Friday, December 1, 2017

Physics of Justice League

During Thanksgiving break, I watched the movie, Justice League. There was one scene in the movie that stood out to me. In this scene, The Flash is able to run up a wall vertically. I began to wonder whether this was physically possible even assuming that his super power existed. To begin James Kakalios, who wrote The Physics of Superheroes, first looked at Newton’s third law, which states that for every action there is an equal and opposite force. When you walk, there is an equal and opposite force on your foot parallel to the ground which is friction. Without it, it would be impossible to walk or run. In calculating the friction force of an object moving parallel to the ground, the force is equal to μmg. The object’s weight, mg, attributes to the force. It is pushing down on the surface and gravity is perpendicular to the ground. Moving back towards the example of the Flash running up the wall, he is moving vertically and no component of his weight acts on the wall. Because of this, there is no friction between him and the wall. Therefore, it would not be possible for the Flash to run vertically up the wall in the same way that we run on the ground perpendicular to the force of gravity.
Another thing that stuck out to me was Superman’s ability to fly. Superman’s abilities stems from the fact that he was born on a different planet, where its gravity is much greater than that of the earth. This results in Superman having much more strength on earth. However, this cannot explain why he is able to “fly”. Kakalios suggests that the most realistic approach is that Superman must instead leap at high speeds in order to travel through the air from one place to another. After calculating the speed, the distance he is able to leap can also be calculated. I first calculated the time that Superman would be in the air given the fact that he was only able to leap 660 feet, or  approximately 200 m, according to Kakalios. Assuming no air resistance, his time spent in the air from the ground to his maximum height is equal to the time from the maximum height to the ground.
y = vit + ½at2
200 m = (½)(9.8 m/s2)t2
t = 6.4 s
ttot = 2(6.4 s) = 12.8s
Now that we have the time, we can also calculate his initial and final velocity. Beginning at the top of his jump, his initial velocity is zero.
vf  = vi + at
vf = 0 m/s + (9.8 m/s2)(6.4 s)
vf = 63 m/s
If 63 m/s was his final velocity before landing, his initial velocity starting from the ground must have also been 63 m/s to achieve his maximum height. If Superman uses this velocity at a 45° angle, he will be able to the largest distance he possibly could with that velocity. I calculated his y-component velocity which was:
63 m/s (sin 45°) = 44 m/s
The time he’d spend in the air was then calculated.
vf = vi + at
44 m/s = 0 m/s +9.8t
t = 4.5 s
Ttot = 12.8 s
Using his total time airborne, we can then calculate how far he can travel in the x-direction. Assuming no air resistance, his velocity remains constant in the x-direction and his acceleration is zero.
x = vxt
x = (63 m/s)(cos 45°)(12.8 s)
x = 560 m

Even if an alien who lived on a planet with a much larger gravity, he would not be able to fly as Superman does because it would defy laws of physics. In the real world, however, he would technically be able to have the ability to jump at much greater lengths and heights and appear to be in “flight”. This is due to the fact that he came from a planet where the gravity is much greater and his body would be able to exert a much larger force on earth where the gravity is vastly lower.

Kakalios, James. The Physics of Superheroes. New York: Penguin Group, 2009.

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