3
$\begingroup$

I have no real knowledge of physics beyond the very basics of classic Newtonian mechanics, but as far as I understand, when a particle moves closer to the speed of light, its "relativistic mass" becomes greater, which means it requires more and more energy to further accelerate it.

I was watching an anime called "One Punch Man", and in this anime there is a character, with absurd strength, that physically jumps from the Moon back to Earth. Watching that scene made me curious about the relativistic effects, and requirements, of such feat.

Considering that the Moon is +/- 1 lightsecond away from Earth, and that the trip took just a few seconds (10 seconds maybe?), it's safe to assume that the character achieved "relativistic speeds", right?

So, assuming that the character weights 70 kilograms, the trip took 10 seconds (from the characters perspective), and the character "crashes" into earth (does not de-accelerate upon entrance):

  1. What speed did the character achieve?

  2. How much energy did the character "consume" to perform the jump?

  3. If the 10 seconds were measured from the perspective of someone on earth, how long did the trip took for the character?

$\endgroup$
  • $\begingroup$ The character would get quite hot hitting the atmosphere of Earth at a good fraction of c $\endgroup$ – zeta-band May 3 at 19:22
  • 1
    $\begingroup$ @zeta-band - Yes, but with enough push-ups, sit-ups and squats one can shrug off re-entry plasma too. It is that kind of anime. $\endgroup$ – Anders Sandberg May 3 at 20:55
  • 1
    $\begingroup$ @zeta Perhaps, but at .1c, he'll pass through the atmosphere in a fraction of a millisecond. I'm concerned about the pressure his feet exert on the ground when he takes off and lands. $\endgroup$ – PM 2Ring May 3 at 21:17
3
$\begingroup$

The average velocity is given by the same formula from Newtonian physics, so $v=\frac{1 ls}{10 s}=0.1 c$. The Lorentz factor corresponding to this velocity is $\gamma=\frac{1}{\sqrt{1-0.1^2}}=1.005$, so relativistic effects are not that great.

By energy "consumed," I'm guessing you mean the kinetic energy they have, which is also equal to the work done to reach that energy. The relativistic formula for this is $KE=(\gamma-1)mc^2$, so substituting in the values you gave yields $KE=3.169*10^{16} J$

For the last part, you are asking for something called the proper time, which is given by the time divided by the Lorentz factor, so $\tau=\frac{t}{\gamma}=9.95 s$, so they experience a very small amount of time dilation.

$\endgroup$
  • $\begingroup$ Thank you physics wizard! You even managed to untangle the mess in my second question. $\endgroup$ – Trauer May 3 at 22:52

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.