are
sufficiently far enough from the Earth. On the other hand, I don't
think that most people are interested in a formula for distance to the
horizon from an "altitude" of several Earth radii.
--
Brian Tung <brian@isi.edu>
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at http://astro.isi.edu/reference/faq.html |
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| S y z y g y |
| Posted: Sun Feb 18, 2007 6:12 pm |
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Actually, this is an approximation. It works well when h is
small compared to R but the exact formula is:
Dist = sqrt(2Rh + h*h)
You can see how much error is in the approximate formula by
assuming you are 25,000 miles above the surface of the
Earth. The approximate formula gives a result of 14,100
miles. Well, with a little thinking you can convince
yourself that if you are 25000 miles above the surface of
the Earth (say 29000 miles above the center) the horizon is
about 29000 miles away too, certainly no where near 14,100
miles.
The exact formula gives 28,723 miles from the horizon.
On the Earth the horizon is one mile away if you are 8
inches high.
- Bill
"Brian Tung" <brian@isi.edu> wrote in message
news:er8tf4$h4f$1@praesepe.isi.edu...
Quote: canopus56 wrote:
What is the visible distance from the horizon on the
Moon and Earth
and for any given planet of radius R, how can this be
computed? -
For any airless, spherical world of radius R, an observer
at altitude h
(h << R) can see to a distance of sqrt(2Rh). Thus, for
instance, the
Moon has a radius of 3,500 km, more or less. An observer
at an altitude
of 10 m (that is, 0.01 km) can see to a horizon distance
of
sqrt(2*3,500*0.01) = sqrt(70)
or about 8.3 km. On the Earth, atmospheric refraction
extends this
distance, and per a suggestion from Andrew Young, the
formula is closer
to sqrt(2.3Rh).
--
Brian Tung <brian@isi.edu
The Astronomy Corner at http://astro.isi.edu/
Unofficial C5+ Home Page at
http://astro.isi.edu/c5plus/
The PleiadAtlas Home Page at
http://astro.isi.edu/pleiadatlas/
My Own Personal FAQ (SAA) at
http://astro.isi.edu/reference/faq.html |
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| canopus56 |
| Posted: Sun Feb 18, 2007 8:59 pm |
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On Feb 17, 11:58 pm, b...@isi.edu (Brian Tung) wrote:
Quote: canopus56 wrote:
snip> > What is the visible distance from the horizon on the Moon and
Earth
Quote: and for any given planet of radius R, how can this be computed? -
Brian replied:
For any airless, spherical world of radius R, an observer at altitude h
(h << R) can see to a distance of sqrt(2Rh). <snip
Thanks for the replies Brian. Young's construction was just what I
was looking for.
http://mintaka.sdsu.edu/GF/explain/atmos_refr/dip.html
- Kurt |
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| TMA |
| Posted: Mon Feb 19, 2007 3:16 am |
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"Brian Tung" <brian@isi.edu> wrote in message
news:eraki8$bro$1@praesepe.isi.edu...
Quote: Bill wrote:
Actually, this is an approximation. It works well when h is
small compared to R but the exact formula is:
Dist = sqrt(2Rh + h*h)
I did state h << R as a condition for the validity of the formula.
Incidentally, although the formula you cite is more accurate for an
airless world, the Earth isn't airless. The recommendation from Andrew
Young is to pretend that the Earth's radius is about 15 percent larger
than it actually is. This correction probably goes away once you are
sufficiently far enough from the Earth. On the other hand, I don't
think that most people are interested in a formula for distance to the
horizon from an "altitude" of several Earth radii.
is this a derivation from radian measure along an arc? |
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| Starboard |
| Posted: Mon Feb 19, 2007 8:43 am |
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Guest
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Quote: Dist = sqrt(2Rh + h*h)
is this a derivation from radian measure along an arc?
Nope. Just a little Pythagorean Theorem (a^2 +b^2 = c^2).
To help visualize it, begin with a circle of radius R. That radius
becomes one leg of a right triangle and the radius + h (h is the
height above the surface) becomes the hypotenuse. Let x be the unknown
leg of the triangle (the distance to the horizon).
So:
a^2 +b^2 = c^2
R^2+x^2=(R+h)^2
Solve for x
x^2=R^2+2Rh+h^2-R^2
x=sqrt(2Rh+h^2) or x=sqrt(2Rh+h*h)
Errol |
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