gravity at the center of a planet

beilzabob · Jul 18, 2005

Hypothetically if you could drill a hole through the center of a planet ;wouldn't an object dropped in settle at the very center floating from the gravity pushing on it in all directions? Also is gravity a constant force from entering the pull of the earth to the center or does it change as you get closer to the center of the mass?Gravity must be directly related to mass or is related to compression of space trying to return to it's normal space without an object of mass occupying it.

najab · Jul 18, 2005

Eventually, yes. It would oscillate back and forth for a long time first though.

lunatio_gordin · Jul 18, 2005

sort of like a rubber band snapping into place. it actually extend beyond it's stable point and wobbles some. how much it wobbles would depend on speed, no?

Saiph · Jul 19, 2005

well...if the hole was a vacuum, and no air interfered...i.e. no drag...the thing would oscillate back and forth forever. If you throw in the air (drag forces)...it'll eventually settle to the center. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"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" -- The Tenth Doctor, "Blink" </div>

alokmohan · Jul 19, 2005

I also think so.

MBA_UIU · Jul 19, 2005

We must remember that the center of mass in a sphere of equally density is not located in its center of the sphere, but at a depth roughly equal to 2/3rds of its circumference. We can demonstrate this using Pi r^2 Example: Diameter 24” = Radius of 12” using Pi r^2 we get 3.1415*12*12= 452sqi Diameter 12” = Radius of 6” using Pi r^2 we get 3.1415*6*6=113 Has we can see if we trim off one half the diameter we are left with just 25% of its original area. Diameter 17” = Radius of 8.5” using Pi r^2 we get 3.1415*8.58.5=227sqi which is roughly one half of our original 24” diameter sphere. This gives us the location of the true center of mass which is 7” below its outer surface. Why is this important? Because the question asked what would happen if we dropped the ball through the sphere. Each of your answers was close, but they all assumed that the center mass of a sphere was in fact in its center. If you could drill a hole through the earth and remove the earths tilt, rotation and heat from the equation then, depending on the balls weight and altitude, the ball would do one of three things: first if the ball was dropped from the very opening of the tube it would drop to a distance of roughly ½ to the center of the sphere then be drawn back to the center of mass. If its initial distance started at equal to the distance of the opening to the center of mass it should reach the center of the sphere and come to a stop, becoming trapped in the center due to equal pulls. If it is dropped from a distance of that is greater then twice the distance of the opening from the center of mass then it should have enough momentum to go completely through the center and go out the other side. Now for the proof experiment, what you will need is round 2 magnets of equal strength that have an equal sized hole in the middle, a clear piece of plastic tubing and a steel ball that will fit within the tubing. Place the tube through the <div class="Discussion_UserSignature"> <img id="268587ce-7170-4b41-a87b-8cd443f9351a" src="http://sitelife.space.com/ver1.0/Content/images/store/6/8/268587ce-7170-4b41-a87b-8cd443f9351a.Large.jpg" alt="blog post photo" /> </div>

Saiph · Jul 19, 2005

That...is completely wrong. First, center of mass should be over 3 dimensions, not 2. Second, the way you calculate it is given here [ur]http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html[/url] At the center there is equal pulls from all sides. As such, there is no net force and thus no acceleration, and nothing causing it to stop there. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"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" -- The Tenth Doctor, "Blink" </div>

labguy · Jul 19, 2005

<blockquote>In reply to:<hr />That...is completely wrong. First, center of mass should be over 3 dimensions, not 2. Second, the way you calculate it is given here [ur]http://hyperphysics.phy-astr.gsu.edu/hbase/cm.html[/url] At the center there is equal pulls from all sides. As such, there is no net force and thus no acceleration, and nothing causing it to stop there. <hr /></blockquote>You are correct, the center of mass and center of gravity are at the center for most astronomical (spherical) objects. From: http://en.wikipedia.org/wiki/Center_of_gravity it states that: "Most astronomical objects are radially symmetric, causing both the center of gravity and the center of mass to coincide at the center of the sphere."

Saiph · Jul 19, 2005

Right, the only reason a sphere would have a center of mass that is not in (or very close in reality) the geometric center, would be due to significant non-uniformities in density. I.e. it's got a really dense chunk somewhere, that isn't mirrored elsewhere. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"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" -- The Tenth Doctor, "Blink" </div>

silylene old · Jul 19, 2005

Thanks Saiph. I too was just about to write a big objection to that wrong explanation about center of mass when I saw you had it all correctly covered. <div class="Discussion_UserSignature"> <div class="Discussion_UserSignature" align="center">- - - - - - - - - - - - - - - - - - - - - - </div><div class="Discussion_UserSignature" align="center">I really, really, really miss the "first unread post" function. </div> </div>

MBA_UIU · Jul 20, 2005

I was trying to be very simplistic in my approach and I see that it came back to bite me in the hinny. Remember I said this was shown to us in high school and I was only 14 years old so I really didn’t question this. Now then… in answer to the question I did some research and found that Newton worked on this question in 1665. The answer appears here http://hep.ph.liv.ac.uk/~green/mechanics/lectures/l15notes.pdf Scroll down to page 4-5 <div class="Discussion_UserSignature"> <img id="268587ce-7170-4b41-a87b-8cd443f9351a" src="http://sitelife.space.com/ver1.0/Content/images/store/6/8/268587ce-7170-4b41-a87b-8cd443f9351a.Large.jpg" alt="blog post photo" /> </div>

Saiph · Jul 20, 2005

is it the "shell theorem" where any matter that is further from the center than you are, in a radially symetric system, produces no net gravitational (or electrostatic) force. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"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" -- The Tenth Doctor, "Blink" </div>

MBA_UIU · Jul 20, 2005

basically it states that there is no change in gravity at any depth <div class="Discussion_UserSignature"> <img id="268587ce-7170-4b41-a87b-8cd443f9351a" src="http://sitelife.space.com/ver1.0/Content/images/store/6/8/268587ce-7170-4b41-a87b-8cd443f9351a.Large.jpg" alt="blog post photo" /> </div>

najab · Jul 20, 2005

<blockquote>In reply to:<hr />basically it states that there is no change in gravity at any depth<hr /></blockquote> Wrong. Consider the situation at the top of the hole and then halfway down. At the top of the hole the entire mass of the planet is "down" from the ball, so it accelerates at 'g'. Halfway down the hole, there is mass "above" the ball so it tries to pull the ball back out of the hole. This is exactly balanced by the same amount of mass on the other side of the centre still trying to pull the ball "down" (green areas) so they cancel each other out. Effectively then, the ball only "feels" the attraction of the smaller sphere that's still "downhill" from it.

MBA_UIU · Jul 20, 2005

here is Newton's work <div class="Discussion_UserSignature"> <img id="268587ce-7170-4b41-a87b-8cd443f9351a" src="http://sitelife.space.com/ver1.0/Content/images/store/6/8/268587ce-7170-4b41-a87b-8cd443f9351a.Large.jpg" alt="blog post photo" /> </div>

MBA_UIU · Jul 20, 2005

newton con... <div class="Discussion_UserSignature"> <img id="268587ce-7170-4b41-a87b-8cd443f9351a" src="http://sitelife.space.com/ver1.0/Content/images/store/6/8/268587ce-7170-4b41-a87b-8cd443f9351a.Large.jpg" alt="blog post photo" /> </div>

MBA_UIU · Jul 20, 2005

newton continued... <div class="Discussion_UserSignature"> <img id="268587ce-7170-4b41-a87b-8cd443f9351a" src="http://sitelife.space.com/ver1.0/Content/images/store/6/8/268587ce-7170-4b41-a87b-8cd443f9351a.Large.jpg" alt="blog post photo" /> </div>

Saiph · Jul 20, 2005

Well, that is indeed the correct material, your synopsis of it <blockquote>In reply to:<hr /> basically it states that there is no change in gravity at any depth<hr /></blockquote> Was innaccurate. If you are in a hollow sphere, then yes, there will be no change. however, if you are in a sphere of uniform (or even radially symentric) density, there is still a change as you increase in depth. The specific section of importance here, is the part about someone in a mine, not feeling the gravity of the earth in the area above him anymore. That means he feels less gravity. As he descends, there is more material above him, and more material that no longer contributes, thus indicating even less net force due to gravity. <div class="Discussion_UserSignature"> --------------------SaiphMOD@gmail.com -------------------"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" -- The Tenth Doctor, "Blink" </div>

pu_aero · Jul 20, 2005

The solution is within your post: " Force within shell: F = -dU/dh = 0" If you are within a spherical shell, the net force is zero. Simple as that. Please note that the analysis you posted is for a spherical shell, which is hollow by definition, rather than a solid sphere, as Saiph pointed out. To analyze a solid sphere you can split it into two sections if you are inside it: 1. The thick spherical shell "above you;" that is, the shell defined between the radius from the center of the earth at which you currently are and the surface of the earth. 2. The sold sphere below you, with radius defined by your current position. Part #1, the shell, exerts no force on you since you are inside it, viz. the analysis you posted. Since you are outside Part #2, it is like you are on the surface of a smaller planet. It exerts a gravitational force pulling you toward its center. As you go closer to the center of the earth, more of it is part of #1, exerting no net force, and less of it is part of #2. So the gravitational force pulling you to the center decreases as you go toward the center. This would indeed produce an oscillation, as many of the posters to this thread have indicated. (neglecting air resistance)

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