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Thread: Is There Any Bias for 'g' at Earth's Surface?

  1. #1 Is There Any Bias for 'g' at Earth's Surface? 
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    Newton claimed he had proved using calculus that objects at or near the surface of a gravitating body such as the Earth do not experience any bias in the gravitational pull resulting from the particles of mass directly beneath the observer’s feet being that much closer than the particles at the exact centre of the gravitating body.

    The equation GM/r² dictates that the gravitational pull is inversely proportional to the distance squared, so any particle of mass directly beneath the observer’s feet would have a disproportionally greater effect than the same sized particle at the exact centre of the Earth or on the other side of the planet. But Newton stated that calculus revealed that any effect such as this is completely cancelled out.

    However trying to picture this effect in my mind’s eye I can’t see how this effect could be cancelled out.

    It is experimentally provable that g = 9.8 m/sec² at the Earth’s surface. Also the equation GM/r² completely balances at the Earth’s surface and when the appropriate values and units are used, produces the figure of 9.8 m/sec² for g. However if any gravitational bias were to occur as a result of the observation outlined above, it ruins everything. The escape velocity would change, the orbits of Earth bound satellites would also change and it could even affect the orbits of the planets.

    Has anyone ever seen Newton's proof by calculus on this question and how does it work exactly? I have not been able to trace any mention of it on the internet to date.


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  3. #2 Re: Is There Any Bias for 'g' at Earth's Surface? 
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    Quote Originally Posted by galexander
    Newton claimed he had proved using calculus that objects at or near the surface of a gravitating body such as the Earth do not experience any bias in the gravitational pull resulting from the particles of mass directly beneath the observer’s feet being that much closer than the particles at the exact centre of the gravitating body..
    Never seen any such "proof," and it wouldn't stand up anyhow. The Earth's surface for example is made up of gravitational anomalies from crustal density differences.


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  4. #3 Re: Is There Any Bias for 'g' at Earth's Surface? 
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    Quote Originally Posted by galexander
    Newton claimed he had proved using calculus that objects at or near the surface of a gravitating body such as the Earth do not experience any bias in the gravitational pull resulting from the particles of mass directly beneath the observer’s feet being that much closer than the particles at the exact centre of the gravitating body.

    The equation GM/r² dictates that the gravitational pull is inversely proportional to the distance squared, so any particle of mass directly beneath the observer’s feet would have a disproportionally greater effect than the same sized particle at the exact centre of the Earth or on the other side of the planet. But Newton stated that calculus revealed that any effect such as this is completely cancelled out.

    However trying to picture this effect in my mind’s eye I can’t see how this effect could be cancelled out.

    It is experimentally provable that g = 9.8 m/sec² at the Earth’s surface. Also the equation GM/r² completely balances at the Earth’s surface and when the appropriate values and units are used, produces the figure of 9.8 m/sec² for g. However if any gravitational bias were to occur as a result of the observation outlined above, it ruins everything. The escape velocity would change, the orbits of Earth bound satellites would also change and it could even affect the orbits of the planets.

    Has anyone ever seen Newton's proof by calculus on this question and how does it work exactly? I have not been able to trace any mention of it on the internet to date.
    Newtonian gravity is completely described by the inverse square law, which applies to point bodies. Calculus is the tool by which that law is extended to larger bodies, The result is that for uniform spheres the gravitational force between a sphere of mass and a sphere of mass is where is the universal gravitational constant and is the distance between the centers of the spheres.

    The derivation is a straightforward integral involving vector calculus. You can probably find it either worked out or as an exercise in a text on advanced calculus.

    Now recall for a mass undergoing acceleration and compare with the gravitational force to conclude that the gravitational acceleration on from is just For a body at the surface of the Earth this turns out to be . No bias is involved, nor has any been measured in numerous experiments, although there are local variations in the acceleration due to variations in the density of the Earth. These variations, called harmonics, are considered in the guidance algorithms for ICBMs and other applications requiring extreme accuracy.

    Newton's theory is used routinely in the calculations of orbits and trajectories of spacecraft, and is extremely accurate. No bias is required for any "surface effect".
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  5. #4 Re: Is There Any Bias for 'g' at Earth's Surface? 
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    Quote Originally Posted by DrRocket
    Newton's theory is used routinely in the calculations of orbits and trajectories of spacecraft, and is extremely accurate. No bias is required for any "surface effect".
    Depends what you mean by "extremely accurate." Due to crustal density changes that 9.8 m / s^2 is +- ~0.02 at earth's mean radius. That of course gets added to difference in "r" because our planet isn't precisely spherical.

    Point being there are differences in gravity due to local differences in mass from what's under your feet, which is what he's asking about--they just aren't by much.
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  6. #5 Re: Is There Any Bias for 'g' at Earth's Surface? 
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    Quote Originally Posted by Lynx_Fox
    Quote Originally Posted by DrRocket
    Newton's theory is used routinely in the calculations of orbits and trajectories of spacecraft, and is extremely accurate. No bias is required for any "surface effect".
    Depends what you mean by "extremely accurate." Due to crustal density changes that 9.8 m / s^2 is +- ~0.02 at sea level. That of course gets added to difference in "r" because our planet isn't precisely spherical.

    Point being there are differences in gravity due to local differences from what's under your feet, which is what he's asking about--they just aren't by much.
    Which is what comes up in the gravitational harmonics, but is not a factor in spacecraft trajectory calculations (I don't consider RVs as spacecraft).[The harmonics between Vandenberg and Kwajalein are very well known.]
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  7. #6 Re: Is There Any Bias for 'g' at Earth's Surface? 
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    Quote Originally Posted by DrRocket
    Which is what comes up in the gravitational harmonics, but is not a factor in spacecraft trajectory calculations (I don't consider RVs as spacecraft).[The harmonics between Vandenberg and Kwajalein are very well known.]
    You're vastly oversimplifying the problem because they not only account for it when they need precision but have to include the moon, centripetal accelerations from the launch point etc. And I fear neither of us is really answering his question. You dismiss it entirely which I think doesn't address his question one whit while you probably think I introduce unnecessary complications.

    "The equation GM/r² dictates that the gravitational pull is inversely proportional to the distance squared, so any particle of mass directly beneath the observer’s feet would have a disproportionally greater effect:"
    The answer is it does, but the relative % of mass in close proximity under his feet is so relatively small compared to the entire mass of the planet that it's gravitation effect is a tiny amount. Even crustal anomalies account for less than a 0.3% change of gravity at equal radius from the planet's center.
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    "The equation GM/r² dictates that the gravitational pull is inversely proportional to the distance squared, so any particle of mass directly beneath the observer’s feet would have a disproportionally greater effect:"
    The answer is it does, but the relative % of mass in close proximity under his feet is so relatively small compared to the entire mass of the planet that it's gravitation effect is a tiny amount. Even crustal anomalies account for less than a 0.3% change of gravity at equal radius from the planet's center.
    As I understand it, that is taken account of in the calculation. The inverse is Newton's shell theorem, where there would be a zero nett gravitational effect on a body within a cavity in a uniform sphere, no matter how near the edge you come. It simply cancels out. With a body on the surface of a uniform sphere, the lines of force add together with a resultant direction of gravitational force originating at the dead centre of the sphere. Or so I think.
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  9. #8 Re: Is There Any Bias for 'g' at Earth's Surface? 
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    Quote Originally Posted by Lynx_Fox
    Quote Originally Posted by DrRocket
    Which is what comes up in the gravitational harmonics, but is not a factor in spacecraft trajectory calculations (I don't consider RVs as spacecraft).[The harmonics between Vandenberg and Kwajalein are very well known.]
    You're vastly oversimplifying the problem because they not only account for it when they need precision but have to include the moon, centripetal accelerations from the launch point etc. And I fear neither of us is really answering his question. You dismiss it entirely which I think doesn't address his question one whit while you probably think I introduce unnecessary complications.

    "The equation GM/r² dictates that the gravitational pull is inversely proportional to the distance squared, so any particle of mass directly beneath the observer’s feet would have a disproportionally greater effect:"
    The answer is it does, but the relative % of mass in close proximity under his feet is so relatively small compared to the entire mass of the planet that it's gravitation effect is a tiny amount. Even crustal anomalies account for less than a 0.3% change of gravity at equal radius from the planet's center.

    Let's try this again. The Newtonian equation is all that is needed for all mechanics involving gravity that does not require general relativity, period.

    If you want to handle missile guidance it is all that is needed given knowledge of the masses involved. The necessary knowledge involved for very accurate targeting of ICBMs requires knowing the deviation of the Earth from uniformity over the flight path. Things like "centrepetal force", "coriolis force" etc. are part of the general theory of mechanics but have nothing to do with gravity. All of that stuff is included in the guidance and control algorithms, along with some estimation algorithms designed to handle noise and uncertainties (see optimal control and Kalman filter). They are also included in artillery fire control calculations (less the control theory stuff except for guided rounds).

    is all that is needed for Newtonian gravity. It is extremely accurate and all that is used in practice in all missile guidance applications and the vast majority of work in astronomy, always to be understood in the context of classical mechanics, whether one uses Newton's formulation or the variational approach of Lagrange or Hamilton.

    There is no additional bias for "surface effects". It is all in . Remember that this is for point particles and is extended to other bodies through calculus.

    "Newton's all you need" -- The Beatles (or something like that)
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    Discounting local gravitational anomalies as well as the moon is why ballistic trajectories of 1960s missiles were later found to be hundreds of miles off their intended impact points. Lunar and local anomaly fields are now accounted for in all missile trajectory calculations of more than a couple hundred miles.
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  11. #10  
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    Quote Originally Posted by Lynx_Fox
    Discounting local gravitational anomalies as well as the moon is why ballistic trajectories of 1960s missiles were later found to be hundreds of miles off their intended impact points. Lunar and local anomaly fields are now accounted for in all missile trajectory calculations of more than a couple hundred miles.
    I have no specific knowledge why 60's missiles would have been off target by hundreds of miles, but it would take more than local gravitational anomalies. Despite the name, long range missiles are not really ballistic. They are guided throughout powered flight. Only after propulsion has ceased is the trajectory ballistic.

    Current MIRV'd systems are sufficiently accurate, and the gravitational harmonics well-enough understood that hitting targets in the lagoon at Kwaj is par.

    The time frame of the 60's was one of great progress in both rocketry and control theory. The earlier servomechanism theory of Weiner, Bode and Nyquist gave way to the modern state space and optimal control methods of Kalman. Lefshetz and Pontryagin. Electronics was developed to implement the theory. Thrust vector control methods for rockets -- liquid injection and swiveling nozzles -- were developed. I find those factors far more credible in avoiding errors of the order of hundreds of miles than any refinement in handling local gravitational anomalies.

    But local anomalies can be important, if a bit more subtle. If you recall the first moon landing, Armstrong shut down the autopilot and landed manually. He was forced to do that because an unanticipated local variation in the gravity of the Moon resulted in an instability in a Kalman filter in the guidance system of the lander.
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    Wow, this is good stuff.
    I have found this snippet which may be of some interest to the the OP. This is a link to the Goce satellite`s view of Earths gravity in high definition.
    http://news.bbc.co.uk/1/hi/sci/tech/8767763.stm
    .
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  13. #12  
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    Quote Originally Posted by DrRocket
    He was forced to do that because an unanticipated local variation in the gravity of the Moon resulted in an instability in a Kalman filter in the guidance system of the lander.
    Probably crap. He was forced to do so because the autopilot was heading them for a boulder field. If you have information to the contrary please provide a citation. Or were you trying to be humourous and commenting on the fact that the size distribution of lunar regolith is Gaussian?
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    I once tought of a theoretic experimental setup related to this. Just checking...

    It goes like this :

    Put four dominostones or whatever cubic things of same sizes same material on a row in line.
    Take the second stone from one end away.
    This leaves two stones A and B directly placed to each other and one stone C seperate.

    C experiences two forces from A and B if A and B are considered to be seperate stones.
    If A and B are considered to be one piece (melted/tipped together with the tip of a welder at one or two points if they where made of a material that can be welded) C would experience only one force coming from objekt (AB).

    Doing the math with Fz=M*G/r^2 results in the conclusion that Fz would change simply by melting the stones A and B together without any change of positions, distances or masses...

    If the gap is 1 and each stone has a width 1 and - supposed - a mass 1 :
    For this ab-c arrangement :

    Fz (a-c) => 1*1*G / 3^2 = 0,111 G (N)
    Fz (b-c) => 1*1*G /2^2 =0,25 G (N)
    Total Fz for C from one direction these add up : Fz= 0,3611 G (N) .................

    g=0,3611 m/s^2

    A and B as one objekt (literally same size as A and B together or or A and B considered) :
    Fz(ab-c) => 2* 1G/2,5^2.= 0,32 G (N)........

    g=0,32

    Results in different Fz,s and g for C.
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    Ah!
    I am a newbie student for science.
    And I just got a question after reading this post.

    What if the distance between the two bodies is 10^-99999999999999999 and their mass is very large?

    and what if the two bodies are just attached to each other so that there will be no distance between them.
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    For B and A gravitation has opposite direction in this situation. With no other influences implied A and B never come loose so the welding makes (or should make) no difference to the other situation. B is pulled towards C stronger then A but for A the gravitation from B adds up to the gravitation from C working in same direction.

    Therefor gravitation for A would still be bigger in the same direction as for B. A is supposed to follow B in whatever direction.....according to the gravitational formula...thus does....according to the same formula.

    With earth calculus can rule this difference out more or less or maybe even total for some topological reason. The half nearest would have a stronger force but there is a difference for convergency related to distance also.

    If each half would be thought of as consisting of dominostones or dices, the effective vertical pull from the half nearest by is decreased because of this stronger convergency due to shorter distance.

    Similar as when two persons pull a car forward with two ropes. The bigger the angle between the ropes the more the resultant in forward direction decreases. (res. Fz = Fz * cos alpha). That way the components perpendicular to the vertical direction rule each other out because of opposite direction.

    This happens also for this example...somewhat and maybe someone can calculate it to all fit...calculus can make miracles happen sometimes. In that case I have to change the domino stones for three lengths of copper wire all laying in one line. Or the other extreme : three huge thin plates with huge surface area for relatively short thickness and distances.
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    Quote Originally Posted by vilyanurchandra
    Ah!
    I am a newbie student for science.
    And I just got a question after reading this post.

    What if the distance between the two bodies is 10^-99999999999999999 and their mass is very large?

    and what if the two bodies are just attached to each other so that there will be no distance between them.
    I guess that would be a black hole.
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    Quote Originally Posted by Harold14370
    Quote Originally Posted by vilyanurchandra
    Ah!
    I am a newbie student for science.
    And I just got a question after reading this post.

    What if the distance between the two bodies is 10^-99999999999999999 and their mass is very large?

    and what if the two bodies are just attached to each other so that there will be no distance between them.
    I guess that would be a black hole.
    So that mean , Newton's law is applicable only in normal conditions and not in extreme conditions?
    If so , Then may be g=GmM/r^2 might not be applicable in the surface of sun?
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    Quote Originally Posted by vilyanurchandra
    So that mean , Newton's law is applicable only in normal conditions and not in extreme conditions?
    If so , Then may be g=GmM/r^2 might not be applicable in the surface of sun?
    Newton's law works just fine on the surface of the sun. The distance between the center of mass of the sun and an object on the surface of the sun is not zero. It's equal to the radius of the sun. The only way the distance between objects can be zero would be if they were compressed to a point and did not occupy any volume in space. That could only happen in a black hole.
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  20. #19  
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    Must be extremely hot to work there for Newton's law...

    I rephrased my former post but to add :

    It,s not possible to think of Fz as a pullforce from two sides that are ecqual according to action is reaction.

    In the example of the dominostones :

    If Fz Ab acting on C is the same as Fz C acting on B (as one force between them)
    g experienced by C : g=Fz/Mc (Mc being C,s own mass)

    For AB : g=Fz/Mab.

    As Mab is 2*Mc : g=Fz/2Mc.

    Hence g is twice as big for C then for AB so they can,t have the same accelleration to each other....maybe different accelleration to a point in the middle somewhere ?

    This is type of thinking how you would have to calculate for a torsionbalance to determin G there is a lot of similarity. Therefor it,s not just wild imagination with absurd examples or the torsionbalance is absurd also and G and Fz=M1M2G/r^2...it,s all absurd then.

    One g is one direction the other the other, the parts fall to each other.
    But if the pullforce would be symmetric to both sides g can,t be if only when they are added up : g ab+gc=g as for what we see as accelleration between the two.

    This conflikts with Fz=M1M2G/r^2 if it,s a symmetric pullforce. Each part still experiences a different acceleration to a point chosen somewhere in between.

    g is normally a lokal meassurement but G seems chosen for the point where they would collapse, sort of a mediation. It can,t be a universal value then but local...in this case ...somewhere in the middle ?

    For a torsionbalance that means not much difference in locality as it,s a small device but on a bigger scale this would be significant.

    G was an unknown for Newton (I don,t even know if he considered it to be a universal constant) but it still is. It,s a huge letter for a tiny number and no one really knows what it is or where it is coming from. There is a set of units put behind to fit Newtons formulations and the meaning comes from these formula,s then while at the same time these formula,s come from this constant.

    I,m not saying that these formula,s are bull or something or that there is no G. But is it really certain to be a universal constant without changing lokally.

    If it does and G lokally on earth is bigger (also becomes bigger with less hight) then there would be mass missed in the universe...then such science fiction ideas come up as "dark matter" to compensate.
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