# Is the tide on Earth caused by curvature of spacetime

The tide on Earth appears absolutely whenever the moon is overhead. Is that tide caused by spacetime, re-curvature in space or attraction gravity?

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–  Qmechanic Aug 17 '13 at 12:06
BTW--The tide also appears when the moon is underfoot (as it were). This is obvious when you understand the physics and may help you in establishing that understanding. –  dmckee Aug 17 '13 at 21:32

Well, basically gravity is a consequence of curved spacetime!

Space-time is distorted by the presence of mass. This results in any other mass in the vicinity being attracted to it. This attractive force is called gravity. So, basically gravity is caused because masses curve space-time.

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Actually the rubber sheet or mattress analogy, though very common, is terrible! See physics.stackexchange.com/q/7781, physics.stackexchange.com/q/16925 and physics.stackexchange.com/q/3009 –  Michael Brown Aug 17 '13 at 14:09
Well it may not be true, but it works something like that - where there's mass, other mass will be attracted. For a layman to understand I think it's quite a good way to show it. –  mikhailcazi Aug 17 '13 at 14:27
Anyway, since it is NOT true, I'll remove that part. –  mikhailcazi Aug 17 '13 at 14:31

First,tides can also occur when the moon is directly below us.Tides occur at an interval of 12 hrs so twice a day.

Now,we don't need curvature of space to explain tides on earth.They can simply be explained by Newtonian gravity.

Consider the situation as shown in the figure.

m stands for some small mass relative to $M_1$ and $M_2$ which represents say the oceans.

Lets calculate the difference between force of moon acting at centre and acting at edge on $m$(This gives us the tidal force)

$F_m=(\frac{GM_2m}{(d-R_1)^2} - \frac{GM_2m}{d^2})\hat{r}$

where

$d=$distance between earth and moon,

$\hat{r}$ is the radius vector from the earth towards the moon.

Now expanding the first term as a Taylor series and neglecting terms of order three and higher($d>>R_1$):

$F_m=(\frac{GM_2m}{d^2}+\frac{GM_2mR_1}{d^3} - \frac{GM_2m}{d^2})\hat{r}$ $=\frac{GM_2mR_1}{d^3}\hat{r}$ Now this force points towards the moon(from our definition of $\hat{r}$) and therefore the water slightly moves towards the moon.

If we were to repeat our calculation for the mass $m$ on the other side of earth,we would observe this "tidal force" to point away from the moon which makes $m$ go away from the earth.This leads to the familiar shape of earth's oceans.

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Indeed this demonstrates the basic mechanism. Then put the system in motion (earth's rotation and moon's orbit) and it no longer lines up properly: en.wikipedia.org/wiki/Lunitidal_interval. Then add solar tidal force. –  Steve Jessop Mar 11 at 11:42