Well, I ask the question because apparently I am going to aim for NaNoWriMo.

I am calculating the density of a planet with 1.8 Earth masses and 0.9 Earth radii, but last I checked, the calculations gave me 0.59 g/cm3, but I checked with the not-so astronomical masses, and then I had 3.21 g/cm3.

Now, that is the only thing I could solve but.....how do I calculate circumference?

If you could also review my math and let me know if it is accurate, I feel like I am doing something wrong.

  • 1
    $\begingroup$ As it stands, this question looks very broad to me. You should maybe focus on one part, and leave the rest to a follow-up question. Incidentally, you'll notice that I changed the formatting of your question, feel free to revert it or modify it. $\endgroup$ – clem steredenn Sep 1 '15 at 19:57
  • 1
    $\begingroup$ Your density calculation is not quite right yet. For 1.8 Earth masses in 0.9 Earth radii, you should get a value of around $13.6 g/cm^3$, which is denser than lead . . . perhaps feasible if the core is made of very heavy metals. As bonus, gravity at the surface would be a heavy $2.2g$. $\endgroup$ – Neil Slater Sep 1 '15 at 20:09
  • $\begingroup$ Given that the density of earth is $5.51 g/cm^3$, making your planet smaller than earth and small than earth means the density should be higher. Note, that you already have set the circumference by defining the radius ($C = 2r\pi$), at least at first pass $\endgroup$ – thurizas Sep 1 '15 at 20:13
  • $\begingroup$ Given that force due to gravity on an object is $F = mg$ and it is also given by $F = GM_pm/r_p^2$. Combining these two, we find that $g=GM_p/r_p^2$. Given that you know the mass and radius of the planet, you can calculate $g$. $\endgroup$ – thurizas Sep 1 '15 at 20:23
  • $\begingroup$ Alright, gentlemen. Fixed it. $\endgroup$ – Future Historian Sep 2 '15 at 14:10

You started to go wrong when you didn't convert between units; fractions of Earth radii do not translate into cubic centimeters without additional math. It's an easy thing to forget to do, but here's all the work done out with that in mind.

It's worth noting that planets are oblate spheroids, not spheres. All measures provided are therefore approximations.


These numbers will be used throughout this answer.


Earth's mass is approximately $5.972 × 10^{24}$ kilograms.
The mass $m$ of your planet is approximately $(1.8)(5.972 × 10^{24})= 1.07496 × 10^{25}$ kilograms.


Earth's radius is approximately $6,378$ kilometers.
The radius $r$ of your planet is approximately $(0.9)(6,378)=5740.2$ kilometers.


$v=(4/3)\pi r^3$
$v= (4/3)\pi (5740.2)^3$
$v= (4/3)(189,138,993,249)\pi$

$v\approx792,263,562,265$ cubic kilometers

Gravitational Constant

The gravitational constant $G \approx 6.674 × 10^{−11}$


$d = m/v$
$d = 1.07496 × 10^{25}/ 7.92263562265 × 10^{11}$
$d\approx1.3568212 × 10^{13}$ kilograms per cubic kilometer
$d\approx1.3568212 × 10^{16}$ grams per cubic kilometer
$d\approx1.3568212 × 10^{1}$ grams per cubic centimeter

$d\approx13.6$ grams per cubic centimeter, exactly what @NeilSlater got in the comments.

Surface Gravity

(The radius has been converted to meters here)
$g = (GM)/r^2$
$g = ((6.674 × 10^{−11})(1.07496 × 10^{25}))/5,740,200^2$
$g = (6.674 × 1.07496 × 10^{14})/3.2949896 × 10^{13}$
$g = 71.7428304/3.2949896$
$g \approx 21.7733101191$ meters per second$^2$

$21.7733101191/9.807$ (Earth gravity) $\approx2.2201805$ times Earth grav.


$c=2\pi r$
$c=2\pi (5740.2)$
$c=2\pi (5740.2)$
$c\approx36,066.7403003$ kilometers

  • $\begingroup$ You might want to add some of the info of this article $\endgroup$ – Feyre Dec 24 '16 at 11:07

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.