PDA

View Full Version : Physics questions. Do we weigh less at night time. And would the moon have a tide



Tom Servo
2010-Jul-21, 04:54 AM
Ok I have two physics questions. Really thinking deep about this stuff for no reason and wanted someone with more knowledge to confirm or correct my thoughts.


First question.

Do we weigh more during the day time than we do at night time.

I think because of the earth rotating around the sun this would put a flinging force on us at night and a pushing force on us during the day.
Probably very insignificant but can't help to wonder exactly how much of a difference this might make.


Second question.

If the moon had oceans would it also have a tide like the earth.
If I understand it correctly the earths tides are created as the gravitational forces from the moon pulls at the earth as it rotates around us.

However since the moon always faces the earth I would think this force would not work the same way for a lunar ocean.

Can someone correct me if I am wrong.

I appreciate all answers.

HenrikOlsen
2010-Jul-21, 05:00 AM
You're thinking tidal effects?
First thing to realize is that there are two high tides every day, one when the moon is above our heads and one when the moon is below our feet.

A very simplified way to look at it is that when the moon is above our head, we're closer to it than the earth's center is, so we're attracted more and when it's below our feet we're farther away and is therefore attracted less.

First question.

Do we weigh more during the day time than we do at night time.

I think because of the earth rotating around the sun this would put a flinging force on us at night and a pushing force on us during the day.
Probably very insignificant but can't help to wonder exactly how much of a difference this might make.
Tidal effect from the sun, that makes you very slightly lighter at noon and midnight and slightly heavier at dawn and dusk.


Second question.

If the moon had oceans would it also have a tide like the earth.
If I understand it correctly the earths tides are created as the gravitational forces from the moon pulls at the earth as it rotates around us.

However since the moon always faces the earth I would think this force would not work the same way for a lunar ocean.

It has something like tides, but since the earth doesn't move when seen from the surface of the moon, they're stationary bulges instead of moving waves, the moon is slightly bulging towards us on the visible side and bulging slightly away from us on the far side.

Can someone correct me if I am wrong.
You thinking was ok, but it was based on the wrong idea about tides, so the results were wrong.

Hornblower
2010-Jul-21, 09:54 AM
The Moon would have rising and falling tides from the Sun's component, with two high tides each month. This would be superimposed on the prolate spheroid imposed by the Earth.

Ivan Viehoff
2010-Jul-22, 12:09 PM
You're thinking tidal effects? First thing to realise is that there are two high tides every day, one when the moon is above our heads and one when the moon is below our feet.
Actually it is more complicated than that. The earth experiences slightly fewer than 2 moon tides per day ("semidiurnal component"), but also experiences one sun tide per day ("diurnal component"), and the magnitude of the former is only about 2-3 times the magnitude of teh second. See http://en.wikipedia.org/wiki/Tides, especially Fig 8 showing how the separate semidiurnal (moon) and diurnal (components) can add up to something really odd at some places, albeit that a marked semidiurnal tide is the experience of most places. Also because of the water sploshing through complex shaped basins in some areas, there are also places where what happens is dominated by the secondary effect of that sploshing, for example an area in the middle of the North Sea which has practically no tide at all, because the local tides rotate about it.

It is not just the sea that has tides. The earth's rock also has a tide. At typical land locations, you move up and down about 50cm twice a day because of tides. At CERN, for example, it wouldn't matter to them much, I suspect, if they were just moving up and down, but the shape of the ring is also distorted by the tidal effect: I think it changes the diameter of the ring by about 5mm, which is quite far enough that the particle beams would completely miss their collision if they didn't adjust the controlling fields for this effect.

The moon keeps its same face towards the earth at all times, with only slight wobbles, so there is no material earth tide on the moon. As has been explained, there is a sun tide.

IsaacKuo
2010-Jul-22, 04:04 PM
Actually it is more complicated than that. The earth experiences slightly fewer than 2 moon tides per day ("semidiurnal component"), but also experiences one sun tide per day ("diurnal component"),
Any particular point on Earth experiences a semi-diurnal solar tide just like it experiences a semi-diurnal lunar tide (but with slightly different periods).

[edit: to simplify things.]

HenrikOlsen
2010-Jul-22, 05:26 PM
The moon keeps its same face towards the earth at all times, with only slight wobbles, so there is no material earth tide on the moon. As has been explained, there is a sun tide.
But there is a stationary bulge on the near and far sides due to the same effect.

Ken G
2010-Jul-22, 10:31 PM
The earth experiences slightly fewer than 2 moon tides per day ("semidiurnal component"), but also experiences one sun tide per day ("diurnal component"),


Any particular point on Earth experiences a semi-diurnal solar tide just like it experiences a semi-diurnal lunar tide (but with slightly different periods).

In case it is not clear from the correction offered by IsaacKuo to the idea that there is any such thing as a "Sun tide", the Sun does not cause a separate tide from the Moon (from the above, one might think you get four tides a day, two from the Sun and two from the Moon!). Instead, the Moon and the Sun get together and decide on the shape of Earth's ocean, and mostly what the Sun does is enhance or weaken the bulges the Moon is making, not create additional bulges. In particular, the shape of the equipotential in the plane of the Sun, Earth, and Moon is always an ellipse-- two bulges. The fact that this plane is not necessarily the equatorial plane causes much complications as the Earth rotates through the bulges, and the fact that the ocean does not immediately conform to the equipotential but instead shows a time lag as it flows toward the equipotential wells also causes complications. Then there's the shape of the coastlines interacting with those flows, and you end up with a spectacularly complicated problem that pretty much requires local observations to make any sense of!

karadan
2010-Jul-23, 11:55 AM
But there is a stationary bulge on the near and far sides due to the same effect.

I have a stationary bulge on my near side.. :)

Sorry...

Sticks
2010-Jul-24, 02:24 PM
Moved to Q&A

cjameshuff
2010-Jul-24, 06:35 PM
The moon keeps its same face towards the earth at all times, with only slight wobbles, so there is no material earth tide on the moon. As has been explained, there is a sun tide.

There is a slight Earth tide on the moon. The moon's orbital distance varies from 363104 km to 405696 km, so the tidal forces that cause that bulge periodically strengthen and weaken. This tide is both weak and slow, with a period of 27 days.

grapes
2010-Jul-24, 09:17 PM
Actually it is more complicated than that. The earth experiences slightly fewer than 2 moon tides per day ("semidiurnal component"), but also experiences one sun tide per day ("diurnal component"), and the magnitude of the former is only about 2-3 times the magnitude of teh second. See http://en.wikipedia.org/wiki/Tides, especially Fig 8 showing how the separate semidiurnal (moon) and diurnal (components) can add up to something really odd at some places, albeit that a marked semidiurnal tide is the experience of most places. Isaac and Ken have pointed out that the diurnal component is not the sun tide, but I browsed through that wiki article to see if it did say that, and I also found this:
The sun also exerts a (less powerful) gravitational attraction on the earth which results in a secondary tidal effect.Of course, the sun exerts a much more powerful gravitational attraction on the earth than the moon does. It's tidal effect is smaller because it is so far away, and the differential is smaller. As the wiki page says later:
The solar gravitational force on the Earth is on average 179 times stronger than the lunar, but because the Sun is on average 389 times farther from the Earth, its field gradient is weaker.


There is a slight Earth tide on the moon. The moon's orbital distance varies from 363104 km to 405696 km, so the tidal forces that cause that bulge periodically strengthen and weaken. This tide is both weak and slow, with a period of 27 days.I know, by that, that you mean a changing tide. Since the tidal force is proportional to mass (earth is about 81 times the moon mass) and radius (the moon's radius is about 1/4 the radius of the earth*), the earth's tidal force on the moon is about twenty times (81/4) larger than the moon's on the earth. And bsides the difference in orbital distance, the moon librates while it changes distance, changing its orientation with respect to the earth.

* If the moon were the same density as the earth, and it has 1/4 the radius, it would be 1/43 or a sixty-fourth the mass. It is less dense, on average.

pzkpfw
2010-Jul-24, 09:48 PM
I think because of the earth rotating around the sun this would put a flinging force on us at night and a pushing force on us during the day.
Probably very insignificant but can't help to wonder exactly how much of a difference this might make.

There is a real "flinging" effect - but this is from the rotation of the Earth around its own axis and isn't a day-night thing. e.g. A person standing on scales at the equator will show slightly less weight than a person doing the same thing at one of the poles (assuming the Earth is a perfect sphere*). e.g. Rocket launch sites (and directions) are chosen to take this into account also.

(*There'd also be an effect from the Earth bulging at the equator, due to the same rotation, making the person farther away from the centre of mass of Earth.)

Ken G
2010-Jul-25, 01:37 AM
Since the tidal force is proportional to mass (earth is about 81 times the moon mass) and radius (the moon's radius is about 1/4 the radius of the earth*), the earth's tidal force on the moon is about twenty times (81/4) larger than the moon's on the earth.
And even that only ratios the differences in the tidal acceleration-- probably the most faithful measure of the strength of the tidal force is the distance by which it lifts a given equipotential (i.e., the "size of the bulge"). That means you also have to ratio to the surface gravity (so a factor of 1/5 in the denominator), to get the fractional change in surface gravity, which is a factor of 100 larger for Earth-on-Moon than Moon-on-Earth (so that tells you the Moon is 100 times more skewed into a "football" shape than is the Earth), and then multiply by the 1/4 radius factor to get the scale of the lifting that results. So that net comparison is actually about a factor of 25. That formula is M2/R4, where M is the Earth mass in Moon units (81), and R is the Earth radius in Moon units (4). That would compare the rise in "Moon oceans" if the Moon were not tidally locked (how much less it really is because you only get the effect at a much lower order is a whole other calculation!)

grapes
2010-Jul-25, 03:59 AM
So that net comparison is actually about a factor of 25. That formula is M2/R4, where M is the Earth mass in Moon units (81), and R is the Earth radius in Moon units (4). That would compare the rise in "Moon oceans" if the Moon were not tidally lockedCute. Of course, my formula (M/R) would be the same as yours, if their densities were the same. :)

The moon is less dense, so tides can affect it more.
(how much less it really is because you only get the effect at a much lower order is a whole other calculation!)Lower order? What do you mean by that?

Ken G
2010-Jul-25, 04:26 AM
Cute. Of course, my formula (M/R) would be the same as yours, if their densities were the same.Interesting point-- I can't see any deeper reason for it, but perhaps there is one. I think it must also relate to the "Roche distance", which is the distance that a moon can come to a planet, say Saturn, before it gets ripped up by the tidal stresses-- and that depends on the density comparison of the moon and the planet, but not the size comparison.

Lower order? What do you mean by that?I just mean the factor of 25 is for the Moon's oceans if it wasn't locked-- since it is locked to the Earth, we'd only get a fraction of that, basically three times the fraction of the change in the Moon's distance. Since that distance changes by plus or minus 1/16 or so, we're talking about a 1/5 effect compared to what a rotation through the bulges could do. Say, that's not so small-- it's still 5 times Earth tides! Unless I've overlooked something, "oceans" on the Moon would suffer monthly tides that were 5 times the height of tides in Earth oceans.

grapes
2010-Jul-25, 07:36 AM
I see what you getting at. The eccentricity is only 1/18 though, so 3/18 or 1/6, of 25 is about four times. Still the lunar tides on earth are only about a meter, top to bottom--exclusive of other resonances. 'Course, there'd be resonances on the moon, right? Well, not really, because this thing we're talking about is very slow.

Ken G
2010-Jul-25, 02:45 PM
You're right about the 1/18, my estimate was too rough. And yes, at a month period, we should expect the oceans to fill their equipotentials I should think, so no "Bay of Fundy" on the Moon! Unless the "oceans" are of thick lava or something, that would be a more serious kind of high tide!