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Homework Help: Gravitational potential energy

  1. Sep 18, 2010 #1
    1. The problem statement, all variables and given/known data

    Derive the gravitational potential energy from Newton's law.

    2. Relevant equations

    [tex]
    \mathbf{F} = -G \frac{m_1m_2}{\left|\mathbf{r}\right|^3} \mathbf{r}
    [/tex]

    [tex]
    W = \int_A^B \mathbf{F} \cdot{} d\mathbf{r}
    [/tex]

    [tex]
    \Delta{}U = W
    [/tex]

    3. The attempt at a solution
    I can use the scalar version of the law, but I want to do it with the vector form.
    So, first some easy substitution:

    [tex]
    W = \int_A^B \mathbf{F} \cdot{} d\mathbf{r}
    = -Gm_1m_2 \int_A^B \frac{\mathbf{r}\cdot{}d\mathbf{r}}{\left|\mathbf{r}\right|^3}
    = Gm_1m_2 \int_A^B \frac{r_xdr_x + r_ydr_y + r_zdr_z}{\left|\mathbf{r}\right|^3}
    [/tex]

    [tex]
    = -Gm_1m_2 \left\{
    \int_A^B \frac{r_x}{\left|\mathbf{r}\right|^3} dr_x +
    \int_A^B \frac{r_y}{\left|\mathbf{r}\right|^3} dr_y +
    \int_A^B \frac{r_z}{\left|\mathbf{r}\right|^3} dr_z
    \right\}
    [/tex]

    Now take one integrand, for example the x part,

    [tex]
    \int_A^B \frac{r_x}{\sqrt{r_x^2 + r_y^2 + r_z^2}^3} dr_x =
    \int_A^B r_x \left( r_x^2 + r_y^2 + r_z^2 \right)^{-\frac{3}{2}} dr_x
    [/tex]

    and perform substitution:

    [tex]
    t = r_x^2 + r_y^2 + r_z^2
    [/tex]
    [tex]
    dt = 2r_x dr_x
    [/tex]
    [tex]
    \frac{dt}{2} = r_x dr_x
    [/tex]

    That yields:

    [tex]
    \frac{1}{2} \int_A^B t^{-\frac{3}{2}} dt_x =
    \left[ -t^{-\frac{1}{2}} \right]_A^B =
    \left[ -\frac{1}{\sqrt{r_x^2 + r_y^2 + r_z^2}} \right]_A^B =
    \left[ -\frac{1}{\left|\mathbf{r}\right|} \right]_A^B =
    -\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)
    [/tex]

    Doing the same for other two integrands and plunging them back:

    [tex]
    W = -Gm_1m_2 \left\{
    -\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)
    -\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)
    -\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)
    \right\}
    [/tex]

    And finally

    [tex]
    W = 3Gm_1m_2 \left(
    \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|}
    \right)
    [/tex]

    Now if we take B at infinity, the potential energy will be

    [tex]
    U = - 3G\frac{m_1m_2}{\left|\mathbf{r}\right|}
    [/tex]

    which is three times as large as it should be. Don't tell me, let me guess. There's an error somewhere up there. *sigh*
    Any help would be highly appreciated.
     
    Last edited: Sep 18, 2010
  2. jcsd
  3. Sep 18, 2010 #2
    This is where you were wrong:
    t is a multi-variable function, therefore: [tex]dt = 2r_xdr_x+2r_ydr_y+2r_zdr_z[/tex]

    A suggestion: [tex]2\vec{r}d\vec{r} = d(\vec{r}^2) = d(r^2)[/tex]
     
  4. Sep 18, 2010 #3
    Thanks, hikaru!
    I knew I should have read that book about multivariable calculus.
     
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