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Gravity - Why do we Use Radius of Earth in Calculating Surface Gravity

  1. Dec 31, 2013 #1
    Newton showed that F=G(m1m2)/r2

    When calculating the force of gravity on the surface of the Earth, why do we use the radius of the Earth for our r value? Of course, the above Newtonian equation suggests we should, but it seems rather counter-intuitive.

    I must admit to not having taken a relevant course since an Honors Physics course in high school. If we were ever given a logical explanation (with minimal calculus required) for this, I cannot recall it. I have also posed this question to a friend of mine who, while not a physics major, attended a small, but well respected, science and engineering school in Southern California. He could not recall an answer to this question, though he did remember learning about Shell Theorem. Despite my limited understanding of Calculus, Shell Theorem does make some degree of logical sense.

    In any case, my question is this: Since the mass directly under my feet contributes significantly more gravitational force than the equally dense mass on the opposite side of the Earth (in a spherically symmetrical Earth where density only varies by depth), shouldn't the center of gravity be somewhat closer to me than the center of the Earth?

    If I were to compose a fictional earth consisting of tens of thousands of spheres, of say 100km radius, would I not find that the sphere directly below my feet with radius=100km contributes many magnitudes more force than the sphere opposite it with radius= approx 12,500km? By my calculations the nearer 100km radius sphere would exert over 15,000 times the force of the sphere on the opposite side of Earth.

    I have been unable to find a satisfactory answer to this quandary (based on my limited knowledge of both physics and calculus).

    I have tried a variety of sources, but keep getting answers similar to this one from, groan, Wikipedia:

    "If the bodies in question have spatial extent (rather than being theoretical point masses), then the gravitational force between them is calculated by summing the contributions of the notional point masses which constitute the bodies. In the limit, as the component point masses become 'infinitely small', this entails integrating the force (in vector form, see below) over the extents of the two bodies.
    In this way it can be shown that an object with a spherically-symmetric distribution of mass exerts the same gravitational attraction on external bodies as if all the object's mass were concentrated at a point at its centre.[3] (This is not generally true for non-spherically-symmetrical bodies.)"


    It seems logical that if the two masses are sufficiently far apart, r would approach the distance from their centers, and yet on the surface of the Earth where the distance is relatively short, my brain cannot comprehend how this is so.

    Is someone able to explain the above in a manner that makes logical sense to someone with my limited math and science skills and is consistent with Newton's Law of Gravitation? How would you demonstrate this to an intelligent Undergraduate who is unable or unwilling to do the vector math that appears to be required.

    Thanks and I have already learned a great many things from the superb minds on this forum.
     
  2. jcsd
  3. Dec 31, 2013 #2

    mfb

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    You are right with the observation that this is not trivial. It can be calculated, however, that the formula gives the result if (and only if...) you use the radius of earth as value. To do that, you have to integrate over the whole mass of earth.
    This is true for all objects with a spherical symmetry and forces that scale with 1/r^2 (for point-masses).

    Per mass, it contributes more, but there is also more mass far away than mass directly below your feet.

    There are many spheres more than 10000km away from you, and just one close to your feet.
     
  4. Dec 31, 2013 #3

    PeroK

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    It's a good question. The answer lies in calculus. If you calculate the combined force of gravity of a sphere, then it's the same as though all the mass were at its centre. Calculations that show this can be found here:

    http://hyperphysics.phy-astr.gsu.edu/hbase/mechanics/sphshell.html

    Personally, I think this is a good example of the power of mathematics. You can calculate a result that it difficult or impossible to imagine intuitively without the maths.
     
  5. Dec 31, 2013 #4

    PeroK

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    First, if the Earth were a cylinder, with you at one end, this would be true - for the reasons you've worked out. The gravity of the cylinder would not average out at the half-way point of the cylinder. But, somewhere closer to you.

    But, the Earth is spherical: so most of the Earth is more than the radius away, because it's off to one side. If you take a distance of r to the centre of the Earth and map out the chunk of the Earth that is less than r away, then that will only be about 1/4 of the mass, roughly. Maybe less. Can you imagine the shape of that portion of the Earth that is within its radius?

    So, most of the mass is more than r away. When you do the maths, it all works out that the force averages out at r.

    Does that make sense?
     
  6. Dec 31, 2013 #5
    Thanks for your response. I certainly agree that, in my example using the 100km sphere, that this sphere represents only a fraction of the Earth's mass. Maybe a better example would be to envision standing on the North Pole. Assuming a perfect sphere, there would be equal matter above the equator and below it. Surely the matter above the equator, and thus closer to my position, contributes a greater gravitational pull on me than the matter below the equator, simply due to its proximity. I may be misunderstanding your explanation (or, equally likely, I failed to pose my question appropriately).
     
  7. Dec 31, 2013 #6
    Aha! Now I am getting closer to understanding, though I hesitate to say I am thoroughly convinced. Are we sure that 1/4 or less of a sphere's mass is within 1 radius of a position on its surface? Or is this purely due to Earth's increasing density with depth?

    Please see my embarrassingly crude drawing:

    http://imageshack.com/a/img600/928/md07.png [Broken]
     
    Last edited by a moderator: May 6, 2017
  8. Dec 31, 2013 #7
    The more I think about my crude drawing above, maybe my mistake is simplifying the Earth to 2d instead of 3d? Is this my error?
     
    Last edited: Dec 31, 2013
  9. Dec 31, 2013 #8

    PeroK

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    Your diagram is good and helps explain it. More of the Earth's mass is further than r away.

    I only guessed at 1/4 of the mass being within r. You could work it out with a bit of calculus though.
     
  10. Dec 31, 2013 #9

    phyzguy

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    In your example where you are standing at the north pole, it's true that the mass above the equator exerts a significantly greater pull on you than the mass below the equator. The point is that when you add up the pulls from the mass above and below the equator, the resulting pull is the same as if all of the mass were located at the center. If you don't believe it, you will simply have to learn the calculus necessary to do the calculations.
     
  11. Dec 31, 2013 #10
    Do I believe it? I should. It's Newton after all right? Do I understand it? Not exactly. I feared when I wrote this I would have to take a couple courses in calculus to fully understand. Thanks to peroK for the link to these calculations. Alas, the math is far above me and I may have to resign myself to not understanding (or going back to school :cry:)

    Yet I guess I am holding on to hope, no matter how unfounded, that there is some logical explanation that can get through my evidently very thick skull and which relies on some mathematical explanation short of calculus.

    How do high school physics teachers impress this sort of thing on their students?

    Let's say I am willing to accept a liberal approximation. Could I assume the Earth is made up of many smaller spheres (ignoring density variation for a moment)? If I assumed the Earth had a radius of 5 mini-Earths, couldn't I just calculate the force of all of these and sum them? Would I expect that the sum of these forces would be approximately equal to the force calculated using the radius of the larger sphere? And as I increased the number of mini-Earth's for which I did this I would get a better and better approximation? If so, does anyone have a link to a diagram or calculations of such?

    Thanks again. You have all been helpful in guiding my understanding and I suppose I am inching closer. Maybe this thread will ultimately assist someone else with a similar intellect to mine.
     
  12. Dec 31, 2013 #11

    phyzguy

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    This is exactly what calculus does. You split the Earth up into small volumes and calculate the force due to each of these small volumes and add them all up. Calculus allows you to calculate this sum in the limit of having an infinite number of these volumes, each one of which is vanishingly small.
     
  13. Dec 31, 2013 #12

    sophiecentaur

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    I wonder why you are finding the need for such a do-it-yourself method for working this out. As soon as you get to learn about Integral Calculus you will see how these problems are approached and you will learn the 'rules'. If you feel the motivation to get this sorted out then you really should look into Calculus first. I think you'd have to be quite a superbrain to get this stuff right from scratch, all by yourself.
     
  14. Dec 31, 2013 #13
    Fair enough. I think I may just have to push this problem out of my head. I can multiply 4 and 5 digit numbers quickly in my head, but for some reason have been unable to grasp Calculus. If I had had a Calculus teacher like phyzguy, who explained Calculus through tangible real-world examples, I might have continued to study math and science, instead of Economics. Oh well, the world can never have enough economists right? :wink:
     
  15. Dec 31, 2013 #14

    sophiecentaur

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    Doesn't Economics use mathematical models, involving Calculus?
     
  16. Dec 31, 2013 #15
    Lots of economic modeling does require Calculus of varying degrees. Some of these, like calculating slopes, I am able to do without problems. But what does it say about our educational system when I was able to get a BS in Economics from one of the top 20 public universities in the US, without being properly taught the key concepts in Calculus? Well I am sure this board doesn't need a discussion on my perception or our country's educational failures.

    Thanks again to all those that gave their input. I continue to find fascinating threads as I browse this forum. I am currently reading Brian Greene's latest book so you can probably count on seeing another thread or two from me (despite his unique ability to explain complex concepts in a way even I can understand).
     
  17. Dec 31, 2013 #16

    sophiecentaur

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    I think everyone has a bit of guilt about the dreaded Integral Calculus. There is no end to how hard it can get. It is worth while getting into it to some level and just 'believe' what the man says when it gets too hard. And there's always Mathematica when it gets too difficult.
     
  18. Dec 31, 2013 #17
    Gnaman,
    I can see where you would have a difficulty in all of this.

    For bodies far from each other one can consider the bodies to be point-like.

    For one body much larger than the other, one can consider the larger to have a gravitational field of attraction. Arranging F/m = GM/r^2, and then stating that g = F/m, one obtains the acceleration due to gravity of a small object towards the larger.

    For extended bodies, such as real planets and stars, this works some of the time but not all of the time depending upon the level of precision ones needs for a calculation.

    Take two bodies the same size and mass ( two earths if you like ) seperated by a distance d. An outside observer would observe the two bodies attracting each other to a location halfway between the two, known as the centre of mass, COM, of the two body system, with a force on each calculated by the Universal Law of Gravitation.

    Similarily, you, standing on the earth, and the earth, also have a COM, but since the earth is so much larger than you, the displacement of the COM from the geometrical centre of the earth is negligable.

    For extended bodies there exists something called tides that involve the parts of the objects nearer each other having a greater attraction to each other than the parts of the objects farther away from each other. You should be familiar with the visible daily fall and rise of the oceans and lakes caused by the moon from this effect. Satellites ( not being point like objects but having some dimension ) orbiting the earth also suffer from this effect, and calculations of their orbits needs to consider this factor.

    You mentioned a model of the earth as being made up of balls and performing the calculation as the balls are made smaller and smaller. Knowledge of calculus is not needed if you do this iterative process and one could set it up on something such as Excel. All calculus does is do the whole iteration in one step.

    You seem to have lightly tredded upon some of these issues.
     
  19. Jan 1, 2014 #18

    mfb

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    It is important to model the earth as sphere, right - for a disk, the result is different.
    Only 31% of the volume is closer than R, and a significant fraction of this is "sidewards" - so its contribution to the force downwards is smaller than that of mass at the center.

    The mass distribution of the earth is not homogeneous, but that does not matter - it has a spherical symmetry (to a very good approximation), and you can consider every spherical shell individually - with the same result, you can use the radius of earth for each shell, and therefore for earth as a whole.
     
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