What is the distance from standing point to the horizon?
466473
The answer to this all depends on how high your head is off the ground.\n\nThis is an excellent page where you can type in how tall you are, and it will tell you what the straight line distance to the horizon is, and the curved distance.\n\nhttp://newton.ex.ac.uk/research/qsystems/people/sque/physics/horizon/
33578
You would be at 40 46.12'S, 106 41.52'E which is in the Indian ocean. The closest land mass is 700 Miles North East (Western Australia).  \n\nThere is almost Nothing near this location but water. \nPerhaps an uncharted Island?\n\nCheck it out with google earth.\n\nYou can use the measure tool to find the point where the straight line will ROTATE like a spinning compass because the line always shows the shortest distance.\n\nAs the crow flys (or swims in this example), these two points are 12,414.94 Miles Apart.
79282
the average distance between the Earth and the Sun (approximately 150 million km or 93 million miles).
611133
Curved. It is the last of the land that the vision can see lining with the sky. Since earth is spherical, the horizon has to be curved.
575563
If it's perfectly flat (Remember the earth is curved), you can see forever.. Stars are a LONG way away, and there is a plane (ok an infinate numbers of planes) that connect you and that star.. Which is the same as "perfectly flat" ground.\n\nNow, Seriously:\nFor someone who is 5' 4" (eyes 5 feet from the gorund) you can see: 3.0097474977147174 miles.\n\n-HtJ
786773
Their distance from the Sun
969415
It's a geometry problem.  Suppose you're high up in a building on the beach overlooking the ocean.  You're at height h above sea level.\n\nNow draw a circle which represents the earth, with radius r.  Your distance from the center of the earth is r+h.  The horizon is some distance away, x.  Your line of sight to the horizon is <tangent> to the radius of the earth, forming a right triangle, where the hypoteneuse is r+h.\n\nUsing the Pythagorean Theorem,\n\nr^2 + x^2 = (r+h)^2 = r^2 + 2rh + h^2\nx^2 = 2rh + h^2\nx = sqrt(2rh + h^2)\n\nThat's the exact answer,  But for practical purposes, 2rh is <much, much> bigger than h^2.  To see this, suppose you're 500 feet in the air.  That's about a tenth of a mile, and using miles, h^2 = 0.1^2 = 0.01.  On the other hand, the earth's radius is about 4000 miles, so 2rh = 2 x 4000 x 0.1 = 800.  This is 80,000 times bigger than h^2.\n\nSo we can safely ignore the h^2, and say x = sqrt(2rh).  That can be broken down further.  Again using miles (you can also do this in kilometers),\n\nx = sqrt(2rh) = sqrt(2r) sqrt h = sqrt(8000) sqrt h = 90 sqrt h\n\n(rounding off, since sqrt 8000 = 89.44; also, h must be in miles).\n\nUsing the formula x = 90 sqrt h, suppose you're in an airplane 4 miles up in the air (21,000 feet).  Then the horizon is 90 sqrt 4 = 180 miles away.\n\nOkay, now let's work it out for h in feet instead of miles.  We'll go back to the formula x = sqrt(2rh), and use r = 3960 miles (a more accurate figure).  Then\n\nx = sqrt(2 x 3960 x h / 5280) = sqrt(3h/2) = 1.225 sqrt h\n\nwhere h is in feet and x is in miles.\n\nTwo examples:  (1) For h = 6 feet, use x = sqrt(3h/2) = 3 miles.  (2) For x = 100 feet, use x = 1.225 sqrt h = 12.25 miles.\n\nTo recap:  For h in miles, use x = 90 sqrt h.  For h in feet, either use x = sqrt(3h/2) or x = 1.225 sqrt h.\n\nIn all cases, x will be in miles.  And of course, all of this can be done using meters and kilometers.
912921
altitude = height = 552km\nAngle of dep = 39.7\n\nTangent of Angle = Alt/ Radius\n\nRadius = Altitude/ Tan(39.7) = 552/tan(39.7) I think... p.s not really sure
179969
Up to the horizon...!
430656
1.Average distance:384,400 km\n2.Maximum Distance(because its orbit is elliptical or oval in shape):405,696 km\n3.Minimum distance:363,104 km
523034
The first answer *edit* was modified so now what I say here makes no sense - but her website confirms that I am right (5km = ~3 miles)*. If we want to know the distance to the horizon from a vantage point on a sphere, the only thing that matters is the radius of the sphere and the height of the vantage point.\n\nSo, assume you are standing with your eyes at 5'5" looking out at a calm ocean. You will typically be able to view the sea surface at 3 arcminutes, or roughly 5 kilometers, out to sea. Beyond that you would still be able to see ships and whatever else sticking above the water quite easily, but upon close inspection realise they appear deeper in the water than they actually are.\n\nAlso note that light bends around mass, so you might be able to "see" farther than your actual vantage point, but I'm not doing that calculation.\n\nI assume that people who have values higher than mine are thinking about higher vantage points than mine, because I think my estimations are close and I'm positive that my logic is right. Correct me if my numbers are off, but I'd like to know why.
857627
The black hole will have the same gravitational force on distant objects.  The event horizon is where light cannot escape.  The distance of the event horizon is approximately 2 km (1.24 miles) for every solar mass.  So it would have to have a lot of solar mass to engulf things around it.
6601
Acording to Kepler's first law of planetary motion, all objects obit other objects in eliptical paths.  This means that the distance from the earth to the sun is not constant.  The average distance is 149600 km or 92 957.1304 miles.  This means that it would take 1239.42 hours or 51 Days 15 hours.
1134341
The answer would depend upon the observer's height, or more accurately, the height of the observer's eyes from sea level.  For simplicity's sake, let's use the numbers 1.5m for eye height, 6,375,000m for the radius of Earth, and of course, assume that the planet is a perfect sphere, our observer has perfect eyesight, etc.\n\nThe line of the observer's vision when it meets the horizon forms a tangent to the surface of Earth.  If you draw a line from this point to the center of the planet, and another line from the observer to the center, you form a right triangle.  One leg is 6,375,000m and the hypotenuse is 6,375,001.5m.  So the other leg can easily be computed using the Pythagorean Theorem:\n\nd = sqrt[ (6,375,0001.5)^2 - (6,375,000)^2 ] = 4373.21\n\nSo to our observer, the horizon is about 4.37km away.\n\nNote that this is the limit of vision for objects with zero height, e.g. something painted on the ground (assuming again our observer has perfect eyesight).  S/he can see objects even further away if they do have height.  For example, two observers with identical height will be able to see one another at twice the distance, though it will only be one another's foreheads poking a tiny bit above the horizon.\n\nLarge objects such as buildings, or ships sailing in, may be seen at a much greater distance, and this distance is also made much greater if the observer moves to higher ground.  For example, even going to the top of a small, 10m building (only a single storey) will extend the horizon to 12.1km, and standing atop a 50m cliff extends it to 25.6km.
55393
It is going side to side...just rmember what the horizon looks like...vertical is up and down.
643004
At an altitude of 2,938 meters, the straight-line distance to the horizon would be 193.615 km.\n\nSee the equation to solve on this website --> http://newton.ex.ac.uk/research/qsystems/people/sque/physics/horizon/
1017869
S=1.317h 1/2  S is the distance from to the horizon, h is how much above sea level. The AMS states that the horizon is at minimum 3 miles away.   So for a person whose eye level is 5 ft above sea level, the horizon is 1.317m(5ft)(1/2) which equals 3.29 miles