In the case of black holes, lights are pulled into by great gravitational force exerted by the dying stars (which forms into a black hole). If light has no mass, how is light affected by gravity?
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Photons are not “massless”, they simply have no ‘rest mass’. Mass is simply energy at rest – energy traveling towards the future slower than other energies – and the photon, being as fast as possible, is therefore said to contain no ‘rest mass’. The photon never stops and never loses momentum – all of its energy is in its momentum and it has no rest mass.
Because of this fact, the photon is affected by the curvature of spacetime, thus it’s affected by gravity.
This is also why it is said that mass CANNOT travel at, or faster, than the speed of light. Mass would simply have to no longer be mass to travel that fast, as mass is an equivalent measure of how *not-*fast energy is moving.
You can’t see this too well in the equation E=mc^2 because contrary to popular belief, that equation has been simplified and is much bigger than that.
E=√((m^2 • c^4 ) + (p^2 • c^2 )) is the full equation. This states that the relativistic energy, E, of a moving object is a function of its mass, m, and its momentum, p (as well as the speed of light, c). If you set p=0 and simplify, you’ll get back to the usual E=mc^2. The above equation doesn’t roll off the tongue quite as easily as E=mc^2 but with the momentum term included, it tells a fuller story about how relativity works.
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