In the context of this:

Our goal in this subsection will be to obtain $r$ as a function of $θ$, for a gravitational potential. The gravitational potential energy between two objects, of masses $M$ and $m$. In the present treatment, let us consider the mass M to be bolted down at the origin of our coordinate system. This is approximately true in the case where $M\gg m$, as in the earth-sun system.

I have seen this symbol ($\gg$ or $\ll$) multiple times in physics problems, but I have no clue what it means. I have heard that it's a limit statement of some sort, but I just want to double check how it works.

  • 3
    $\begingroup$ $\gg$ - Far greater than. $\endgroup$ – exp ikx Apr 19 at 1:50
  • 2
    $\begingroup$ FYI, you can typeset it using $\gg$ for $\gg$ and $\ll$ for $\ll$. And you can do e.g. $\ggg$ too with $\ggg$ $\endgroup$ – innisfree Apr 19 at 3:10

$`` >> "$ or $`` \gg "$ means "much greater than".

$`` << "$ or $`` \ll "$ means "much less than".

Typically, if $x \gg y ,$ then the following are considered to apply:$$ \begin{alignat}{7} y & \ll x \tag{1} \\[10px] x + y & \approx x \tag{2} \\[10px] x - y & \approx x \tag{3} \\[10px] \frac{y}{x} & \approx 0 \tag{4} \\[10px] \left|\frac{x}{y}\right| & \approx \infty \tag{5} \end{alignat} $$It's a bit of a fuzzy, approximate logic, so it should be used with care. Still, it's often convenient to say that one value is so much larger/smaller than another that we can simplify functions involving them without knowing their exact values.

Some care should be taken with signs. For example, it's technically true that $-{10}^{10} \ll 1 \,,$ though $\left| -{10}^{10} \right| \gg \left|1\right| \,,$ which might cause some confusion. In the case of mass, as given in the question statement, this is dodged as mass is always positive.


It means "much greater than". $M \gg m$ means that $M$ is so large compared to $m$ that $M$ is essentially fixed and any backreaction of $m$'s motion on $M$'s motion is neglected. This is, of course, an approximation. Making (valid) approximations is crucial in physics because it is extremely hard to find exact solutions to most problems. Often, numerical simulations are performed to get closer to the exact physical phenomenon.

We don't even have an exact solution for the 2-body problem in GR. Approximations are exceptionally important for operational reasons.


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.