So I gather the way you (and Vera Rubin) calculate a galaxy's mass is by measuring a star's orbital velocity $v$ and its distance $R$ from the galactic center, and then plugging them into this equation derived from Newton's second law:


($G=6.67\times10^{-11}\ \mathrm{m^3\ kg^{-1}\ s^{-2}}$. Units of $v$ and $R$ are km/s and km, respectively)

The value obtained for $M_\mathrm{gal}$ this way famously disagrees with the value you would obtain by measuring the brightness of the galaxy, leading to the dark matter theory.

I was playing around with some solar system data and found that if I calculate the mass of the sun by plugging this data into the equation above the result is too low by 9 orders of magnitude. $\sim 1.98 \times 10^{21}\ \mathrm{kg}$ instead of the actual solar mass (according to Wikipedia) of $1.98 \times 10^{30}\ \mathrm{kg}$.

Is there some "dark vacuum" in the solar system? Or where have I gone wrong in the calculation?

Are there any datasets of $v$ and $R$ for stars in a particular galaxy out there that can be downloaded? I've heard that thousands of galaxies have now been observed to have stars "going too fast." Has any of that data been made available?

  • $\begingroup$ Side note: the relation you cited is for circular orbits (a good approximation for many stars in the galaxy) orbiting a spherically symmetric mass. That many galaxies look like disks should give you pause when applying the formula. $\endgroup$ – user10851 Aug 26 '14 at 8:25
  • 1
    $\begingroup$ Other side note: Historically, it was very easy to determine relative distances $R$ and velocities $v$ in the solar system. It was much harder but doable to get absolute values for $R$ and $v$. It was harder still to get $GM_\odot$ independently, so really we know this last quantity only because of applying the formula you have. (And the most difficult measurement of all is $G$ itself, which is why we know $GM_\odot$ better than $M_\odot$.) $\endgroup$ – user10851 Aug 26 '14 at 8:31
  • $\begingroup$ Ben, you made a typical freshman mistake in treating G as a number rather than as a quantity with dimensions. If you take the dimensionality of G into account, you'll arrive at 2*10^30 kg for all of the planets. Aside: Chris White is spot on in his comment about the gravitational parameter $\mu_{\odot}=GM_{\odot}$ versus the naive $G*M_{\odot}$. While astronomers can determine the Sun's gravitational parameter $\mu_{\odot}$ to ten places of accuracy, scientists know G (and hence $M_{\odot}$) to a paltry four places of accuracy. $\endgroup$ – David Hammen Aug 27 '14 at 6:38


($G=6.67\times10^{-11}\ \mathrm{N\cdot(m/kg)^2}$. Units of $v$ and $R$ are km/s and km, respectively)

You gave $G$ in MKS, then: $R$ and $v$ are m, m/s, $1\ \mathrm m= \frac {1}{10^3}\ \mathrm{km}$, that's why you got a wrong result: $10^3 \times (10^3)^2 = 10^9$ that's the order of magnitude you are missing $$ 1.5\times10^{11} \times(3\times10^4)^2/(6.6\times10^{-11}) = 2\times10^{30}\ \mathrm{kg}$$


I think you are doing your math incorrectly if you get $10^{21}\ \mathrm{kg}$.

$$M = \frac{Rv^2}{G}$$ Let's try Jupiter from your reference. $$M = \frac{(778 \times 10^6\ \mathrm{km}) (13.1\ \mathrm{\frac{km}{s}})^2}{G}$$ $$M = \frac{(7.78 \times 10^{11}\ \mathrm m) (1.31 \times 10^4\ \mathrm{\frac{m}{s}})^2}{6.6743 \times 10^{-11}\ \mathrm{\frac{m^3}{kgs^2}}}$$ $$M = \frac{1.34 \times 10^{20}\ \mathrm{\frac{m^3}{s^2}}}{6.6743 \times 10^{-11}\ \mathrm{\frac{m^3}{kgs^2}}}$$ $$M = 2.00\times 10^{30}\ \mathrm{kg}$$


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.