In order to perform Superman's iconic Earth-reversing flight maneuver seen in the 1978 film, The Man of Steel would have to fly at a whopping 660 million miles per hour.
This figure, which equates to an angular velocity of 46.296 radians per second (98 percent the speed of light), was calculated by a group of physics students from the University of Leicester (via Nerdist).
There is however, a caveat, as Superman's size plays a factor. In order have an impact on the planet, he'd have to increase his mass 13.7 million times over. By flying at near light speed, he'd build up a large amount of energy, which in turn would result in Kal-El having a larger relativistic mass, making it theoretically possible... sort of.
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This figure, which equates to an angular velocity of 46.296 radians per second (98 percent the speed of light), was calculated by a group of physics students from the University of Leicester (via Nerdist).
There is however, a caveat, as Superman's size plays a factor. In order have an impact on the planet, he'd have to increase his mass 13.7 million times over. By flying at near light speed, he'd build up a large amount of energy, which in turn would result in Kal-El having a larger relativistic mass, making it theoretically possible... sort of.
Continue reading…
Continue reading...