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The moon orbits Earth at about $380,\!000 \,\mathrm{km}$ away from it, at around $3,600 \,\mathrm{km}$ an hour.

I was thinking, with light traveling at $300,\!000 \,\mathrm{km/s}$, how close to earth (probably in the $\mathrm{nm}$ range is my guess) would light have to be to Earth to orbit it?

Update: After reading the below answers, here's my reasoning to why I thought it would be in the nanometre range.

I thought that if light was as close as possible to Earth (like, a planck length away or something), Earth's gravity would make it hit Earth immediately, but I forgot that the pull of gravity gets weaker when one is further away from Earth (i.e. at the surface of Earth is still somewhat "far away").

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You can take the Newtonian expression for the orbital speed as a function of orbital radius and see what radius corresponds to an orbital speed of $c$, but this is not physically relevant because you need to take general relativity into account. This does give you an orbital radius for light, though it is an unstable orbit.

If the mass of your planet is $M$ then the radius of the orbit is:

$$ r = \frac{3GM}{c^2} $$

where $G$ is Newton's constant. The mass of the Earth is about $5.97 \times 10^{24}$ kg, so the radius at which light will orbit works out to be about $13$ mm.

Obviously this is far less than the radius of the Earth, so there is no orbit for light round the Earth. To get light to orbit an object with the mass of the Earth you would have to compress it to a radius of less than $13$ mm. You might think compressing the mass of the Earth this much would form a black hole, and you'd be thinking on the right lines. If $r_M$ is the radius of a black hole with a mass $M$ then the radius of the light orbit is $1.5 r_M$.

So you can only get light to orbit if you have an object that is either a black hole or very close to one, but actually it's even harder than that. The orbit at $1.5r_M$ is unstable, that is the slightest deviation from an exactly circular orbit will cause the light to either fly off into space or spiral down into the object/black hole.

If you're interested in finding out more about this, the light orbit round a black hole is called the photon sphere, and Googling or this will find you lots of articles on the subject.

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Re: the first sentence. What's wrong with $r=GM/c^2$, as calculated by setting circular acceleration at speed $c$ and radius $r$ equal to Newtonian gravitational acceleration? –  Chris White May 5 at 9:56
    
So, basically speaking, with a sphere the size of Earth and the weight of Earth, I can't get light to orbit? Whoa. And I thought it would be in the nanometre range... –  think123 May 5 at 10:47
    
"there is an stable orbit for light once you take general relatitivity into account" - did you mean unstable? –  user2357112 May 5 at 11:36
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@think: If gravity was that strong nanometers above the earth's surface, bacteria would be crushed. As would be the soles of your feet, quickly followed by the next few layers of cells. You'd end up as a puddle. The mistake you're making is that gravity varies with the distance to the center of the earth, not the distance to the surface. (There's no anti-gravity underground either!) –  MSalters May 5 at 13:46
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@think123: Assuming near-perfectly spherical bodies, gravity doesn't care where their surfaces are. Only their mass and the position of the center of mass matter. If the Earth were suddenly compressed to a black hole, nothing much would happen to the Moon and the artificial satellites. –  JohannesD May 5 at 15:13
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I'm going to propose an alternate answer. The title of the question stipulates that the sphere should be the same SIZE and mass as the Earth. @John Rennie's answer above assumes that the Earth can be compressed to an infinitely small sphere.

In this case, there is no radius at which light has a closed orbit. As @Keshlam said, closer to the centre, most of the Earth's mass is equally distributed in all directions and the gravitational forces on an object will cancel out. The strongest gravitational forces are somewhere close to the surface of the Earth, and it should be pretty obvious that light here is not trapped by gravity.

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Shouldn't the answer just be the Schwarzchild radius? Anything inside and light can't escape. Anything outside and it will escape if pointed away from the event horizon. So it seems a beam of light just at the Schwarzchild radius would forever orbit at that distance.

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If you are a physics nerd (and why else would you be reading this :-) the reasons why your suggestion isn't the case are absolutely fascinating. It's because the radial coordinate $r$ that we use to measure distance from the centre and the time coordinate $t$ change as you approach the event horizon. Sadly you can't really appreciate what goes on without a lot of effort to learn the basics of GR, but believe me it's worth the effort! –  John Rennie May 6 at 6:03
    
@JohnRennie interesting. My intuition would be that only light pointed perfectly away from the black hole would be able to escape just outside of the event horizon and you'd have to be a bit farther away before you could have light pointed horizontally and orbit. It sounds like I'm also wrong but for a different reason. –  Brandon Enright May 6 at 6:11
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There is a way to get a beam of light to orbit the Earth, but it would require special material (probably hasn't been made yet). You would need to construct a rod of this material to circle the Earth. This material would need to transmit light without absorbing it (unlikely) and dense enough to have a very high index of refraction. It needs to slow the speed of light to equal the speed to orbit the Earth.

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As stated, the question presupposes that there is a distance (from the surface of the mentioned sphere) that would cause a beam of light to circle the given sphere. This is a false (or invalid) assumption for any "valid" answer. If the use of fiber-optic cable is allowed, then any cable length that circles the sphere, would be an answer. If we are interested in "bending" the light beam (by gravity) so that it has the same radius as the mentioned sphere, we would need something like the mass/density of a neutron star, so the mass/density of the earth will not do. If we are allowed to use the mass of the earth as a "point mass" (radius not maintained), then we obtain the answer (13mm) already provided.

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