For example, if a planet is considered as a point mass, then an object inside the planets' gravitational field experiences an equal attraction from all of the planets' mass.

However if the planet is not considered as a point mass, i.e is considered as having a diameter, then does this mean that an object inside the gravitational field experiences a greater attraction to some parts of the mass than others (more attraction to closer parts of the planet)?

Does this therefore mean an object experiences a different force of attraction depending on whether the planet is considered as a point or spherical mass, as my physics textbook does not make this clear?

  • 2
    $\begingroup$ You are correct, it does experience a greater force, gram for gram, from the parts of the planet that are closer but there are more parts that are farther away. If you do the integration you will find that the net result is the same as treating the whole mass as being a point mass at the centre, $\endgroup$ – M. Enns Jul 27 '16 at 15:48
  • $\begingroup$ Ah ok then, so that's why we don't really worry about distinguishing the two in cases like this. Thanks a lot :) $\endgroup$ – John Jul 27 '16 at 15:50
  • $\begingroup$ Ya, it's very convenient. $\endgroup$ – M. Enns Jul 27 '16 at 15:51
  • 3
    $\begingroup$ Note that this works only when the mass distribution is spherically symmetric. Our actual planet is not, and NASA and others have mapped out the variation of g. $\endgroup$ – Jon Custer Jul 27 '16 at 16:19
  • $\begingroup$ Related: physics.stackexchange.com/q/177462/2451 and links therein. $\endgroup$ – Qmechanic Jul 27 '16 at 19:56

The gravitational potential field can be found by a full volumetric integration from the overall volume, or planet, or whatever: $$ \Phi(\mathbf r) = -G\int_V\frac{\rho(\mathbf r')}{|\mathbf r - \mathbf r'|}dV $$

This just comes from having a distribution for $M$ in the potential formula: $$ \Phi = \frac{GM}{r} $$

Also, its easy to see that, if you are integrating over a ball, then the gravitational field reduces exactly the a point. This is actually the shell theorem. Once integration is completed, we can calculate the gravitational field: $\mathbf g = -\nabla\Phi$, and then calculate the gravitational force in a point object of mass $m$: $\mathbf F = m\mathbf g$.

Now, for a general shaped object, the gravitatonal field can be calculated by doing the well known multipole expansion. Basically, in the denominator of the integral, we have: $$ |\mathbf r - \mathbf r'|^{-1} = \left(r^2 + r'^2 - 2\mathbf r\cdot\mathbf r'\right)^{-1} = r^{-2}\left(1 + \frac{r^2}{r'^2} + \frac{2\mathbf r\cdot\mathbf r'}{r}\right)^{-1} $$

We can use the Taylor expansion around $x=0$ of the function $(1+x)^\alpha$. $$ \left(1 + x\right)^\alpha = 1 + \alpha x + \frac{1}{2!}\alpha\left(\alpha-1\right)x^2 + \frac{1}{3!}\alpha\left(\alpha-1\right)\left(\alpha-2\right)x^3 + \dots $$

Setting $\displaystyle x = \frac{r^2}{r'^2} + \frac{2\mathbf r\cdot\mathbf r'}{r}$ and $\alpha = -1$, we have until second order: $$ |\mathbf r - \mathbf r'|^{-1} \approx \frac{1}{r}\left(1 - \frac{r^2}{r'^2} + \frac{2\mathbf r\cdot\mathbf r'}{r} + \dots\right) $$

You can continue to expand in as many orders you like. Substituting in the potential, we shall have: $$ \Phi(\mathbf r) = -\frac{G}{r}\int_V\rho(\mathbf r')dV + \frac{G}{r^2}\mathbf r\cdot\int_V\mathbf r' \rho(\mathbf r')dV + \dots $$

We then define: $$ M = \int_V\rho(\mathbf r')dV,\quad\quad \mathbf p = \int_V\mathbf r' \rho(\mathbf r')dV,\quad\quad Q_{ij} = \int_V\rho(\mathbf r')\left(r_i' r_j' - r'\delta_{ij}\right)dV $$

Where $M$ is the gravitational monopole moment (the usual pontual mass..), and $\mathbf p$ is the gravitational dipole moment, and $Q_{ij}$ is the quadrupole tensor moment. Notation $Q_{ij}$ means that $Q$ is a matrix, and the $ij$ component of the matrix is $Q_{ij}$. Also, $\delta_{ij}$ is the components of the identity matrix. That is $\delta_{ij} = 1$ if $i = j$, and zero otherwise. Notice all of this is not complicated. I just made a taylor expansion and inserted it in the potential.

Field of $M$ falls in $r^{-1}$, and field for $\mathbf p$ falls in $r^{-2}$. The term where field falls in $r^{-3}$ is called gravitational quadrupole term. And so on. As you can see, for a object near enough a non-spherical object, it will experience the monopole field (the usual $GM/r$ field), plus, will also experience the contribution of the dipole field. Higher order contributions disappear faster, thus are hardly felt by faar objects.

If you take a closer look, you will identify the gravitational dipole moment to simply be $M\mathbf r_{cm}$. Thus, directly related to the center of mass. Therefore, can one chose a the frame of reference exactly in the center of mass, because then the dipole moment will vanish, and thus the next contribution in the expansion will be the quadrupole term.

In case of Earth.. It spins. Thus, it is not fully spherical: earth is a little bit oblate. Taking the rotational axis to be the $z$ axis, such that $\theta$ is the azimuthal angle, we can calculate the potential in spherical coordinates: $$ \Phi(r) = -\frac{GM}{r} - \frac{GQ}{4r^3}\left(3\cos^2\theta - 1\right) $$

Where $Q$ is the (scalar) gravitational quadrupole moment. Since Earth is oblate, then $Q < 0$. Also, satellites near Earth will experience the quadrupole term, and they have to take it into account. The quadrupole term (in earth case) is responsible for precessing the near satellites around the polar axis.

Thus, in satellite orbiting Earth, when it is close enoguh, the quadrupole contribution will cause the satellite:

  • Precession. There will be a force pushing satellite to the equator, away from the poles. Ie, a force in azimuthal direction.
  • The greatest field in azimuthal direction will appear in $45$ degree angle of inclination of orbit.
  • At the equator and at the poles, the field in azimuthal direction is zero.
  • The field pushing satellite to earth (the radial field) is higher in the equator.
  • The field pushing satellite to earth (the radial field) is lower in the poles.
  • $\begingroup$ I might be worth mentioning that the integral for a bal is also called the shell theorem. $\endgroup$ – fibonatic Jul 27 '16 at 16:43
  • 1
    $\begingroup$ Thank you for the extensive answer! Although this is quite a bit above my level for now, I'll definitely look at this in the future. $\endgroup$ – John Jul 27 '16 at 16:49
  • $\begingroup$ @John Its really not that complicated. Probably the hardest part here is when the taylor expansion was done and inserted in the potential. Well.. you welcome. =). $\endgroup$ – Physicist137 Jul 27 '16 at 17:08
  • $\begingroup$ We haven't covered gravitational monopoles/dipoles or even what the upside down delta is haha. I don't think that's taught until university anyway in the UK. $\endgroup$ – John Jul 27 '16 at 17:16
  • 1
    $\begingroup$ @ja72 Well.. there is poisson equation...: $\nabla^2\Phi = 4\pi G\rho$. $\endgroup$ – Physicist137 Jul 27 '16 at 18:47

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.