Consider a sphere of radius R. Now a sphere of radius R/2 with its centre at a distance of R/2 from centre is cut from the bigger sphere so that there is a hole of radius R/2. Now is there any gravitational field inside the cavity. According to me it's not possible because if we consider a gaussion surface enclosing the hole since there is no mass enclosed there will be no gravitational field. But the correct answer is there will be gravitational field inside the cavity and it will be uniform. First explain to me the correct answer and also if you can, please caluculate the gravitational field.
122142
First, a gaussian surface enclosing a space with no mass inside only means that the total flux going into the surface is the same as that going out. Of course if it were possible to assume the flux density over the surface were all outward (or inward) then it would be zero.\n\nI was able to find the gravitational field in the cavity based on your hint that it was uniform (otherwise I wouldn't have bothered to try):\n\nThe gravitational field produced by a sphere of uniform density rho at a radius r inside the sphere is radially downward and has magnitude F = Kr where K = G(4/3)pi rho = (GM/R^3). This follows from the fact that the field is the same as if all the mass from 0 to r (e.g. rho * (4/3) pi r^3) were concentrated at the center. Of course the mass above r doesn't contribute. This can also be written as a vector equation F = -Kr where r is the radius vector from the center to the point.\n\nNow consider any point P within the cavity. Let r be the vector displacement from the center of the big sphere to P. Let S be the vector displacement from the center of the big sphere to the center of the cavity (|S| = R/2).\n\nThe field at P is the same as the field produced by a solid sphere of radius R minus the field produced by the smaller (removed) sphere of radius R/2. The displacements of P from the centers of the two spheres are respectively: r and r-S, so the total field is: \n\nF = -Kr-(-K(r-S)) = -KS\n\nIn other words the gravitational field everywhere in the cavity is parallel to the axis connecting the center of the cavity to the center of the big sphere, and has the same magnitude (KR/2) as the surface gravity of a solid sphere of radius R/2. Another way of putting it is that the gravity is the same as it would be at the center of the cavity if there were no cavity.
188986
Yes. Any hollowed out space can be called a cavity.
597069
Ok, first things first.  This situation cannot occur.  Such a string does not exist and probably cannot exist.\n\nA black hole is actually not a "hole" so to speak.  It is a super-dense core.  The denser an object is, the stronger the force of gravity that this object exerts on all objects around them.  That means that even you or I exert gravity on everything around us.  However, our own gravititational force is so weak that we don't even perceive it.\n\nNo one that we know of has gone near a black hole.  The nearest known black hole is 1600 light-years away.  One light year is approximately 6 trillion miles.  So the nearest black hole (that we know of) is approximately 1600 times 6 trillion miles away, or 9.6 quadrillion miles away.\n\nAnyway, I don't want to complicate it too much.  This is stuff you would learn in a high school or college physics class, and my guess is that you have not taken such a course yet in your life.
185472
Let me ask you something.  Since density is defined as the mass divided by the volume, how did you determine the mass of a black hole to be 10^13 kg/m^3?  Since the volume of a black hole is 0 (a singularity is a 0 dimentional point with no volume), then the mass divided by the volume (0) equals infinity.  Therefore, a black hole has an infinite volume, which is a heck of a lot larger than 10^17kg/m^3.\n\nPS Densiest isn't a word, and even if it were, you'd be spelling it wrong.  Grow a brain before you start talking down to people.
73728
Yes.\ngravity is inversely proportional to radius squared, and radius is inversely proportional to the curvature of the same surface, so your statement is true.
629405
I suppose you do know this\nIn mathematics, a hypersphere is a sphere which has dimension 3 or higher\nThe set of all points on this sphere has dimension n − 1, so it is called the (n − 1)-sphere and is denoted . It may be written as (x1,x2,...,xn) where\n\nThe hyperdimensional volume of the space which a (n − 1)-sphere encloses (the n-ball) is\n\n    _\nV=II^n/2. R^n/Γn/2=1  II-pie\n   \n\nIf the dimension,  , is not limited to integral values, the hypersphere volume is a continuous function of  with a global maximum for the unit sphere in "dimension" n = 5.2569464... where the "volume" is 5.277768...\n\nThe hypercube circumscribed around the unit n-sphere has an edge length of 2 and hence a volume of 2n; the ratio of the volume of the hypersphere to its circumscribed hypercube decreases monotonically as the dimension increases.\n\n\nHEY I KNOW ONLY THE VOLUME
369316
Unknown. Without an adequate theory of quantum gravity, this object that GR predicts would be a singularity cannot be described. At quantum mechanical scales, pure General Relativity is not a realistic depcition of space-time.
475187
According to Newtonian gravity, gravitational force is -GMm/r^2.  Any mass experiences a gravitational force toward every other mass in the universe, proportional to the product of their masses, and inversely proportional to the square of their distance.  This means my body experiences a certain gravitational force from the sun, no matter what else is happening in the universe.  That says it would be impossible to shield me from the sun's gravity.\n\nTo find the net gravitational force on me, simply do a vector sum of all the gravitational forces on me from all other masses in the universe.  Simple to say, not so simple to do.\n\nThat sum is zero at the center of mass of the universe, wherever that is.  Unless the expansion of the universe is perfectly uniform, that point will shift with time.  I could find other points where the sum is zero.  I move away from that center of mass but toward a large nearby mass, just close enough that the pull of that nearby mass exactly balances the pull of the rest of the universe.  These will be single points, not regions.  At these points, I still can't say there's no gravitational force acting on me.  But I can say there are countless gravitational forces acting on me, but their sum is zero.\n\nLet's say I tell you I found such a zero-sum spot, about 1/3 of the way between our sun and Alpha Centauri, and I take you there.  How will you know I am telling you the truth?  The principle of equivalence says that there's no test you could make to tell whether the net gravitational force there is zero, or whether you are just accelerating according to the effect of the nonzero net force on your mass.\n\nThe relativistic warping of spacetime in the presence of gravitational mass is more a way of representing or measuring that force, rather than explaining it.  Neither Newton nor Einstein could tell us why gravity exists or how it works.  If spacetime is not warped in your region, it doesn't mean the gravitational field is zero, it means it is uniform.  And again, how would you tell whether spacetime is warped in your region?\n\nI've tried for decades to think of a gravity shield or a gravity insulator.  If I do find one, watch for my Nobel Prize in physics, and for me to quickly surpass Bill Gates on the wealth scale.
214843
Well, as far as we know, all the matter that gets sucked in gets crushed into a single point with zero dimensions. As in, a point with no length, width, anything. This is called a singularity.\n\nOr, another theory is that there are wormholes. If you get sucked in a black hole and survive (heh...not likely), you get thrown into a different part of the universe, or another universe, or another dimension, or whatever. However, the wormhole theory is losing support as of now.
301138
The easy answer is no. The pieces that the sphere would have to be split into would not be physically possible. The proof itself depends on the axiom of choice. In fact, there are models of set theory that don't have the axiom of choice where the Banach-Tarski paradox cannot happen. This shows that an infinite number of choices is needed, which demonstrates the physical impossibility.
148129
Electric field:\n\nE=(1/4*pi*epsilon0)*(q/r^2)--> [spherical shell at r>Radius]\n\nE=0 --> [spherical shell at r<Radius]\n\nE=(1/4*pi*epsilon0)*(q/r^2) --> [spherical distriburion at r<=Radius]\n\nor E=(q/4*pi*e0*R^3)*r [uniform charge]\n\n-----------------------------------------------------\nPotential field:\n\nit doesn't have a formula but u can find it with this integral:\n\nv=(1/4*pi*e0)S(dq/r)-> S is integral!
179495
If the spheres are of same volume, their apparent loss of mass will be the same.\n\nIf the spheres are of same mass, Aluminium sphere's apparent loss of mass will be more.
277495
Remember the fact that what they say( that outside our universe is the void, which is absolutely nothing, NO SPACE OR AREA) is only a strongly held VIEW, or CONCEPTION. As far is science is concerned, there is no evidence that there is " no matter beyond the area within which the universe is existing ".\n             Suppose you have balloon. Imagine it to be the universe, then, people living inside the balloon, cannot acually guess about the fact there is no matter beyond the balloon, unless and until they are being confronted with the FATE of the ballon. Remember that the universe is like a balloon, continously expanding, so, there is a high possiblity that it might all actually begin to compress after some billion years or something. \n            What I meant to say is that, there is no solid method to find out about the matter beyong the universe, as the FATE of universe alone will answer these questions. OR, there is a lot of matter beyond the universe, just as air beyond the balloon, may be the expansion of the UNIVERSE would NOT allow us to determine whether or not, ther is matter beyond the universe, as for now, we cannot say anything.\n          Hence, what I have said above is also a conception, a theory, a prediction, which MAY be TRUE.\n\nDa Vinci.
595355
Consider the graph of effective potential (i.e. that including the angular momentum contribution) vs. radius 'r' for a test particle in orbit around a Schwarzschild hole. i.e. for timelike geodesics.   The slope of this graph at any value of r can be regarded as the force required to keep an object in orbit at that radius.  At r=3M, the slope can be zero (if the angular momentum's right), but for smaller r (i.e. approaching the Schwarzschild radius r=2M from larger values of r), this slope increases rather sharply (as r->0 the r^-3 term dominates).  Now, this is still FINITE, not infinite at r=2M (so the thread posted by the Canadian guy has incorrect info) but it can still be very large...  So, to keep your arm intact, you'd have to keep your whole arm more or less in orbit at a succession of raidi approaching r=2M.  \n\nThis then becomes a medical question: how strong is your arm?  If it's not strong enough, your hand breaks off (or something) and the lost bits freely fall into the hole.  If your arm is "strong enough", however... hmmm.  That's a good question!\n\n...Oh, and tell your ship's captain: "Don't forget to compensate for the decay of our orbit due to the gravitational radiation we're emitting!"
580182
It is quite possible for the universe to be both finite and without a boundary. No wall, but still it doesn't go on forever. Here's a way of visualizing it. Imagine that nomatter which direction you choose, you will eventually get back to your starting point if you go far enough in a 'straight' line. Impossible? Free your mind. It happens in two dimensions on the surface of a sphere. While it currently looks like the universe really is infinite, such three dimensional spheres were models used for finite universes until the data showed they are probably wrong.
103197
I think,to a good approximation, at the centre you would not feel an appreciable gravitational field from the mass of the earth. \n\nIt's a nice result in physics that if you're on the interior of a spherical shell of mass then you feel no gravitational force. You can think of the centre as being surrounded by layer after layer of spherical shells of radius ~0 up to the radius of the earth.\n\nIf you do the calculation, once you start going interior to the earth the gravitational attraction varies linearly with distance from the centre-going from 9.81m/s^2 at the surface to 0 m/s^2 at the centre.\n\nOutrageous!
360820
Consider the frustum of the cone a surface point of the sphere, then  it's height is given by 0<h<2R (R the radius of the sphere), and because it is inscribed, the base is a planar cut of the sphere, so its radius is given by Sqrt[h(2R-h)] and so the volume of the cone is 1/3*pi*(h(2R-h))h, or VC=pi/3*(2Rh^2-h^3).  Taking the derivative and setting equal to 0, we get 4Rh-3h^2 = 0, or h=4R/3.  The Volume is then (plugging in the formula), VC=32piR^3/81, or 8/27 = (2^3)/(3^3) of the volume of the sphere.\n\nSteve