We know that the Vis-Viva equation is given by:

$v=\sqrt{G M\left(\frac{2}{r}-\frac{1}{a}\right)}$

If I plug-in the values for Earth (the mass is obviously that of the Sun):

$G=6.6743 \times 10^{-11} \mathrm{~m}^3 \mathrm{~kg}^{-1} \mathrm{~s}^{-2}$

$M_\circ=3.955 \times 10^{30} \mathrm{~kg}$

$a=149.60\times 10^9 \mathrm{~m}$

and I calculate it for $r=a$ I get a result that looks quite false:

$v=\sqrt{G M\left(\frac{2}{r}-\frac{1}{a}\right)}=1.33\mathrm{~km}\mathrm{~s}^{-1}$

when the average orbital speed for Earth is known to be $30\mathrm{~km}\mathrm{~s}^{-1}$.

I don't get it... Did I type a value wrong in my calculator or is it just that I applied the Vis-Viva equation wrong?

Any help is appreciated.

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    $\begingroup$ Uhoh mentioned you had the wrong mass for the sun and I went to look at that. According to nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html it's about 1.98x10^30 kg, as Uhoh showed. But if you google "What is the mass of the sun?", the top of the page is big bold letters that say "3.955 × 10^30 kg", your exact value. So I think Google is somehow screwy and returning nearly twice the correct value. $\endgroup$ Dec 8, 2023 at 15:47
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    $\begingroup$ Hilariously, directly below that, Google helpfully shows that value as also "1.988 M☉", acknowledging that it just told you the mass of the sun is 1.988 solar masses. $\endgroup$ Dec 8, 2023 at 15:49
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    $\begingroup$ @DarthPseudonym Wow, that's messed up. $\endgroup$ Dec 9, 2023 at 2:37

1 Answer 1


With $r = a$ for a circle, you're (correct!) equation simplifies to

$$ v = \sqrt{\frac{GM}{a}}$$

which using a different mass for the Sun than you used, gives the right answer. I used

G = 6.674E-11
M = 1.9885E+30
a = 150E+09

with the same units as yours, and got v = 29744.7 m/s.

The fairly small difference for the Solar mass can't account for your very large numerical error; I think you also made some arithmetic error somewhere.

However: If you forgot three zeros somewhere, your wrong solar mass plus a factor of $\sqrt{1/1000}$ does indeed give 1326.5 m/s!!

orbital speed of the earth quickie

note: In the future, use the standard gravitational parameter (the product $GM$) instead of $G$ and $M$ separately. We can measure solar system velocities and distances to great accuracy, and those give us the $GM$ product to high accuracy. But measuring either one separately is a lot harder, so the uncertainty on those numbers is much larger.

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    $\begingroup$ Great answer! When I did some orbital trajectory calculations, I always used G and M as separate variable. The standard gravitational parameter definitely makes things a lot easier and helps me avoid simple mistakes. $\endgroup$ Dec 8, 2023 at 10:36
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    $\begingroup$ @TheRocketfan We only know G to 4 and a bit digits. So it's always better to use the gravitational parameter, if you have that option. This page has Solar System GM values that are consistent with JPL DE 440: ssd.jpl.nasa.gov/astro_par.html $\endgroup$
    – PM 2Ring
    Dec 9, 2023 at 5:05
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    $\begingroup$ @PM2Ring indeed! Where to find the best values for standard gravitational parameters of solar system bodies? $\endgroup$
    – uhoh
    Dec 9, 2023 at 5:32

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