It is difficult to determine has the mass of Betelgeuse is poorly constrained. Estimates range between 8 and 20 times the mass of the sun. For this mass range, the diameter would have to be between 47 and 118 kilometers (29 and 73 miles) for the whole mass of Betelgeuse to form a black hole.
To calculate the required size, we can use the equation for escape velocity, v = sqrt((2GM)/r), where G is the gravitational constant, M is the mass of Betelgeuse, and r is its radius. Setting the escape velocity equal to the speed of light (c), we would need the radius to be smaller than the Schwarzschild radius (2GM/c^2). For Betelgeuse, this would require the radius to be smaller than about 3 kilometers (1.9 miles).
The escape velocity of a black hole is equal or greater than the speed of light, so light cannot escape
Because every matter has a mass which tries to stop the matter from gaining the equal or higher velocity as light.
A black hole is a region in space-time with very strong gravitational pull that even light cannot escape from it. The ESCAPE VELOCITY is greater than SPEED OF LIGHT.
At the point of the event horizon, the pull of gravity is so strong that the escape velocity reaches the speed of light. As such, when light reaches this point, it cannot escape.
The light takes about 640 years to reach Earth, as Betelgeuse (a red supergiant star) is about 640 light years from Earth.
The escape velocity of a black hole is equal or greater than the speed of light, so light cannot escape
No light is the fastest possible. Nothing escapes the velocity of light. ==================================== Sure. The escape velocity at some appropriate distance from a black hole is equal to 'c'. That's why the hole is black. The exact value of the distance depends on the hole's mass.
The word "black" aptly describes the inability of light to escape - all light and matter that passes the event horizon can only do so in one direction, falling in. The reason is, the escape velocity inside the event horizon is greater than the speed of light, the event horizon itself being the boundary at which the escape velocity is equal to that speed. Outside that horizon, the escape velocity is less than the speed of light, hence it would be possible for light and objects moving at speeds approaching that of light to escape.
Not at all. It would take an infinitely large mass to produce an infinite escape velocity, and no such infinite mass exists. Furthermore, the escape velocity for any object is the same no matter what is trying to escape, so light does not have its own escape velocity. This question presumably concerns black holes. Light does not escape from black holes because the escape velocity is greater than the speed of light. The speed of light is not infinite, it is 300,000 kilometers per second.
It is called the Schwarzschild radius
No. The escape velocity of a black hole is greater than the speed of light.
Because charge particles produces magnetic field which causes electromagnetic force that's why moving charges move with the velocity equal to the velocity of light.
The Schwarzschild radius is a concept related to black holes. Given a body it is the radius such that, if all the mass of the body were squeezed (uniformly) within that sphere, then the escape velocity at the surface of the velocity would be equal to the speed of light.
Betelgeuse is approximately 600 light years from Earth.
Betelgeuse is about 640 light-years from the sun.
Light gases; Gases that are light weight will more easily reach escape velocity than heavier gases.
By definition, the event horizon is a boundary of a black hole at which escape velocity reaches "c", the speed of light. Hence, the event horizon defines a boundary, within which, events can't affect an outside observer; neither light nor matter can escape.