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Difference between G and gravitational accleration

  1. Dec 20, 2013 #1
    Hello,

    F=G m1.m2/r^2

    whereas g=Gm/r^2

    Now, I found reading that, the Mass of Earth, radius from the center r and obviously the gravitational constant is always constant and hence is grouped under a single constant 'g'.

    Hence: F=mg, where g=GM/r^2

    Now, while calculating gravitation potential energy we do:

    U=mgr, so U=M(Gm/r^2)r , where r = is the distance and hence forth, do the calculation......


    Somewhere else is is written g=Gm/r^2 is 'gravitational acceleration'.

    Which one is the case?

    If you can please explain it.
     
  2. jcsd
  3. Dec 21, 2013 #2

    Simon Bridge

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    The "r"'s in the relations you list each mean different things.

    Define
    R: the radius of the Earth,
    r: distance from the center of the Earth
    h: distance from the surface of the Earth
    ... so that h=r-R

    Then, the acceleration due to gravity (by F=ma) is: $$a_g=\frac{GM}{r^2}$$
    It is usually easier to measure distances from the surface of the Earth ... so we can rewrite that: $$a_g=-\frac{GM}{(R+h)^2} =\frac{GM}{R^2(1+\frac{h}{R})^2}\\ \qquad =g(1+h/R)^{-2}$$ ... where ##g## is defined by: $$g=\frac{GM}{R^2}$$
    If ##h <\! < R##, then ##(1+h/R)\approx 1## making ##a_g\approx g##, so ##g## is the (gravitational) acceleration of objects close to the surface of the Earth ...

    For gravitational potential energy: $$U=-\frac{GMm}{r}=-\frac{GMm}{R+h} = -\frac{GMmR}{R^2(1+h/R)} = -\frac{mgR}{1+h/R}$$ ... so for h<<R, the gravitational potential energy for a mass ##m## is ##U=mgR##

    This is for ##U(r)=0## when ##r=\infty## - which is where the minus sign comes from.
    But the choice of ##U=0## is arbitrary - if we choose ##U=0## when ##r=R## then we can find an expression for ##U## in terms of ##h##. When you crunch those numbers you get: ##U(h)\approx mgh: h<\! <R## ... you should be able to show that now.
     
  4. Dec 21, 2013 #3
    Ok, let me approach it in this way:

    F=ma, then a=F/m. From this equation, a =GM/r^2. I might be asking you the wrong answer: Is it that F=GM?
     
  5. Dec 21, 2013 #4

    Simon Bridge

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    No. You are still not being careful about defining your variables.

    Close to the surface of the Earth, the force of gravity is F≈mg, where g=GM/R^2 ... the approximation is so good that for most purposes we can write F=mg.

    Because F=ma, as well, you can write (F=mg)=ma therefore you see that a=g.

    This applies only close to the surface of the Earth ... i.e. where ##|r-R| <\! < R##.

    If this is not the case then the acceleration will be some fraction of g.

    ---------------------
    I have been working in magnitudes here - force is actually a vector.
    Newton's law should be ##\sum \vec F = m\vec a##
    ... and the force of gravity close to the surface of the earth is ##\vec F_g=-mg\hat{k}##
    (where ##\hat{k}## is a unit vector pointing upwards).

    For an object moving under gravity alone - $$\sum \vec F = \vec F_g = m\vec{a}\\ \implies \vec{a}=-g\hat{k}$$ ... which may make more sense.
     
    Last edited: Dec 21, 2013
  6. Dec 21, 2013 #5
    Ok, let me try to put it correctly.

    F=mg. Putting it into the equation F=G m1m2/r^2, we get mg=Gm1m2/r^2, so g=Gm/r^2.

    Is that so that we get it?
     
  7. Dec 21, 2013 #6

    Simon Bridge

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    Nope - you don't seem to be paying attention: I have already answered this question.

    By definition: $$g=\frac{GM}{R^2}$$... where ##G## is the gravitational constant, ##R## is the mean radius of the earth at the equator, and ##M## is the mass of the Earth.

    In general $$\frac{GMm}{r^2}\neq mg$$... where ##r>R## is any distance from the Earth.

    The complete equation for gravity due to the Earth is:
    $$\vec F=\left\{ \begin{array}{rll}
    -\frac{GMmr}{R^3}\hat{r} & = -\frac{mg}{R}\vec{r} &: r<R\\
    -\frac{GMm}{R^2}\hat{r} & = -\frac{mg}{r}\vec r &: r=R\\
    -\frac{GMm}{r^2}\hat{r} & = -\frac{mgR^2}{r^3}\vec{r} &: r > R\end{array}\right.$$... where ##\hat{r}=\vec r/ r## is a unit vector pointing in the same direction as ##\vec{r}##.

    In terms of magnitudes - it's simpler:
    $$F=\left\{ \begin{array}{cl}\frac{mgr}{R} & :r<R\\ mg & :r=R\\ \frac{mgR^2}{r^2} & : r>R\end{array}\right.$$
    ... so now: what was your question?
     
    Last edited: Dec 21, 2013
  8. Dec 23, 2013 #7
    Nothing. Understood. Thank you very much. Got clear.
     
  9. Dec 23, 2013 #8
    I think you answered your own question. G is a constant of matter, sort of like the charge on an electron or similar. g is the acceleration produced on another object by a certain amount of matter m, at a distance r. In other words, g is (amount of mass × the G constant per unit mass) / distance2. G is a factor in g, but not vice versa. Look at:
    http://en.wikipedia.org/wiki/Gravitational_constant
    http://en.wikipedia.org/wiki/Gravity_of_Earth

    And as Simon said, be careful when reading gravitational force equations in random texts. Some of them are written in reference to "inertial space", some in reference to one of the bodies (either its center or its surface). It may not be obvious, until you have looked at a lot of these, which is which.
     
  10. Dec 23, 2013 #9
    Hello tfr000,

    Thank you very much for the response.

    The links are really helpful.
     
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