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Relationship between gravitational field strength and potential

  1. May 31, 2014 #1
    Hi, I am a bit confused with the relationship between gravitational field strength and gravitational potential. As far as I know, gravitational field strength is defined as:

    g=[itex]\frac{F}{m}[/itex]=[itex]\frac{GM}{R^{2}}[/itex]

    and gravitational potential is defined as:

    V=[itex]\frac{-GM}{R}[/itex]

    Now if I differentiate V with respect to R, I get:

    [itex]\frac{dV}{dR}[/itex]=[itex]\frac{GM}{R^{2}}[/itex]=g

    However, the formula booklet that I'm using at school suggests that

    [itex]\frac{-ΔV}{ΔR}[/itex]=g

    Why doesn't my derivation above agree with the formula in the formula booklet? (i.e. why are the signs different?) A friend of mine suggests that I cannot use differentiation to replace the symbol Δ, but I see no reason why it is so.
    Thank you very much!
     
  2. jcsd
  3. May 31, 2014 #2

    vanhees71

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    2016 Award

    I don't know, at which level you are studying Newtonian gravity, but it is very important in any case to have the right concepts about the different quantities in question.

    First of all the gravitational force is a very fundamental concept, which cannot be derived somehow from simpler principles. It was Newton's ingenious insight to conclude from the empirical knowledge about celestial mechanics, i.e., the orbits of the planets around the Sun that there is a fundamental force acting between two point masses which is inversely proportional to the square of the distance of the bodies and their masses. It is always directed along the straight line connecting the two bodies and is always attractive. It is very important to keep in mind that a force is always a vector quantity.

    For the gravitational force on a mass [itex]m_1[/itex] exerted by another mass [itex]m_2[/itex], located at the origin of your reference frame you obtain
    [tex]\vec{F}_1=-\gamma \frac{m_1 m_2}{r^2} \frac{\vec{r}}{r},[/tex]
    where [itex]\vec{r}[/itex] is the position of body 1.

    Further the gravitational force from several bodies on body 1 is additive, i.e., the superposition principle holds. Thus for a point particle with mass [itex]m_1[/itex] the gravitational force due to a mass distribution [itex]\rho(\vec{x})[/itex] (where [itex]\rho[/itex] is the mass density) you get
    [tex]\vec{F}_1(\vec{r})=-\gamma m_1 \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{\rho(\vec{x})(\vec{r}-\vec{x})}{|\vec{r}-\vec{x}|^3}.[/tex]
    The force is always proportional to the mass of the "test body" 1. Thus you can introduce the gravitational field a
    [tex]\vec{G}(\vec{r})=\frac{\vec{F}_1}{m_1},[/tex]
    which is independent from the mass of the test body.

    It is also immediately clear that the gravitational field is conservative, i.e., there exists a scalar potential for it such that the gravitational field is the negative gradient of this scalar potential:
    [tex]\vec{G}(\vec{r})=-\vec{\nabla} V(\vec{r}).[/tex]
    For a single point mass in the origin of the coordinate system you have
    [tex]\vec{G}(\vec{r})=-\gamma m_2 \frac{\vec{r}}{|\vec{r}|^3}.[/tex]
    It is easy to see that
    [tex]V(\vec{r})=V(r)=-\frac{\gamma m_2}{r}, \quad r=|\vec{r}|.[/tex]
    That's easily seen by taking the derivatives carefully. Since the potential only depends on the distance we have
    [tex]\vec{\nabla} V(r)=V'(r) \vec{\nabla} r.[/tex]
    Now
    [tex]r=\sqrt{x^2+y^2+z^2} \; \Rightarrow \; \frac{\partial r}{\partial x}=\frac{x}{\sqrt{x^2+y^2+z^2}}=\frac{x}{r}[/tex]
    or analogously for the other two components
    [tex]\vec{\nabla} r=\frac{\vec{r}}{r}.[/tex]
    Further we have
    [tex]V'(r)=+\frac{\gamma m_2}{r^2}.[/tex]
    This finally leads to
    [tex]\vec{G}=-\vec{\nabla} V(r)=-\frac{\gamma m_2}{r^2} \frac{\vec{r}}{r}=-\frac{\gamma m_2 \vec{r}}{r^3}.[/tex]
    With the more general case of a mass distribution you can use the superposition principle also for the potential. This leads to
    [tex]V(\vec{r})=-\gamma \int_{\mathbb{R}^3} \mathrm{d}^3 \vec{x} \frac{\rho(\vec{x})}{|\vec{r}-\vec{x}|}.[/tex]
    You can as well bring this law to a local form by noting that this is the solution of the Poisson equation,
    [tex]\Delta V(\vec{r})=4 \pi \gamma \rho(\vec{r}).[/tex]
    You can solve it by finding the Green's function of the Laplace operator.

    This whole analysis ist prefectly analogous to the treatment of electrostatics. So you find most of the math about this subject well explained in textbooks on electromagnetism, e.g., Griffiths or Jackson.
     
  4. May 31, 2014 #3

    Nugatory

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    Staff: Mentor

    You've already lost a minus sign. Force is a vector, and it points in the direction of decreasing R, so [itex]F=\frac{-GmM}{R^{2}}[/itex]
     
    Last edited: May 31, 2014
  5. May 31, 2014 #4
    That's because the formula you quoted for the field [itex]g = \frac{F}{m}[/itex]=[itex]\frac{GM}{R^{2}}[/itex] is just the formula for the magnitude of the field. Note that the field is a vector and the formula gives us a scalar. the full formula is [tex]g = - \frac{GM\vec R}{R^{3}}[/tex]
     
  6. May 31, 2014 #5
    Thank you very much for your reply, vanhees71. However, I am only a high school student, thus can only understand a bit of basic calculus and a bit of Newton's law of gravitation. But I will surely take a look at your post again when I start learning those advanced mathematics :)

    I actually had the same feeling that it should be something like this. Like you said, the force is a vector and points towards the big mass, thus it must be negative; and since g=[itex]\frac{F}{m}[/itex], the gravitational field must be negative as well (as it also points towards the big mass). However, if the force is indeed negative, then shouldn't the definition of the gravitational potential energy be:

    W = [itex]∫^{R}_{∞}[/itex] F dr = [itex]∫^{R}_{∞}[/itex][itex]\frac{-GMm}{r^{2}}[/itex]dr = [itex]\frac{GMm}{R}[/itex]

    instead of this definition:

    W = [itex]∫^{R}_{∞}[/itex][itex]\frac{GMm}{r^{2}}[/itex]dr = [itex]\frac{-GMm}{R}[/itex] that we always use?
     
  7. May 31, 2014 #6

    jtbell

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    Staff: Mentor

    In general, the definition of potential energy W at point P due to a conservative force F is:
    $$W = - \int^P_{P_0} {\vec F \cdot d\vec r}$$
    Point P0 is the "reference point" where W equals 0 by definition. Note the minus sign!

    In this case, ##\vec F = - \frac{GMm}{r^2} \hat r## and ##d\vec r = dr \hat r##, so
    $$W = \int^P_{P_0} {\frac{GMm}{r^2}dr}$$
    The reference point P0 is at r = ∞, so if point P is at r = R, then
    $$W = \int^R_{\infty} {\frac{GMm}{r^2}dr} = - \frac {GMm}{R}$$
     
  8. Jun 1, 2014 #7
    Yes, this was really my source of confusion. It seems I forgot that the definition of potential energy created by a conservative force needs to be changed as well (i.e. a negative sign must be added); because otherwise, the resulting potential energy would be positive, which could not be right. So thank you very much, jtbell :)
     
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