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Proof of newtons gravitational law?

  1. Dec 5, 2007 #1
    proof of newtons gravitational law!?

    hi every one,
    i am new in this forum n here is my first question that really bothers me for a long time we always take for granted the formula given by newton F=G*(M1*M2)/R^2 do u have any proof that is really satisfing.n i dont understand why earth is on its orbit if moon exerts on it a great deal of accelaration as every partical do.i know i am missing some little part but i cant figure it out.besides according to newton, every partical exerts a force on each other so how newton compute every forces on universe as he calculate the net force on a body? i need immediate help for this!
  2. jcsd
  3. Dec 5, 2007 #2


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    Who is the "we" that takes it for granted?

    Newton's law is empirical, we use it because it works. If it did not work it would not be used, what more proof do you want?
  4. Dec 5, 2007 #3


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    Again, who is the "we" who ares taking this for granted? The reason it's called "Newton's law" is that Newton proved, about 400 years ago, that, in order for the planets to obey Kepler's laws, the force causing them to accelerate and deccelerate along their orbits must be of that form. The proof is not difficult and many Calculus texts include a section on it.
    Because the mass of the sun is so much greater than that of the planets one can pretty much ignore the effect of the other planets in calculating orbits (except for planets that come reasonably near Jupiter. In that case, corrections using Newton's law of gravitation can be made.) Because of the "r^2" in the denominator, it is not necessary to account for "every forces on universe"- the stars, for example, are much too far away to affect planetary orbits.
  5. Dec 5, 2007 #4


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    If you would like your own proof, it is not a difficult thing to prove. You can calculate, for example, your weight on the moon given the moon's size and mass, then compare that weight to what people say the weight difference is (you weigh 1/6 as much on the moon).

    Now obviously you can't go to the moon and carry a scale with you, but you can assume that since we successfully landed a number of spacecraft on the moon that we knew how big the engines needed to be to stop the spacecraft from crashing into it.
  6. Dec 6, 2007 #5
    how can u state that we can neglect them? to be precise we must put in account all of them are'nt we?besides my question was about the roots of the formula there he states that this force exists between every body n if we want to calculate net force on a body we must calculate all of them.n other thing that as he derive this formula he only uses earth and an apple i mean only two body. he dont put in account other bodies like sun or planets.

    i think to myself that every other mass forces in this universe exerting zero net force on apple the only force remains is earth's. is it some thing like that?:)
    Last edited: Dec 6, 2007
  7. Dec 6, 2007 #6
    No this is not so. The stars and other distant celestial bodies are so far away that the r^2 term in the equation becomes huge thereby making the force almost imperceptible.
    Why don't you try a rough calculation of how much acceleration a star say,100 light years away would cause to you, sitting on earth?
  8. Dec 6, 2007 #7


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    Calculate the gravitational force that the sun exerts on the earth. Then calculate the gravitational force that (for example) Mars exerts on the earth when it is at its closest distance. Compare the two.

    You are correct that in principle we have to include the gravitational effect of every object in the universe when calculating (for example) the Earth's motion. Nevertheless, in practice we get a good approximation to the correct result by including only the Sun, a better approximation by including the Moon also, and a still better approximation by including some of the other planets (at least Jupiter and perhaps also Mars and Venus). For very precise calculations one does in fact include these effects.

    Most practical calculations in physics involve approximations and simplifying assumptions, which depend on how precise the result needs to be for the purpose at hand.
  9. Dec 6, 2007 #8
    In his formula, there's the term distance R between the two masses. We are so close to Earth compared to other celestial bodies so we can not feel them, so does the apple. But, satellites' trajectories are to be calculated with certain consideration of Jupiters gravitational force, for example.
  10. Dec 6, 2007 #9
    i see. here i want to say that to measure newtons gravitional constant G, we need kind of condition that only two body of intrest must be put in a space that completely isolated of other masses (by the way of gravitional effects of these masses) .only in this environment this formula can be experimented correctly so can we say that constant G is impossible to calculate.if it is not tell me how to approximate ?! n how newton come to make such an experiment?

    n according to your sayings can we say that earth is moving in its orbit in a determined interval of distance please say yes!:)
    please some body master of this issue help me!
  11. Dec 6, 2007 #10


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    And you've hit a nail on its head.

    G is perhaps that natural constant which is hardest to measure to extremely high precision.

    However, some headway can be made (i.e, its first decimals might be calculated correctly) by Cavendish-like experiments, and others.
  12. Dec 6, 2007 #11
    As Arildno mentioned the Cavendish experiment, you can see that the value of G can be calculated quite correctly. Anyway, all experiments will have certain uncertainty. Let say you measure the lengh of a rod and you can not have the absolute value, let alone other sophisticated ones such as G , speed of light, electron charge/weight etc..
  13. Dec 6, 2007 #12

    D H

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    Bingo. The http://physics.nist.gov/cgi-bin/cuu/Value?bg|search_for=gravitational+constant" (relative error =6.8e-10) or time (relative error ~ 1e-12, depending on which unit of time one is talking about).

    What's striking is that we know the product GM of some bodies such as the Sun and Earth to a very high precision. For example, we know http://ssd.jpl.nasa.gov/?constants" to more than ten significant digits (relative error=6e-11).
    Last edited by a moderator: Apr 23, 2017
  14. Dec 7, 2007 #13
    what i figure here that these fundamental formulas so far that physicist discovered are nothing but a rough mathematical interpretation of the observations they made.only scale for phisical good jop is if they r working according to actions we r observing.
  15. Dec 7, 2007 #14
    Find out formulae describing a phenomenon is one thing, understand what exactly the phenomenon is, is another. I am sure, by now no one can explain clearly what causes the gravity.
    By the way, do you have any formulae to present?
  16. Dec 7, 2007 #15
    (we need kind of condition that only two body of intrest must be put in a space that completely isolated of other masses (by the way of gravitional effects of these masses) .only in this environment this formula can be experimented correctly )

    but here u r missing a point , we need the conditions that i told to calculate G but we also need same conditions to derive such a formula like F=G*M1*M2/R^2, from this situation it is impossible to experiment n derive this formula.if it is not please somebody explain n help me i have a great headache because of this!;)
  17. Dec 7, 2007 #16

    D H

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    Okay, if you would quit writing in TXT-speech you might get more people involved in the conversation. This board has a lot of experts in this field who simply don't bother with the posters who write like you do.

    That said, would you translate what you said here into clear English?

  18. Dec 7, 2007 #17
    ok.sorry for my english if it is not so clear, D H. i am going to try to express my problem in a more clear way.
    in the quote above i understand that the condition, needed to acomplish this experiment, is impossible to get. so if newtons law (F=G*M1*M2/R^2) expressing the phenomenon in this impossible condition how can he be sure that in this special experiment conditions, force going to be proportional to the masses and reversely proportional to the distance square.

    i mean the mathematical expression in this formula is totally derived by assuming that only gravitational sourse is earth and he predicted this formula according to the motion of an apple or moon.i am saying that here he neglected all the other masses effecting these masses and derived this formula.so in a space that only these two masses of intrest exist,should not the mathematical expression change.i mean these proportionalities may vanish.namely structure of mathematical expression may decompose. if it will not. tell me why?
    Last edited: Dec 7, 2007
  19. Dec 7, 2007 #18
    I do not really get your point. pls check the Cavendish experiment. He measured the gravity between two lead balls. The two exert to each other in horizontal direction, so the earth's gravity can be no interference because it's perpendicular.
    How about other astronomical objects? They are all around but very far, and their gravity on the 2 lead balls are almost cancel each other. In addition, if they create a small force on ball #1, then the same force will exert on ball#2 : same direction and same magnitude . So the force between the two balls which needs to be measured, are not affected, at least for measurement accuracy. You should think about other errorneous sources such as the scales sensitivity, the convection of the air around and others...which sum up the measurement error.
  20. Dec 7, 2007 #19

    D H

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    In Newton's time, Kepler's laws were ad-hoc laws that explained the motion of the planets. Newton's laws of motion dictates that because the planets must be under the influence of some kind of force. To give it a name, let's call it solar gravity for now. Why call it "solar gravity"? The planets obviously orbit the Sun. The Sun must be doing something, i.e. exerting a force, on the planets to make this happen. What are some other characteristics of this "solar gravity"?

    The mass of a planet is not involved in Kepler's laws. The acceleration that "solar gravity" induces on a planet is solely a function of the distance between the planet and the sun. This means that the "solar gravity" force on a planet must be proportional to the mass of the planet.

    There is no reason to think space has a preferred direction or a preferred plane. If solar gravity also has no preferred direction or plane, then the influence of solar gravity must fall off with distance inversely proportional to the surface area of the sphere with radius equal to the distance in question. This means solar gravity is an inverse square law! So now we have, the solar gravitational force acting on a planet p as [itex]F_p \sim M_p/r_p^2[/itex].

    This expression and Newton's laws of motion generate Kepler's laws. Validation! Newton was able to take Kepler's ad-hoc laws to something much deeper. It's still ad-hoc, but much deeper. Newton wasn't done. The Moon orbits the Earth, and other planets were known to have satellites as well. The Moon's orbit about the Earth and the orbits of the moons of Jupiter and Saturn fit nicely into this scheme. However, the proportionality constant is different for the Moon's orbit about the Earth compared to the Earth's orbit around the Sun.

    It's not too much of a leap to say that the difference in proportionality constants results from the mass of the central body. After all, the force has to go both ways (Newton's third law). So now we have [itex]F_{1\to2} \sim M_1M_2/r_{1\to2}^2[/itex] as a much more universal gravitational relationship. This works not only for the planets about the suns but for various moons around different planets.

    The theory looks very nice and simple. Nice and simple theories often die horrible deaths when confronted with experimental evidence. Not this one. It wasn't until a nearly a couple of hundred years later that experimental evidence hinted something might be a bit wrong. in the precession of Mercury. Early last century, Einstein showed show how to resolve the problem with general relativity. For the most part, the difference between what Newton's law of gravity says a body will do and who general relativity says a body will do is very, very small.
  21. Dec 7, 2007 #20


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    An experiment who'se error margin is a tiny fraction of one percent is not "rough". With what we have, we can predict, to the second, the timing of celestial events htat will happen years in advance.
    Why don't you do what was suggested earlier and actually calculate the magnitude of that effect. First, though, come up with a criteria for what consitutes "rough". Is it 1%? .1%? Then calculate the difference in gravitational force between two lead balls a meter apart and see if it will give nou enough error to consider the experiment "rough".
    Last edited: Dec 7, 2007
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