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This may be a dumb question but here goes nothing...

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CometPhoenix

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If the Earth has high and low tides because of the moons gravity is effecting the Earth, does that mean that to some small degree, that you would weigh less during high and low tides? And why is it that we only experience tides on a 1st and 3rd quarter moon? <div class="Discussion_UserSignature"> <font color="#0000ff"><p style="margin-top:0in;margin-left:0in;margin-right:0in" class="MsoNormal"><font face="Times New Roman" size="3" color="#0000ff">What ever happens, happens/</font><font face="Times New Roman"><font color="#0000ff"><font size="3">Just call me Phoenix</font></font></font></p></font> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>If the Earth has high and low tides because of the moons gravity is effecting the Earth, does that mean that to some small degree, that you would weigh less during high and low tides? And why is it that we only experience tides on a 1st and 3rd quarter moon? <br /> Posted by CometPhoenix</DIV></p><p>&nbsp;</p><p>Yes, you would weigh less.&nbsp; Due to the bulge in the earth making you further from the center of mass would decrease the effects of gravity via the inverse square law.&nbsp; With that said, though, there is no device on earth that could measure this.&nbsp; It would simply be too small to notice.</p><p>We experience 4 tides (2 high tides and 2 low tides) a day due to the earths rotation.&nbsp; You might be thinking of spring tides and neap tides when referring to 1st and 3rd quarter.&nbsp; Full moons and new moons are the spring tides which are the highest tides due to the alignment of the moon and sun.&nbsp; Neap tides are when the moon is at 90 degress to the direction of the sun... the high tides during neap tides are not as high during spring tides becasue the sun and moon are not aligned.</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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Saiph

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as for your weight during tides, the actual surface of the earth experiences a ~1ft shift in height...so yes, your weight fluctuates a little bit.&nbsp; But not much at all (no more than taking a single step on a flight of stairs). <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>
 
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centsworth_II

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<p><font color="#666699"><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>...yes, your weight fluctuates a little bit.&nbsp; But not much at all (no more than taking a single step on a flight of stairs). <br /> Posted by Saiph</DIV><br /></font>I wonder how the change in weight compares to , say, the soughing off of one dead skin cell. <br />Or the loss of one molecule of water from the body.&nbsp; Which would be closer, I wonder. </p> <div class="Discussion_UserSignature"> </div>
 
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Mee_n_Mac

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I wonder how the change in weight compares to , say, the soughing off of one dead skin cell. Or the loss of one molecule of water from the body.&nbsp; Which would be closer, I wonder. <br />Posted by <strong>centsworth_II</strong></DIV></p><p>I'd say 1 skin cell.&nbsp; <u>If</u> the reduction in weight was solely due to rising ~1 foot away from the Earth's center then for a large person (100kg) it's quite a few water molecules.&nbsp; I don't know what a skin cell weighs but from some dubious sources I'm estimating ~1 x 10^-8 grams and so the&nbsp;loss (~10mg)&nbsp;is closer to that weight.&nbsp;</p><p><img src="http://sitelife.space.com/ver1.0/content/scripts/tinymce/plugins/emotions/images/smiley-cool.gif" border="0" alt="Cool" title="Cool" /></p><p>Now the next question is how much less (more ?)&nbsp;would a scale indicate your weight to be when the&nbsp;the Sun is directly overhead (forgetting about any deformation of the Earth) vs at midnight&nbsp;?&nbsp; How about when the Moon is overhead or not ?&nbsp; How about during a solar eclipse ?</p><p>&nbsp;</p><p>(insert comment re: angels dancing on pinhead here)</p><p>&nbsp;</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>
 
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CometPhoenix

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<p>Then why are the tides affected so much if the moon has little to no affect on us?</p> <div class="Discussion_UserSignature"> <font color="#0000ff"><p style="margin-top:0in;margin-left:0in;margin-right:0in" class="MsoNormal"><font face="Times New Roman" size="3" color="#0000ff">What ever happens, happens/</font><font face="Times New Roman"><font color="#0000ff"><font size="3">Just call me Phoenix</font></font></font></p></font> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Then why are the tides affected so much if the moon has little to no affect on us? <br /> Posted by CometPhoenix</DIV></p><p>The oceans are affected more than the earth because the earth is a rigid object.&nbsp; These are also very large objects that have distances large enough between the closest side to the moon and the furthest allowing for a gravitational gradient.&nbsp; A swimming pool in your back yard will not see tidal effects because the gravitational gradient between the depths aren't extreme enough.&nbsp; Therefore, the weight of the water in the pool will remain the same (minus immeasurable effects) because the entire mass of the pool remains at constant distance to the center of mass of the Earth.</p><p>If I was several thousand km tall, the gravitational gradient between my head and my feet would be more extreme.&nbsp; The moon would have more of an effect on me.&nbsp; This effect would stretch me further, pulling more of my mass further from the center of mass of the Earth.&nbsp; At this point, you might see a measurable weight difference when the moon is over my head.&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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CometPhoenix

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The oceans are affected more than the earth because the earth is a rigid object.&nbsp; These are also very large objects that have distances large enough between the closest side to the moon and the furthest allowing for a gravitational gradient.&nbsp; A swimming pool in your back yard will not see tidal effects because the gravitational gradient between the depths aren't extreme enough.&nbsp; Therefore, the weight of the water in the pool will remain the same (minus immeasurable effects) because the entire mass of the pool remains at constant distance to the center of mass of the Earth.If I was several thousand km tall, the gravitational gradient between my head and my feet would be more extreme.&nbsp; The moon would have more of an effect on me.&nbsp; This effect would stretch me further, pulling more of my mass further from the center of mass of the Earth.&nbsp; At this point, you might see a measurable weight difference when the moon is over my head.&nbsp; <br />Posted by derekmcd</DIV><br /><br />Appearently I know nothing about this gravitational gradient. I'm a visual learner so if you can get a picture of this concept and explain it, it would make my day. I've been having a pretty bad week. <div class="Discussion_UserSignature"> <font color="#0000ff"><p style="margin-top:0in;margin-left:0in;margin-right:0in" class="MsoNormal"><font face="Times New Roman" size="3" color="#0000ff">What ever happens, happens/</font><font face="Times New Roman"><font color="#0000ff"><font size="3">Just call me Phoenix</font></font></font></p></font> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Appearently I know nothing about this gravitational gradient. I'm a visual learner so if you can get a picture of this concept and explain it, it would make my day. I've been having a pretty bad week. <br /> Posted by CometPhoenix</DIV></p><p>The forces of gravity extend infinitely.&nbsp; Though immeasurably small, Earth's gravity can be felt on pluto and beyond.&nbsp; As you approach Earth, the force felt get stronger as you get closer.&nbsp; This is the called the inverse square law.&nbsp; Double the distance, the force felt is 4 times as weak... triple the distance, it is 9 times as weak.&nbsp; The near side of the earth is closer to the moon than the far side, so, via the inverse square law, the near side of the Earth will feel a stronger force than the far side.&nbsp; It, essentially, get streched by the gravitational gradient.&nbsp; The gradient is the rate at which the difference in the force is felt.&nbsp; Even you standing here on earth, there is a difference in the force of gravity between your head and feet. </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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CometPhoenix

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The forces of gravity extend infinitely.&nbsp; Though immeasurably small, Earth's gravity can be felt on pluto and beyond.&nbsp; As you approach Earth, the force felt get stronger as you get closer.&nbsp; This is the called the inverse square law.&nbsp; Double the distance, the force felt is 4 times as weak... triple the distance, it is 9 times as weak.&nbsp; The near side of the earth is closer to the moon than the far side, so, via the inverse square law, the near side of the Earth will feel a stronger force than the far side.&nbsp; It, essentially, get streched by the gravitational gradient.&nbsp; The gradient is the rate at which the difference in the force is felt.&nbsp; Even you standing here on earth, there is a difference in the force of gravity between your head and feet. <br />Posted by derekmcd</DIV><br /><br />Ok, so that means that the oceans are effected more simply because it is a liquid and not soild, and it's bigger&nbsp;so more of it gets effected. And on the opposite side of the Earth, the oceans have the opposite effect than on the other side. (high and low tides). I did know about the gravatational gradient, I just didn't know that the gravity difference had a name. <div class="Discussion_UserSignature"> <font color="#0000ff"><p style="margin-top:0in;margin-left:0in;margin-right:0in" class="MsoNormal"><font face="Times New Roman" size="3" color="#0000ff">What ever happens, happens/</font><font face="Times New Roman"><font color="#0000ff"><font size="3">Just call me Phoenix</font></font></font></p></font> </div>
 
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CometPhoenix

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>The oceans are affected more than the earth because the earth is a rigid object.&nbsp; These are also very large objects that have distances large enough between the closest side to the moon and the furthest allowing for a gravitational gradient.&nbsp; A swimming pool in your back yard will not see tidal effects because the gravitational gradient between the depths aren't extreme enough.&nbsp; Therefore, the weight of the water in the pool will remain the same (minus immeasurable effects) because the entire mass of the pool remains at constant distance to the center of mass of the Earth.If I was several thousand km tall, the gravitational gradient between my head and my feet would be more extreme.&nbsp; The moon would have more of an effect on me.&nbsp; This effect would stretch me further, pulling more of my mass further from the center of mass of the Earth.&nbsp; At this point, you might see a measurable weight difference when the moon is over my head.&nbsp; <br />Posted by derekmcd</DIV><br /><br />So wait, you are saying the ocean has tides because the mass of the ocean fluctuates? <div class="Discussion_UserSignature"> <font color="#0000ff"><p style="margin-top:0in;margin-left:0in;margin-right:0in" class="MsoNormal"><font face="Times New Roman" size="3" color="#0000ff">What ever happens, happens/</font><font face="Times New Roman"><font color="#0000ff"><font size="3">Just call me Phoenix</font></font></font></p></font> </div>
 
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derekmcd

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<p><strong>"Ok, so that means that the oceans are effected more simply because it is a liquid and not soild..."</strong></p><p>Yes. </p><p><strong>"...and it's bigger&nbsp;so more of it gets effected."</strong> </p><p>Sort of.&nbsp; Due to the oceans size, the scale of the effect is more pronounced. </p><p><strong>"And on the opposite side of the Earth, the oceans have the opposite effect than on the other side. (high and low tides)."</strong> </p><p>Actually, the opposite side will also have a high tide.&nbsp; Due to the gravitational gradient and the fluidity of the ocean, the ocean forms an an ellipsoid.&nbsp; It is basically stretched in the direction of the moon.&nbsp; With the earth being a rigid body inside this ellipsoid of water, the rigid body will accelerate towards the barycenter of the earth/moon system faster than the water.&nbsp; The earth and moon are actually in free fall around each other.&nbsp; The center point at which they are orbiting is called the barycenter. </p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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CometPhoenix

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<p>This is all so confusing. When I made this topic, I was thinking that tides only occured during certain&nbsp;moon phases, but appearently I knew next to nothing about tides. I had no idea that we have tides all the time/everyday. So wait, would a lake have a tidal effect? What about a small pond?</p> <div class="Discussion_UserSignature"> <font color="#0000ff"><p style="margin-top:0in;margin-left:0in;margin-right:0in" class="MsoNormal"><font face="Times New Roman" size="3" color="#0000ff">What ever happens, happens/</font><font face="Times New Roman"><font color="#0000ff"><font size="3">Just call me Phoenix</font></font></font></p></font> </div>
 
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Saiph

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<p>Hmm...I know you have to be pretty big to get any measurable tidal effects.&nbsp; Remember, even a lake is basically in one small local position under the moon.&nbsp; The entire thing is going to feel the same general effect at any given time.</p><p>I think the great lakes have tides...but I don't know about smaller bodies of water.&nbsp; They just aren't big enough for the moon to drag enough water out of place.&nbsp;</p> <div class="Discussion_UserSignature"> <p align="center"><font color="#c0c0c0"><br /></font></p><p align="center"><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">--------</font></em></font><font color="#999999"><em><font size="1">----</font></em></font><font color="#666699">SaiphMOD@gmail.com </font><font color="#999999"><em><font size="1">-------------------</font></em></font></p><p><font color="#999999"><em><font size="1">"This is my Timey Wimey Detector.  Goes "bing" when there's stuff.  It also fries eggs at 30 paces, wether you want it to or not actually.  I've learned to stay away from hens: It's not pretty when they blow" -- </font></em></font><font size="1" color="#999999">The Tenth Doctor, "Blink"</font></p> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>This is all so confusing. <br /> Posted by CometPhoenix</DIV></p><p>Don't feel bad.&nbsp; It took one of the most brilliant people to ever live to figure them out (Newton).&nbsp;</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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CometPhoenix

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Don't feel bad.&nbsp; It took one of the most brilliant people to ever live to figure them out (Newton).&nbsp; <br />Posted by derekmcd</DIV><br /><br />Ok, here's a hypothetical question: </p><p>Lets say that Europa wasn't ice, but it was liquid water. How big of a tidal effect would Europa experience from the gravity of Jupiter? Wouldn't it just suck it off of the moon? If so, then why does Europa even have ice water?</p> <div class="Discussion_UserSignature"> <font color="#0000ff"><p style="margin-top:0in;margin-left:0in;margin-right:0in" class="MsoNormal"><font face="Times New Roman" size="3" color="#0000ff">What ever happens, happens/</font><font face="Times New Roman"><font color="#0000ff"><font size="3">Just call me Phoenix</font></font></font></p></font> </div>
 
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derekmcd

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Ok, here's a hypothetical question: Lets say that Europa wasn't ice, but it was liquid water. How big of a tidal effect would Europa experience from the gravity of Jupiter? Wouldn't it just suck it off of the moon? If so, then why does Europa even have ice water? <br /> Posted by CometPhoenix</DIV></p><p>&nbsp;</p><p>Europa isn't nearly close enough to have any hypothetical surface water sucked of it.&nbsp; If it was, the entire moon would lose momentum and spiral into Jupiter.&nbsp; That's not to say that the tidal effect on Europa aren't strong.&nbsp; You can see the effects in the cracking of the surface ice.</p> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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Mee_n_Mac

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Appearently I know nothing about this gravitational gradient. I'm a visual learner so if you can get a picture of this concept and explain it, it would make my day. I've been having a pretty bad week. <br />Posted by <strong>CometPhoenix</strong></DIV><br /><br />Here's a site with some good visuals and the math to back them up.</p><p>http://www.rain.org/~mkummel/stumpers/06dec02a.html</p><p>You also might like the visuals at How Stuff Works ...</p><p>http://science.howstuffworks.com/question72.htm</p><p>(<a href="http://www.howstuffworks.com/framed.htm?parent=question72.htm&url=http://riker.ps.missouri.edu/RICKSPAGE/Moon/Tides.html" target="_blank">and the animation linked to from there)</p></a><p>Lastly if you can bear reading the following you can calculate the answers to my questions way above and also find out that the height of a tide is largely dependant on the topography of the ocean bottom and funneling effect of the bay you're in.&nbsp; If the Earth were covered in 1 big ocean the tides would be less than 1 meter in height.&nbsp; Compare that to what's observed in the Bay of Fundy.&nbsp; The author also notes that the Great Lakes here in the US have a tide of perhaps 2".</p><p>http://www.jal.cc.il.us/~mikolajsawicki/Tides_new2.pdf</p> <div class="Discussion_UserSignature"> <p>-----------------------------------------------------</p><p><font color="#ff0000">Ask not what your Forum Software can do do on you,</font></p><p><font color="#ff0000">Ask it to, please for the love of all that's Holy, <strong>STOP</strong> !</font></p> </div>
 
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derekmcd

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Good choices on the links as none of them mention centrifugal force.&nbsp; I'm suprised NOAA still uses this as an explanation for the bulge opposite the moon.<br /> <div class="Discussion_UserSignature"> <div> </div><br /><div><span style="color:#0000ff" class="Apple-style-span">"If something's hard to do, then it's not worth doing." - Homer Simpson</span></div> </div>
 
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