4
$\begingroup$

As the Moon recedes from the Earth, are the tides getting taller or shorter? If the Moon is, someday, twice as far from the Earth, how many high tides will be there be each day?

$\endgroup$

3 Answers 3

1
$\begingroup$

The number of tides stays just about the same, as the earth turns under the tidal bulges. On a water earth we would have two tides per day. Local landforms impact that greatly, varying from place to place. The tidal force is the difference of the moon's gravity on the near and far sides of the earth. It falls off as $\frac 1{r^3}$, so the if the moon were twice as far away the lunar tides would be (about) $\frac 18$ as tall. The solar tides would become the dominant ones, so the total height would not decrease nearly that much.

$\endgroup$
1
$\begingroup$

Just imagine there would be no moon at all. Obviously there would be no tides due to the moon. As Dr. Chuck pointed out in the comments, there would still be tides due to the sun, but the tidal forces of the Sun are only about 46% of the moons tidal forces (taken from german Wikipedia about Tides: https://de.wikipedia.org/wiki/Gezeiten). So the farer the moon is, the smaller the tides get.

The question about the number of tides is a bit trickier: If the moon is twice as far away from earth, the gravitational force between earth and moon is only $\frac{1}{4}$ of todays. Assuming the moon is still in a circular orbit around earth, its centrifugal force therefore might also be only $\frac{1}{4}$ of todays, and since $F_c \propto \omega^2$ its angular velocity $\omega$ might only be half the speed than today.

If we assume conservation of angular momentum $\vec{L}=\vec{r}\times\vec{p}$ instead, its velocity $v\propto p$ has to be only half of todays value. And since $\omega = \frac{v}{r}$ this would mean that its angular velocity is reduced by a factor of 4. However this would not result in a circular orbit and if you do the calculations for an elliptic orbit, you probably end up with the same factor of 2 as before (just guessing here...).

As others pointed out in theirs answers and dmckee in the comments, both will only have a minor effect on the number of tides. Moon orbits earth roughtly once a month while the earth rotates once a day. Since the moon orbits in the same direction as earth rotates, it actually slows down the speed of tides. So a slower rotating moon would actually result in more tides each day, but the effect would be minor.

$\endgroup$
3
  • $\begingroup$ If there were no moon, there would still be tides (albeit smaller) due to the sun. It is the relative positions of the moon and sun that cause the so called spring tides (when moon and sun are aligned) and neap tides (when they are at right angles). $\endgroup$
    – Dr Chuck
    Commented Jan 25, 2016 at 6:12
  • $\begingroup$ The timing of the tides is mostly down to the Earth's rotation. Moving the moon out would barely affect their frequency (which is currently a shade less than twice a day, and would then be a slightly smaller shade less than twice a day). $\endgroup$ Commented Jan 31, 2016 at 21:16
  • $\begingroup$ True, i`ve edited my answer to make it correct. Still kept the calculation, even if its answer is not that important anymore. $\endgroup$
    – Anedar
    Commented Jan 31, 2016 at 21:45
0
$\begingroup$

The tides will get smaller as the moon moves away having less gravitational effect. As long as there are tides there will be two per day because this is based on the Earths rotation not the moons distance. As the Earth spins the tides rise on the side facing the moon and the opposite side because of gravity. Its these tidal forces along with the Earths spin that's causing the moon to recede.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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