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If we look at what the Wikipedia says about gravitomagnetic fields $B$ generated by rotating planets / stars we have: https://en.wikipedia.org/wiki/Lense%E2%80%93Thirring_precession#Gravitomagnetic_analysis

The problem is the resulting measurement units are: $$ m/s^2 $$ (gravitoelectric field).

Or: $$ (kg*m*s^{-2}*m^2*kg^{-2})/(m/s*m^3)*(kg*m^2*s^{-1}) $$

So what is the correct scalar equation for the gravitomagnetic field $$ s^{-1} $$ of a rotating planet / star vs distance?

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    $\begingroup$ Why is this a problem? The gravitoelectric acceleration around the Earth, for example, has a magnitude of $9.8 \text{ m/s}^2$, so the units are what we expect. And the $\vec{B}$ field has the same units if you cancel thing out in your expression above. $\endgroup$
    – Michael Seifert
    Commented Sep 11, 2023 at 1:59
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    $\begingroup$ The measurement unit of the gravitomagnetic field should be: $$ s^{-1} $$. $\endgroup$
    – Phil Bouchard
    Commented Sep 11, 2023 at 2:26
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    $\begingroup$ Not in the conventions used in the Wikipedia article you linked to above. Lower down in that same article, the total force on a test particle is given as$$\vec{F} = m \left( \vec{E} + 4 \frac{\vec{v}}{c} \times \vec{B} \right)$$implying that $\vec{E}$ and $\vec{B}$ have the same units. Other sources might use a convention where that factor of $c$ is folded into the definition of $\vec{B}$ instead, giving it units of $\text{s}^{-1}$ as you expect. $\endgroup$
    – Michael Seifert
    Commented Sep 11, 2023 at 11:20
  • $\begingroup$ Yeah I suspected c might be in the way. Thanks for confirming. $\endgroup$
    – Phil Bouchard
    Commented Sep 11, 2023 at 16:25

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As confirmed by Michael Seifert, you just have to divide by $c$.

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