# Unknown Gravitational Anomaly Observation

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#### Jerry451

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<font size="2"><p>Hello Space.com People,</p><p>I tripped across the "Top 10 Cool Moon Facts" and one of them reminded me of an unknown property of gravity&nbsp;that the world at large still does not know. Basically to add to the list of cool Moon facts this property says that if you bring mass from the Moon to the Earth the departure of the Moon from the Earth is quickened because the force of gravity between the two bodies decreases. The principle is detailed below with the math proof attached. Be my guest and verify it yourself.</p><p>Subject: Gravitational anomaly observation </p><p>Comment:</p><p>&nbsp;</p><p>Something that an instructor said during my undergraduate studies at the University of Alabama got me thinking about the uncoupling of a light wave as it propagates through a gravitational field. From my understanding I expected an effect and went looking for it. I observed that if you fix total system mass and the relative positions of each mass but vary the amount of mass at each point that the total gravitational force of attraction peaks when all of the system masses are exactly equal. Symmetry arguments could be made as to why this is so. This is a previously unrecognized property of gravitational force.</p><p>&nbsp;</p><p>Proof</p><p>LET m + M = K or M = K -m</p><p>F = G/R^2 (m M)</p><p>F = -G/R^2 (m^2 - K m)</p><p>dF/dm = -G/R^2 (2m - K) = 0</p><p>m = K/2</p><p>Why K/2</p><p>* consider m=0 ( or arbitrarily close to zero ) and the force is zero ( or arbitrarily close to zero )</p><p>* Same for m = K or close to K</p><p>So it has to peak somewhere between 0 and K.</p><p>* Paint one blue and one red. Since color doesn't affect gravity, the result should be symmetric in m, M. Simplest symmetric solution is at m = M or m = K/2.</p><p>This leads to the possibility of modulating the strength of gravity between two bodies creating a gravitational wave fluctuation that could be used as a new (covert?) form of communication and the means to produce a gravitational harmonic for use in a reactive propulsion system within natural gravitational fields. Also, how much dark matter mass could be accounted for by this increase of gravitational force due to the attraction of equals?</p><p>Hmmm... What do you think...</p></font> <div class="Discussion_UserSignature"> </div>

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#### DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Hello Space.com People,I tripped across the "Top 10 Cool Moon Facts" and one of them reminded me of an unknown property of gravity&nbsp;that the world at large still does not know. Basically to add to the list of cool Moon facts this property says that if you bring mass from the Moon to the Earth the departure of the Moon from the Earth is quickened because the force of gravity between the two bodies decreases. The principle is detailed below with the math proof attached. Be my guest and verify it yourself.Subject: Gravitational anomaly observation Comment:&nbsp;Something that an instructor said during my undergraduate studies at the University of Alabama got me thinking about the uncoupling of a light wave as it propagates through a gravitational field. From my understanding I expected an effect and went looking for it. I observed that if you fix total system mass and the relative positions of each mass but vary the amount of mass at each point that the total gravitational force of attraction peaks when all of the system masses are exactly equal. Symmetry arguments could be made as to why this is so. This is a previously unrecognized property of gravitational force.&nbsp;ProofLET m + M = K or M = K -mF = G/R^2 (m M)F = -G/R^2 (m^2 - K m)dF/dm = -G/R^2 (2m - K) = 0m = K/2Why K/2* consider m=0 ( or arbitrarily close to zero ) and the force is zero ( or arbitrarily close to zero )* Same for m = K or close to KSo it has to peak somewhere between 0 and K.* Paint one blue and one red. Since color doesn't affect gravity, the result should be symmetric in m, M. Simplest symmetric solution is at m = M or m = K/2.This leads to the possibility of modulating the strength of gravity between two bodies creating a gravitational wave fluctuation that could be used as a new (covert?) form of communication and the means to produce a gravitational harmonic for use in a reactive propulsion system within natural gravitational fields. Also, how much dark matter mass could be accounted for by this increase of gravitational force due to the attraction of equals?Hmmm... What do you think... <br />Posted by Jerry451</DIV></p><p>I think you blew the math.</p><p>The mistake came when you calculated dF/dm and treated K as a constant when in fact K depends on m (hoilding M constant).&nbsp; If you calculate the derivative correctly you find, as you would expect from the original expression for F that dF/dm=G/R^2(M) which is never zero unless M is zero, which it is not.</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 think you blew the math.The mistake came when you calculated dF/dm and treated K as a constant when in fact K depends on m (hoilding M constant).&nbsp; If you calculate the derivative correctly you find, as you would expect from the original expression for F that dF/dm=G/R^2(M) which is never zero unless M is zero, which it is not. <br />Posted by <strong>DrRocket</strong></DIV><br /><br />I think he had it right, given his stipulation that the total mass of the 2 body system remain a constant.&nbsp; I understood that constant to be K not M.&nbsp; M or m were to be the masses of the Earth and Moon or vice versa.&nbsp; He finds an inflection point where M=m and claims it to be a maxima because the 2 endpoints, m = 0 and m = K, yeild a zero result.&nbsp; The claim is true though not proven.&nbsp; Basically what's the maximum product of 2 positive numbers (fractions), that sum to one.&nbsp; </p><p>What all that has to do with the rest of his post remains unknown to me.&nbsp; I suspect if we could modulate gravity short of redistributing mass we could do a lot of things.&nbsp; Maybe we need to ask Obama re: the redistributing scheme ....</p><p>ps - And given the effect we call gravity is the result of a geometry and not a true force, I'll need a bit more of an explanation on the later part of the OP.</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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#### Jerry451

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I think you blew the math.The mistake came when you calculated dF/dm and treated K as a constant when in fact K depends on m (hoilding M constant).&nbsp; If you calculate the derivative correctly you find, as you would expect from the original expression for F that dF/dm=G/R^2(M) which is never zero unless M is zero, which it is not. <br />Posted by DrRocket</DIV><br /><br /><font size="2"><p>I think I understand where some confusion may exist. Let me first put in a couple of parentheses so that there is no confusion in the gravitational force equations.</p></font><p>Proof</p><p>LET m + M = K or M = K -m</p><p>F = (G/R^2) (m M)</p><p>F = -(G/R^2) (m^2 - K m)</p><p>dF/dm = -(G/R^2) (2m - K) = 0</p><p>m = K/2</p><p>Last, technically by definition K is a constant because I have defined above that the total system mass remains unchanged for any set of m + M. The two masses (m & M) are the variables not the total system mass (K)</p> <div class="Discussion_UserSignature"> </div>

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#### Mee_n_Mac

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I think I understand where some confusion may exist. Let me first put in a couple of parentheses so that there is no confusion in the gravitational force equations.ProofLET m + M = K or M = K -mF = (G/R^2) (m M)F = -(G/R^2) (m^2 - K m)dF/dm = -(G/R^2) (2m - K) = 0m = K/2Last, technically by definition K is a constant because I have defined above that the total system mass remains unchanged for any set of m + M. The two masses (m & M) are the variables not the total system mass (K) <br />Posted by <strong>Jerry451</strong></DIV><br /><br />OK, so how does this lead "to the possibility of modulating the strength of gravity between two bodies" by (I assuming the by got left out), well, by any means ?&nbsp; <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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#### emperor_of_localgroup

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I think I understand where some confusion may exist. Let me first put in a couple of parentheses so that there is no confusion in the gravitational force equations.ProofLET m + M = K or M = K -mF = (G/R^2) (m M)F = -(G/R^2) (m^2 - K m)dF/dm = -(G/R^2) (2m - K) = 0m = K/2Last, technically by definition K is a constant because I have defined above that the total system mass remains unchanged for any set of m + M. The two masses (m & M) are the variables not the total system mass (K) <br /> Posted by Jerry451</DIV></p><p><font size="2">Hhhmmm, interesting thought. </font></p><p><font size="2">But if m=K/2, then M is also K/2, meaning equal mass for both objects. Why would two equal masses create zero force at any point between them? I think the problem is with R. R should R(m), a function of varying mass. R can not be a constant if want to keep them in orbit, because with changing m, R must also be changed.</font></p><p><font size="2">Another way to look at this is, if we use changing M as&nbsp; M(m):</font></p><p><font size="2">dF/dm = d(GmM/R^2)/dm = GM/R^2 + (Gm/R^2)(dM/dm) =0</font></p><p><font size="2">This leads to M = -(m)(dM/dm) </font></p><p><font size="2">rewitten as dM/M = -dm/m&nbsp; [this relation is also interesting]<br /></font></p><p><font size="2">By integrating,&nbsp; logM = - logm + C (integration constant, let's write it as logC)</font></p><p><font size="2">log(Mm)=logC</font></p><p><font size="2">M = C/m &nbsp;</font></p><p><font size="2">&nbsp;There's nothing new here except C.&nbsp; If m decreases, M increases as your thought experiment started with.&nbsp;</font></p><p><font size="2">I haven't used calculus for ages, correct me if I have made any mistake. </font></p><p>&nbsp;</p> <div class="Discussion_UserSignature"> <font size="2" color="#ff0000"><strong>Earth is Boring</strong></font> </div>

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#### Mee_n_Mac

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Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Hhhmmm, interesting thought. But if m=K/2, then M is also K/2, meaning equal mass for both objects. <u>Why would two equal masses create zero force at any point between them</u>? Posted by <strong>emperor_of_localgroup</strong></DIV><br /><br />Answering <u>the above</u>, they don't.&nbsp; The slope, dF/dm, when = 0, means an inflection point (assuming the slope was non-zero to start with) and so you have a local maximum or minimum. <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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#### DrRocket

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>I think he had it right, given his stipulation that the total mass of the 2 body system remain a constant.&nbsp; I understood that constant to be K not M.&nbsp; M or m were to be the masses of the Earth and Moon or vice versa.&nbsp; He finds an inflection point where M=m and claims it to be a maxima because the 2 endpoints, m = 0 and m = K, yeild a zero result.&nbsp; The claim is true though not proven.&nbsp; Basically what's the maximum product of 2 positive numbers (fractions), that sum to one.&nbsp; What all that has to do with the rest of his post remains unknown to me.&nbsp; I suspect if we could modulate gravity short of redistributing mass we could do a lot of things.&nbsp; Maybe we need to ask Obama re: the redistributing scheme ....ps - And given the effect we call gravity is the result of a geometry and not a true force, I'll need a bit more of an explanation on the later part of the OP. <br />Posted by Mee_n_Mac</DIV></p><p>You are right.&nbsp; I missed that constraint.&nbsp; I blew it.</p><p>But what is being maximized is the gravitational force being felt by one body as a result of the other at the center of mass of the first body.&nbsp; If you look at the gravitational field at a gread distance from the two bodies, far enough that the distance between them is negligible in comparison and the two bodies are approximable by a point, then there is no effect of the ratio m/M of the gravitational force felt by a third body.&nbsp; So I don't see where this is going.</p><p>&nbsp;</p> <div class="Discussion_UserSignature"> </div>

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#### emperor_of_localgroup

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<p><BR/>Replying to:<BR/><DIV CLASS='Discussion_PostQuote'>Answering the above, they don't.&nbsp; The slope, dF/dm, when = 0, means an inflection point (assuming the slope was non-zero to start with) and so you have a local maximum or minimum. <br /> Posted by Mee_n_Mac</DIV></p><p><font size="2">You are right. It's the maxima thing between two masses with zero slope. If we do a 3D plot with M along x, m along y, and F along z, we'll get a peak point for F. And this peak will occur when m=M=K/2. Question is, SO? </font></p> <div class="Discussion_UserSignature"> <font size="2" color="#ff0000"><strong>Earth is Boring</strong></font> </div>

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