# Energy needed for Superman to take off and fly at the speed of sound

I just watched "Man of Steel", and I'm wondering if my logic is correct.

Let's assume Superman is 80 kg. The energy required to take off from the rest to reach the speed of sound in air (if I neglect the drag) is:

$E_k = 0.5mv^2$ = $0.5\cdot80\cdot340^2$ = $4\times10^6 \ J$.

Also add the potential energy at height $h$, $E_p=mgh = 784h \approx 0.2\times10^6 \ J$ (Let's assume at $h = 300 \ m$)

Total energy is roughly $4.8\times10^6 \ J.$

Superman gains his energy from the sun. Assume solar flux at Earth's surface is $1340 \ W/m^2$ (max), and Superman's surface area is roughly $2 \ m^2$ (calculated using Du Bois formula). Then the maximum energy that can be absorbed by Superman on Earth is $2\cdot1340 = 2680 \ J$ per second. (Solar flux is much less on the surface, but here I used this value anyway.)

Then to take off, he needs to wait:

$(4.8\times10^6)/2680 \approx 1791 \ s \approx 29.9 \min.$

It doesn't seem correct. Please correct me if I am wrong. (This is close to the perfect situation, which neglects many factors that could make the charging time longer. This also assumes his tank is empty. Thanks for pointing it out. If you are interested, please feel free to write down a more realistic estimation.)

• Why do you think it's wrong? – JSQuareD Aug 12 '13 at 12:22
• I think energy requirements shouldn't be your first worry when it comes to realism and superman. It's all just 'comic-book-physics'. – JSQuareD Aug 12 '13 at 12:27
• Why do you feel he doesn't store solar energy? So he doesn't need to wait 30 minutes to take off? – mikhailcazi Aug 12 '13 at 12:41
• ...and that $1340\ W/m^2$ figure is the flux at the top of the atmosphere. Significantly less reaches the surface. There is a reason you don't see solar powered fighter jets. – Michael Brown Aug 12 '13 at 13:53
• I find it amazing that a significant minority of physics students that I teach at the high school level are influenced by comic book heros. Some of my students have told me that it is possible to do Iron Man stunts if you are wearing his suit! It would be nice if students would realize that Hollywood physics is pure fantasy and nothing more. In other words, it's MUCH MORE DIFFICULT to teach students to unlearn a mis-concept than it is to teach the proper concept to a totally unbiased student. – David White Aug 5 '15 at 21:29

actually I'm only a high school student but i hope i can help.

first of all, we should specify how we want the superman to fly. i assume here that it takes off straight upward from earth and then it flies directly parallel to earth's surface.(i have assumed that earth is a plane not a sphere.)

for the take off, we need to dominate earth's gravity. the energy we need is the same as the difference of the gravitational potential energy of the 2 points. (1 on the surface of the earth and the other on the appropriate height, what you assumed to be 300 m ) so we have:

$E=mgh$
$E= 80* 9.8* 300=2.35*10^5 J$

now super man is at the desired height and the only thing we need is to give it a speed of sound. the energy needed is kinetic energy given by:

$k=1/2 *m *v^2$
$k=1/2* 80* 340^2 = 4.62* 10^6 J$

the total energy is :

$E_t=48.55*10^5 J$

till here all your calculations were right. but, the energy that super man absorbs produce a pressure on him and so a force.the force direction is opposite to sun, it means toward earth! so it can't directly use sun's energy to accelerate away from earth or to reach the speed of sound at a direction perpendicular to the radiation.
if we assume super man has some "thing" that can convert radiative energy to mechanical energy (for example by using solar pills,...) then your calculation is right! but it even takes less time for superman to leave earth. because at first it doesn't need to have the hole energy! i think we should calculate the difference of potential energy by unit time. and compare it with the energy received at his body's area. if the energy recieved is larger than the energy required, so it can fly. and otherwise, not.

There is a much bigger problem given conservation of linear momentum: When Superman zooms thisaway, what zooms in opposition thataway? Remarkably, the Book of Mormon seems to address this issue. David O. McKay, 1963; softcover edition published 1976; left side color plate facing p. 397. This is not the Son of Krypton's solution.

## protected by Qmechanic♦Apr 20 '15 at 18:46

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