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around us, most of them are very, very small.

Suppose we have a drinking water glass (just suppose its mass is 10kg)

hence at radius 1.48 E-26 m (far smaller than even the radius of an electron)

there exists a black hole at the centre of mass of the glass.

(by the calculation done using escape speed equation, plugging in the speed of

light as the escape speed, we can get the value of r, in which any particle

even in the speed of light cannot escape from.)

because by calculation, the escape speed;v of any object inside the

radius will be greater than the speed of light hence it cannot escape, even light cannot.

so there is a tiny black hole at the centre of the glass. Am i right? please explain if not.

Thank you

here's the extract of the article in

*http://en.wikipedia.org/wiki/Black_hole*

consider a heavy object of mass M centered at the origin. A second object with mass m starting at distance r from the origin with speed v, trying to escape to infinity, needs to have just enough kinetic energy to make up for the negative gravitational potential energy, with nothing left over:

mv^{2}/2 - GMmr^{-1}= 0

That way, as it gets closer to r=infinity it has less and less kinetic energy, finally ending up at infinity with no speed.

This relation gives the critical escape velocity v in terms of M and r. But it also says that for each value of v and M, there is a critical value of r so that a particle with speed v is just able to escape:

r = 2GM/v^{2}

When the velocity is equal to the speed of light, this gives the radius of a Newtonian dark star, a Newtonian body from which a particle moving at the speed of light cannot escape. In the most commonly used convention for the value of the radius of a black hole, the radius of the event horizon is equal to this Newtonian value.

r_{Schwarzschild}= 2GM/c^{2}