<p><strong>Is there a limit to how fast a pulsar or other large, massive, dense body can spin?</strong></p>
<p>The answer is yes! If it spins faster than a massless satellite in equatorial orbit at zero altitude, then it won't be stable. This is because any particle moving faster would migrate into a higher orbit. And how do we go about calculating the period of such a theoretical satellite? By applying Kepler's third law in this form: <span class="math-container">$M=A^3/P^2$</span>, where <span class="math-container">$M$</span> is the mass of the pulsar, <span class="math-container">$A$</span> is the equatorial radius, and <span class="math-container">$P$</span> is the orbital period. (We also may have to adjust for oblateness).</p>
<p>This is much better said by <a href="https://www.aanda.org/articles/aa/full_html/2009/29/aa11605-08/aa11605-08.html" rel="nofollow noreferrer">Haensel et. al</a> [2008]:</p>
<blockquote>
<p>The frequency <span class="math-container">$f$</span> of stable rotation of a star of gravitational mass
<span class="math-container">$M$</span> and baryon mass lower than the maximum allowable for non-rotating
stars is limited by the (Keplerian) frequency <span class="math-container">$f_{\rm K}$</span> of a test
particle co-rotating on an orbit at the stellar equator.</p>
</blockquote>
<p>So we call <span class="math-container">$f_k$</span> a &quot;Keplerian frequency&quot; because it is the maximum possible rotational frequency for a star, and we calculate it using Kepler's 3rd law.  The Keplerian frequency is important because it helps bound the size and mass of pulsars and other rapidly rotating bodies.</p>

CC BY-SA 4.0
2022-03-24T16:40:36.477Z
2022-03-24T16:40:36.477Z
Connor Garcia
<p>I am assuming that it means the frequency of a planet on its orbit.</p>
<p>But why do we specifically say Keplerian? Is there some other kind of definition of frequency we use for astronomical bodies in space that requires us to emphasise that it is Keplerian?</p>

<orbital-mechanics><keplers-laws>
What is a Keplerian frequency, why do we specifically say Keplerian?
WDUK
# What is a Keplerian frequency, why do we specifically say Keplerian?

(Asked by: username_0 on 2022-03-24)

I am assuming that it means the frequency of a planet on its orbit.

But why do we specifically say Keplerian? Is there some other kind of definition of frequency we use for astronomical bodies in space that requires us to emphasise that it is Keplerian?

**Tags:** orbital-mechanics, keplers-laws

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## 1 answers

### Answer by username_1 on 2022-03-24. Score: 3 | Accepted answer

**Is there a limit to how fast a pulsar or other large, massive, dense body can spin?**

The answer is yes! If it spins faster than a massless satellite in equatorial orbit at zero altitude, then it won't be stable. This is because any particle moving faster would migrate into a higher orbit. And how do we go about calculating the period of such a theoretical satellite? By applying Kepler's third law in this form: $M=A^3/P^2$, where $M$ is the mass of the pulsar, $A$ is the equatorial radius, and $P$ is the orbital period. (We also may have to adjust for oblateness).

This is much better said by [Haensel et. al](https://www.aanda.org/articles/aa/full_html/2009/29/aa11605-08/aa11605-08.html) [2008]:

> The frequency $f$ of stable rotation of a star of gravitational mass
> $M$ and baryon mass lower than the maximum allowable for non-rotating
> stars is limited by the (Keplerian) frequency $f\_{\rm K}$ of a test
> particle co-rotating on an orbit at the stellar equator.

So we call $f\_k$ a "Keplerian frequency" because it is the maximum possible rotational frequency for a star, and we calculate it using Kepler's 3rd law. The Keplerian frequency is important because it helps bound the size and mass of pulsars and other rapidly rotating bodies.


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