1

I did this question: In a sense we all have kinetic energy, even when we are standing still. The earth, with a radius of $6.37×10^6 \mathrm m$, rotates about its axis once a day. Ignoring the earth's rotation about the sun, what is the kinetic energy of a $50 \mathrm{kg}$ man standing on the surface of the earth?

I calculated the kinetic energy using the following: Since $W=mg$ provides centripetal force,
$g = v^2/r$
$v^2=rg$
$0.5mv^2=0.5mrg$

However, the answer given to calculate kinetic energy was to first calculate $v$ was $v = rw =\frac {2\pi r}{1 \ \text{day}}$, and calculate $0.5mv^2$, which gave a totally different answer.

I am not sure why the first answer gives a different answer to the second.

ACB
  • 2,588
Raphael Tan
  • 11
  • 1

3 Answers3

2

The question is badly parsed so I can't exactly understand what you did in the first method, but there's a probable error in that you are equating the centripetal force to the gravitational force. The gravitational force provides some of the centripetal force; the rest manifests as the gravitational attraction we feel to the Earth. If the centripetal force were actually equal to the gravitational force, we would feel weightless (or, as PM 2Ring put it, we'd be in orbit). This naturally requires a much-greater speed. Hence you should get that the second method yields a much smaller kinetic energy than the first.

Allure
  • 23,373
0

Your first method made an incorrect assumption. The gravitational force does not equal the centripetal force required to account for circular motion on the surface of the Earth- it greatly exceeds it. The centripetal force required varies depending on latitude, being greatest at the equator and zero at the poles. Even at the equator, the force required is a tiny fraction of the gravitational force.

Professor Sushing
  • 29,350
0

In a sense we all have kinetic energy, even when we are standing still

Relative to another observer also on earth and stationary, you have zero kinetic energy. So this is not true in such a context.

Ignoring the earth's rotation about the sun, what is the kinetic energy of a 50 kg man standing on the surface of the earth?

Relative to who? You mean for an observer at a fixed point not on the earth?

Since $W=mg$ provides centripetal force,
$g = v^2/r$
$v^2=rg$
$0.5mv^2=0.5mrg$

You cannot equate the man's weight with his centripetal force since they are never equal. You can only do this for objects that are in circular orbits around the earth where the centripetal force does equals gravitational force.

If you use the equation $$v=\omega R$$ where $v$ is the tangential velocity of the man or any point on the surface of the earth relative to an external fixed observer, then you can calculate the angular speed first $$\omega=\frac{2\pi}{T}$$ and use $v$ to calculate the kinetic energy $$E_k=\frac{1}{2}mv^2$$ of the man.

joseph h
  • 31,821