Given is a mass $M$ at the origin of the coordinate system, and another mass $m$ which is locate at $\vec{x}$

I have to show that the force acting on $m$ is a gradient field.

My approach:

I define the vector field as:

$$\vec{F}(\vec{x}) = \frac{mMG}{|x|^2} \cdot \frac{\vec{x}}{|x|}$$

does it help if I parametrize $\vec{x}$ though spherical coordinates? In that case, how would I compute the "backwards" (integration) step to find the potential, $\vec{F}(\vec{x}) = - \vec{\nabla}\phi$ ?


$\require{cancel}$ Yes, you can certainly parametrize the vector ${\bf x}$ in spherical coordinates, in which case

$$ {\bf F} = \color{red}{-} GMm \frac{\hat{\bf r}}{r^2} \tag{1} $$

and the potential is

$$ \phi = -\frac{GMm}{r} \tag{2} $$

which certainly satisfies

\begin{eqnarray} -\nabla\phi &=& -\frac{\partial \phi}{\partial r}\hat{\mathbf r} - \frac{1}{r}\color{blue}{\cancelto{0}{\frac{\partial \phi}{\partial \theta}}}\hat{\boldsymbol \theta} - \frac{1}{r\sin\theta}\color{blue}{\cancelto{0}{\frac{\partial \phi}{\partial \varphi}}}\hat{\boldsymbol \varphi} \\ &\stackrel{(2)}{=}& -\frac{\partial}{\partial r}\left(-\frac{GMm}{r}\right)\hat{\mathbf r} \\ &=& -\frac{GMm}{r^2} \hat{\mathbf r} \\ &\stackrel{(1)}{=}& {\bf F} \end{eqnarray}

  • $\begingroup$ but is it necessary ? is it really that more complicated with normal cartesian coordinates? $\endgroup$
    – Leroy
    May 26 '18 at 20:55
  • $\begingroup$ @Leroy Spherical coordinates exploit the spherical symmetry of the problem, for example, in spherical coordinates it is very obvious that the angular momentum is a conserved quantity and, therefore, the orbit of the mass $m$ restricts to a plane. You can certainly arrive to the same conclusion using rectangular coordinates, but it is not as easy $\endgroup$
    – caverac
    May 26 '18 at 20:59
  • $\begingroup$ I understand, specifically integrating the components of the field on by one is particularly hideous, but could you show the steps you went though to find the potential ? $\endgroup$
    – Leroy
    May 26 '18 at 21:12
  • $\begingroup$ @Leroy Sure, just look at the expression -${\rm d \phi}/{\rm d} r = -GMm/r^2$, integrate once and you get $\phi(r) = -GMm/r$ $\endgroup$
    – caverac
    May 26 '18 at 21:23
  • 1
    $\begingroup$ @Leroy Correct! $\endgroup$
    – caverac
    May 26 '18 at 21:34

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