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As I understand, satellites and the Moon orbiting Earth are in free fall. Isn't the same true for Earth orbiting the Sun?

My question is then: How can the Sun's gravity affect tides? Aren't the molecules in the oceans in free fall relative to the Sun?

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    $\begingroup$ How much do you know about tidal forces? This is important information for any potential answerers. $\endgroup$ – Wouter May 7 '14 at 18:32
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    $\begingroup$ @Wouter well, I'm a biologist, so not much... $\endgroup$ – Dave May 7 '14 at 18:46
  • $\begingroup$ Just imagine Earth and Sun (we always deal with moon to speak about tides, for you its OK) in the empty space. They will be just like approximated billiard balls. Remember earth is not completely hard. It has water on its surface, when earth keeps rotating, the water on the surface, raises and falls due to gravity due to its spinning motion or other consequences as explained by Wouter. $\endgroup$ – Immortal Player May 8 '14 at 2:03
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Tidal forces are residual forces, they are the consequence of gravitational forces acting more strongly on one part of an extended body than another.

Remember that gravity is proportional to $1/r^2$. So one side of the earth (the "nearby" side from the Sun's perspective) feels a gravitational force

$$F_g^{near} \propto 1/(D_{s-e}-R_e)^2$$

where $D_{s-e}$ is the distance between the center of the Sun and the center of the Earth and $R_e$ is the earth's radius.

Contrarily, the "far" side of the earth feels a force

$$F_g^{far} \propto 1/(D_{s-e}+R_e)^2 < F_g^{near}$$

so the far side of the earth is accelerated towards the Sun less than the near side. This residual force (the difference between the force on the near side and that on the far side) is what we call a tidal force.

Notice that this discussion is incomplete, since the gravitational force is not only influenced by distance, it is also proportional to mass (or mass density). Because water and (basically) rocks don't have the same mass density, the effect will be different for the oceans and the 'solid' chunk of 'rock' they 'envelope'. (lots of quotes there) This isn't vital to our discussion here, though, since the tidal forces aren't strong enough to significantly distort the shape of the rigid earth. But it's worth keeping in mind.

Furthermore, note that the tides on earth are mainly due to the gravitational effect of the moon on the earth, not the Sun. The Sun may be a lot heavier than the moon, but the moon is a lot closer. And gravitational forces scale only linearly with mass while they scale quadratically with inverse distance. As a consequence, the Solar tides are about half as large as the lunar tides. (see also this illustrative app and this link for some more information)

Richard Feynman actually briefly addressed the tides in his Lectures on Physics and it's fun and interesting to watch, so here's a link. His discussion of the tides starts at around 25:00.


Explicit calculation and comparison between Solar tides and Lunar tides

Let's quickly crunch some rough numbers, shall we? Of course, we'll need values for a few quantities. (I'll be using SI units since you'll probably be most familiar with those)

$$\begin{align} M_S &\approx 2\times10^{30}\,\text{kg} \\ M_M &\approx 7.3\times10^{22}\,\text{kg} \\ D_{SE} &\approx 1.5\times10^{11}\,\text{m} \\ D_{ME} &\approx 3.8\times10^{8}\,\text{m} \\ R_E &\approx 6.4\times10^{6}\,\text{m} \\ G &\approx 6.7\times10^{-11}\,\text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2} \end{align}$$

With these numbers we can calculate the gravitational acceleration $a=F_G/m$ experienced by the near and far side of the earth. (we don't calculate the force because the affected mass $m$ will in general be different) The difference between the gravitational acceleration for the near and far side is a measure for the strength of the tides. For the Sun we find

$$a_S = GM_S\left(\frac{1}{(D_{SE}-R_E)^2} - \frac{1}{(D_{SE}+R_E)^2}\right) \approx 1\times10^{-6}\,\text{m}\,\text{s}^{-2}$$

and for the moon

$$a_M = GM_M\left(\frac{1}{(D_{ME}-R_E)^2} - \frac{1}{(D_{ME}+R_E)^2}\right) \approx 2.3\times10^{-6}\,\text{m}\,\text{s}^{-2}.$$

So we see from this crude estimate that indeed the Solar tides are about half as large as the Lunar tides.

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  • $\begingroup$ I had thought the diameter of Earth is negligible compared to the distance of the Sun, so we could treat it as point-like... Thanks for the answer! $\endgroup$ – Dave May 8 '14 at 8:23
  • $\begingroup$ Well, you're right, the effect from the Sun is much smaller than that of the moon because it's so much farther away. But because it's also so much heavier, the Sun's effect on the tides is not negligible. I've edited two more links into my second-to-last paragraph which should help you and perhaps even suggest som further reading as well. $\endgroup$ – Wouter May 8 '14 at 11:03
  • $\begingroup$ @Dave: The diameter of the Earth is indeed negligible. For the purpose of calculation the gravitational force exerted by the Sun upon the Earth, you may treat is as point-like. The point is that the Earth is in free fall, so the acceleration exactly accounts for this part of the force. The remainder, however small, will have some effect: If $F_r=F+\varepsilon - F$, the quantity $\varepsilon$ is important even if $\varepsilon \lll F$. $\endgroup$ – Ansgar Esztermann May 8 '14 at 14:13
  • $\begingroup$ I added a quick estimation to show where the statement "the Solar tides are about half as large as the lunar tides" comes from. You see that even though the distance to the Sun is so big, its large mass still makes for a significant tidal effect. $\endgroup$ – Wouter May 8 '14 at 23:01

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