A space ship leaves earth on it's way to the sun. The farther away from the earth the temperatures become freezing.  I want to know how many miles from eath & the sun will the temperatures begin to rise again?
214689
temperature will rise depending on starting temperature, albedo(reflectivity), density, and probably lots of other factors, so there is no set answer.
607171
The sun DOESN'T know the ship is there, and it doesn't have to:  Gravity is a warping or deformation of the four-dimensional fabric of space-time, not an actual pulling.\n\nThe earth travels around the sun because the earth is actually travelling IN A STRAIGHT LINE in its "local" space.  The end result is an ellipse because the space around the sun is literally bent into a curved shape.\n\nSo, your ship can travel faster than light, but it can't travel faster than space -- which is to say, wherever it arrives after its hyperspace trip -- it IS where it IS and if the space is bent the ship will follow the contour of its local space regardless of how far it is from the source of the gravitational spacetime warping.  No information or "graviton" needs to actually travel from the ship to the sun.
490249
Actually, none of the above.  The most significant change is in the thermosphere.  Around 120km to 130km a huge curve on the temperature occurs.  Object falling into our atmosphere do generally burn up once they reach the mesospere, but when rising in altitude, it actually gets cooler, not warmer at all, throughout the mesosphere.  Hope this helps.
313556
ok..lets start\n\njet. power to lift off, cruise, and land with power.\n     highest alt is aprox 42000 ft.\n\nshuttle. no power to lift off, orbit or cruise, and land\n        without power (glide, no power)\n        it uses rockets for lift off, and burns off the rest\n        of its fuel for landing(deorbit), the rest of the \n       operation is glidding.\n        max altitue. aprox 200 miles.\n\n     hope it helps
512059
A NASA site states:\n"The heliopause has never been reached by any spacecraft; the Voyagers may be the first to pass through this region, which is thought to exist somewhere from 8 to 14 billion miles from the Sun. Sometime in the next 5 years, the two spacecraft should cross an area known as the termination shock. This is where the million-mile-per-hour solar winds slows to about 250,000 miles per hour—the first indication that the wind is nearing the heliopause. The Voyagers should cross the heliopause 10 to 20 years after reaching the termination shock."\n\nAlso check APOD 2002 June 24  (APOD = Astronomy Picture Of the Day).  Their best guess is 110 AU (Astronomical Units; 1 AU is the average Sun-Earth distance), in the direction of the apex -- this would make it the shortest distance.\n\nYou ask for average distance.  Using their drawing (if it is to scale) the trailing end of the heliopause appears 7 times further than the front end.  Thus the average is 4 times  (seven plus one, the sum divided by two): 440 AU.\n\nAs the previous person said:  this is a guess.  It is based on an educated guess for the front end (NASA), and a relatively well thought out drawing that, maybe, could represent the quasi-ellipsoidal form of the heliopause surface.\n\nA mile is 1.6 km (8 billion miles is 12.8 x 10^9 km)
521342
I'm not 100 % certain about this, but here goes. \n\nWe have the temperature of the asteroid. This is caused by solar heating, just as global warming on Earh is caused by solar heating. I think we must assume that the asteroid has the same albedo as the Earth. So it absorbs and reflects the same proportion of energy that the earth does. \n\nFind the energy absorbed by the earth per unit area. If the asteroid is at the same radial distance from the sun (same orbital speed) as the Earth, the asteroid will absorb the same energy per unit area, since it has the same albedo.\n\nWe're given the equilibrium temperature of the asteroid. the ratio between the surface temperature of Earth and the asteroid shoud be the same as the ratio between their radial distances from the sun. \n\nOnce you know the radial distance of the asteroid from the sun, you can easily find the orbital period. \n\nI think.\n\nAnyhow, I hope this at least give you a start. Good luck!\n\n30 NOV 06, 1947 hrs, GMT.
442673
The sun emits energy at a rate of P = 4x10^26 W. The amount of power density that reaches the icy planey is then this\n\nPd = P/(4*pi*R^2)\n\nwhere R is the distance from the sun to the planet. This is the total flux of energy emitted by the sun spread over the surface area of a sphere of radius R. The albedo (A) of a planet indicates how much of the incident energy is reflected back into space. The absorbed power density is then\n\nPa = (1 - A)*Pd = (1 - A)*P/(4*pi*R^2).\n\nThe absorbed energy heats the planet and also leads to radiation. The radiated power follows the Stefan-Boltzmann law\n\nPr = e*s*T^4\n\nwhere e is the emissivity, s is Stefan's constant, and T is the temperature of the radiating body. Stefan's constant is s = 5.67x10^-8 W/(m^2*K^4), and for a blackbody, the emissivity is one.\n\nAt equilibrium, the absorbed power and the radiated power are equal\n\n(1 - A)*P/(4*pi*R^2) = s*T^4\n\nwhich can be solved for T, the temperature of the planet.\n\nThe temperature of the planet will impart kinetic energy to gas molecules in the air. The speed of the molecules will be distributed with an rms spedd os\n\nVrms = sqrt(3*k*T/m) = sqrt(3*R*T/M)\n\nwhere k is Boltzmann's constant, R is the universal gas constant, m is the mass of the molecule, and M is the molar mass.\n\nFor a planet the escape velocity for an object is given by\n\nVesc = sqrt(2*G*M/r)\n\nwhere G is the universal gravitation constant, M is the mass of the planet, and r is the planet's radius.\n\nIf the speed of the gas molecules is greater than the escape velocity of the planet, then that planet will not be able to retain that gas at the given temperature.
109812
okay lets clear this matter!  We do have one spacecraft that has left our solar system, and that is Voyager 1.  Voyager 1 gave us a complete picture of our solar system by way of Neptune's Orbit.  Neptune's orbit allowed the spacecraft to venture 6.4 kilometers outside our solar system, and return back.  Voyager 1 was an exploration that gave us vast information about our solar system, but more importantly it gave us the only close up images of pluto we have to date.  However 6.4 kilometers is not enough to give us a picture of the milky way.  So Nasa uses Cosmic Background Explorer's (COBS) Diffuse Infrared Background Experiment (DIRBE) to combine images obtained at several near-infrared wavelengths.  These wavelengths are gathered from light wavelengths from neighboring stars that look directly into our galaxy.  So all images therefore are from a distance looking at our solar system from the outside.
1285
No, the Earth will continue in its orbit around the sun. But as the sun runs out of hydrogen fuel and balloons into a red giant, it will become so huge that Earth will be completely incinerated. Sorry to break it to you.
574679
about 1000 mph.\n\napprox #'s here--earth is 25,000 miles in circuvfrence, 24 hours per day.   this is at equator.\n\ni am thinking, the farther away from equator, the speed can be progressivly slower.
292060
space shuttle burns hydrogen. imagine you're standing under it when it launches. endothermic or exothermic?\n\ncorrect reaction formula was already given.
479067
I think your facts of science need to be updated.\n\nMars orbit is 1.5 AU (about 145 million miles)  from the Sun. \n\nThe Earth is 93 million miles away from the Sun.  At its closest, Mars is about 50 million miles away, and we have sent many probes to Mars without the need for "infinit (sic) energy"\n\nA manned Mars mission is well within human ability.
257741
If I'm reading your description right, I think your first boat is heading up and to the right ("direction N.68 degrees E") making an angle of 68 degrees with north, or 22 degrees with east.  And I think your second boat is headed directly east ("S.90 degrees E").  So I'm assuming that they form a triangle, with one angle being 22 degrees and another one being a 90 degree angle, since we're looking for the point where one boat will be directly south of the other.\n\nThe first boat traveled 12t miles so far.  The second boat left an hour later and traveled 16(t-1) miles.\n\nThe first boat's path turns out to be the hypotenuse of the triangle, and the second boat's path will be one of the adjacent legs to the 22 degree angle.  So we can set up an equation with trig identities:\n\ncos 22 = 16(t-1)/12t\n0.9272 = 4(t-1)/3t\n2.7816t = 4t - 4\n4 = 1.2184t\nt = 3.3 hours\n\nSo after 3.3 hours, one boat will be directly south of the other.\n\nFinding that distance can be done one of several ways: using the Pythagorean theorem, sine relationships, law of cosines, etc.  I'll use sine, since we're already doing trig.  After 3.3 hours, the northernmost boat will have traveled 39.6 miles.  The distance between the boats becomes our side opposite the 22 degree angle.  So,\n\nsin 22 = distance between/39.6\ndistance = 39.6sin 22\n= 14.83 miles\n\nNow if I read the angle wrong, then simply go back and change all the "22" angles to the correct one and recalculate.
110220
It will be from way out on a plane based on our orbital track. I think the intensity of the sun will block out the view of whether or not it goes behind or in front of the sun. The center of the universe, even though it is a non-existent place, for arguments sake, would be moving as everything in the universe is dynamic. (There is no end to infinite, therefore the center cannot be defined.)
51446
I like natural ways answer, but I think direct sun exposure would burn you up... \nSpace walks by astronauts require a thermal suit that regualtes their temperature.  If you had no protective layer like that you would burn up in temperatures of something like 390 K (117C)  Near boiling temperature (around 190 F)!  So yes the sun would keep you warm too warm..  Also the source below may help you, remember that space is essentially a vacuum.
7697
The distance between the earth and the sun is not constant.  The earth's\norbit around the sun is nearly circular, but not perfectly so.  The earth\nis actually closest to the Sun in the northern hemisphere winter, but the average distance between the earth and the sun is 93,000,000 miles.
292244
How hot is concentrated light, what is the temp of the sun?\n\nWELL WE HAVE A ENERGY SHORTAGE, ONLY AT SUNSET.\n\n I read about Solar collection in power plants like Solar One.  In these plants the where they make molten salt, for heat transfer for running steam generator. At the reciever of Solar One plant or concentrated light point temperature reachest approximately 1,050 Fahrenheit.\n    \n      THIS LINK IS ABOUT YOUR SUN AND THE TEMPS ARE HOT.\n         SO HOW MUCH LIGHT CAN WE CONDENTRATE MORE  \n           DEVELOPMENT IN SOLAR SAILS\n\nhttp://hyperphysics.phy-astr.gsu.edu/hbase/hframe.html \n\n Earth's Sun is a medium-sized star which lies on the main sequence with 90% of the known stars. It has a effective surface temperature is 5780 K, putting it in spectral class G2. Its mass is 1.989 x 1030 kg and its mean radius is 6.96 x 108 meters. The mass of the sun is over 99.8% of the mass of the entire known solar system, leading de Pater and Lissauer to refer lightly to the solar system as "the Sun plus some debris". \n\nThe sun radiates energy at the rate of 3.85 x1026 watts. Just outside the earth's atmosphere solar energy is received, assuming normal incidence, at the rate of 1340 watts per square meter. \n \n\nThe orbit of Earth ranges from 1.47 to 1.52 x 1011 meters from the Sun. The average light travel time to the earth is 8.3 minutes. \n\nThe radius of the sun at 696,000 km is 109 times the Earth's radius. Its surface gravity is 274 m/s2 or 28.0 times that of the Earth. Its mean density is 1410 kg/m3 or 0.255 times the mean density of Earth. \n\nThe composition of the sun is 71% hydrogen, 27.1% helium and less than 2% of all other elements. \n\nThe center temperature is modeled to be 15.5 million K. The Sun is fueled by the proton cycle of nuclear fusion.\n\nEscape velocity = 618 km/s\n\nSO