3
$\begingroup$

High and low tides are caused by Moon gravity attracting water. Now there's friction, waves cause erosion, their energy is used in power plants yet the tides work for millions of years and are perceived as a free source of energy.

Now that's impossible - energy can't just appear out of nowhere.

What is the energy source for the high and low tides?

Improve this question
$\endgroup$
3
  • 4
    $\begingroup$ Surely Are tidal power plants slowing down Earth's rotation? covers the same ground pretty exhaustively? $\endgroup$
    – dmckee --- ex-moderator kitten
    Jan 9, 2013 at 4:56
  • $\begingroup$ This is why we always see the same face of the moon: there are no longer tides on the moon. Tidal forces lead to friction, which locks the precessional and rotational rates. $\endgroup$
    – emarti
    Jan 9, 2013 at 5:00
  • $\begingroup$ Also related: the three year old question Where does tidal energy come from?, to which I recently posted an answer. The energy source for the high and low tides is the Earth's rotation, making the Earth's rotation rate slow down. $\endgroup$
    – David Hammen
    Aug 1, 2016 at 10:39

1 Answer 1

Reset to default
6
$\begingroup$

To essentially quote http://en.wikipedia.org/wiki/Tide:

Energy of the Earth is not conserved while energy of the Earth-Moon system must exist. Energy from bodies of water are diminished (by about 3.75 TeraWatts) where about 98% of this energy loss is due to marine tidal movement.

Because energy is lost in the water, this imposes a torque on the Earth which changes its Rotational Kinetic Energy ($KE_{rot} = \frac 12 I \omega^2)$. Because angular momentum is also conserved in the Earth-Moon system, that gradually transfers angular momentum to its orbit (By conservation of angular momentum to the Moon's orbit ($L = I \omega$, Earth spins less, Moon gets pushed further away) The equal and opposite torque on the Earth reduces its rotational velocity, thus lengthening the day by about 2 hours per 600 million years.

$E = \frac {1}{2}( I_{earth} \omega_{earth}^2 + I_{moon} \omega_{moon}^2) = constant$

A torque on the earth causes its angular momentum to decrease

$\mathscr{T} = \frac{dL}{dt} = \frac{ d{I \omega}}{dt}$

But Angular momentum is conserved, so the moon must rotate faster around the earth

$L = I_{earth} \omega_{earth} + I_{moon} \omega_{moon} = constant$

Improve this answer
$\endgroup$
6
  • 1
    $\begingroup$ One should conclude clearly that rotational energy is turned into water and crust kinetic energy. $\endgroup$
    – anna v
    Jan 9, 2013 at 5:26
  • 2
    $\begingroup$ Just to be nitpicky - energy is conserved, always and everywhere. If it is not, the system you are considering is incomplete :) In this case, you considered only the Earth crust/ocean system, whereas in the Earth-Moon system, energy is conserved perfectly fine. $\endgroup$
    – Rody Oldenhuis
    Jan 9, 2013 at 8:18
  • $\begingroup$ @annav: Rotational energy of the Earth is converted into ocean/crust kinetic energy, and heat (friction), and orbital energy of the Moon. $\endgroup$
    – Rody Oldenhuis
    Jan 9, 2013 at 8:26
  • $\begingroup$ @RodyOldenhuis Surely energy is only guaranteed to be conserved in a closed system? $\endgroup$
    – gerrit
    Jan 9, 2013 at 9:22
  • $\begingroup$ @gerrit: That's what I mean, but in different wording. Energy never just disappears; it only "disappears" from systems that do not consider the parts external to that system absorbing the system's energy loss. In short, I think my problem is with Cactus BAMF's wording -- "Energy of the system is not conserved"; the validity of that statement depends on what system you are considering. $\endgroup$
    – Rody Oldenhuis
    Jan 9, 2013 at 10:07

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.