Imagine that the Earth is spinning much faster than real situation and a man on the Earth's surface feels weightless. Suddenly the man decides to jump vertically with respect to the Earth. What will happen to him and the trajectory ignoring all the frictional forces?

Here's what I think. Since the man feels weightless he is just like a satellite in close circular orbit around the planet. So jumping process is just like giving a radially out impulse to a satellite in which the circular orbit becomes elliptical. More about this: https://physics.stackexchange.com/a/70418/192515

But how does the event happen in the man's frame?

At the moment he jumps he feels a sudden momentary weight, and he gets to the highest altitude and then comes down to the same spot he jumps (ignoring coriolis effect which might be too powerful this time?) and touches the ground with the same feeling of weight which will be momentary again.

I'm especially not sure about the second part of my answer. Please correct me if I have any mistakes.

  • 1
    $\begingroup$ How do you know that he comes down to the same spot where he jumped from? $\endgroup$ – harshit54 Oct 6 '18 at 9:28
  • $\begingroup$ Im not sure about it. Thats why I am asking the question. $\endgroup$ – physicsguy19 Oct 6 '18 at 9:30
  • $\begingroup$ It would depend on his time of flight and if it matches the time period of the rotation of earth, only then it will land back at the same spot. $\endgroup$ – harshit54 Oct 6 '18 at 9:32
  • $\begingroup$ Why do you think in the man's frame he should land back on the ground? $\endgroup$ – Aaron Stevens Oct 6 '18 at 10:48
  • $\begingroup$ Sounds like you are asking this question with the person on the equator, which is a special case. Do you want the general answer for any latitude? $\endgroup$ – cms Oct 6 '18 at 12:17

Your first question is basically: what is the new orbital? Since the force is radial the angular momentum remains the same. The energy is increased by $\int \vec F \cdot d\vec{s}$. It is an elliptical orbital with the same angular momentum as the original circular one but higher energy. Its perigee will be at a nonzero height above the surface. In other words, the man does not land.

This can be seen as follows. At the same position the speed in the new orbit must be higher than in the old one, as the total energy increased. Were the perigee at, or below, the surface then this would require a higher angular momentum, which is ruled out.

  • $\begingroup$ So when the man loses the contact (When the force of his push becomes zero), he is at the perigee. And his distance to Earth's surface becomes greater and comes back to the nonzero altitude in a periodic manner? Great answer. Thank you. $\endgroup$ – physicsguy19 Oct 6 '18 at 11:29
  • $\begingroup$ Indeed he is. Good point, @physicsguy. $\endgroup$ – my2cts Oct 6 '18 at 11:31
  • $\begingroup$ One more thing please, when the man loses contact, his tangential velocity also is a little less than before the jump to fulfill angular momentum conservation. Am I right? $\endgroup$ – physicsguy19 Oct 6 '18 at 11:57
  • $\begingroup$ @physicsguy19 Corrected comment. Good point again. If the force is radial, the man must slow down. Thus he should be in frictionless, slipping contact with Earth. If not, the force is not radial, angular is not conserved and he ends up rotating about an axis perpendicular to his spine. $\endgroup$ – my2cts Oct 6 '18 at 15:33

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