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I was using this formula to calculate the orbital period of a satellite in days:

$T = \sqrt{(4\times \pi^2)\times \frac{R^3}{GM_{center}}}$

Where R^3 is the radius of the orbit, or distance of the semi-major axis, G is the gravitational universal constant, and M-center is the mass of the object being orbited.

I'm attempting to calculate the orbital period for a planet that has a diameter of $5,124$ kilometers, or $\frac{804,500,000}{157}$ meters, and a mass of $5.526\times 10^{13}$ Kilograms.

The star, which substitutes M-center, has a mass of 3.978*10^30 Kilograms. The length of the semi-major axis is $2.3AU$, or $3.44\times 10^{11}$ meters.

When substituted, I get:

$T = \sqrt{\frac{(4\times \pi^2)\times(3.44\times 10^{11})^3}{(6.674\times 10^{-11})\times(3.978\times 10^{30})}}$

I simplified this too:

$T = \sqrt{\frac{(1.607094708736\times 10^{33})}{(6.674\times 10^{-11})\times (5.5616^{13})}}$

Which further simplifies too:

T = sqrt(1.607094708736*10^33)/(2.656092*10^20)

Something definitely seems wrong at this point. But, this comes out to:

$T = \sqrt{(6.0505988073304689747192491826337340724643574093066053\times 10^{12})}$

$T = 2.459796\times 10^{6}$

I find it hard to believe that it takes a planet that long to orbit around a star that is only 2.3AU away from it. Obviously, I am doing something very wrong, but I simply do not know where.

Somewhere, I am missing something.

Harsh Kumar
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Colin Stricker
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    No idea how you claim Step 1 "simplifies" to step 2. It does not. The mass of the planet should not feature in this simple calculation at all. – ProfRob May 06 '16 at 06:49
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    First of all the "4 pi squared" inside the SQRT is really just a "2 pi" at the start of the equation. Try using the equation seen in this question and make sure you used the right mass as Rob suggested... – Andy May 06 '16 at 08:22
  • Yes. Rob is correct. For some reason, I put the wrong mass in step two. I'm such a ditz. I tried the equation T = 2pisqrt(a^3/GM), the one that Andy gave in his link. 2.3AU comes out to 3.4410^11, and when cubed that comes out to 4.070758410^34. Substituting, that comes down too 2pisqrt(4.07058410^34/(6.67410^-11)(3.97810^30). That gives 7.77962*10^7. How do I interpret this? Seconds, days? If it's seconds, it comes out to 900.42 days, which seems unreasonable. – Colin Stricker May 06 '16 at 13:47
  • Steps 2 and 3 in your question should have 1.607*10^36 in the numerator, making T the same as in your comment. Also the number of decimal places shouldn't grow; see Signifcant figures. – Mike G May 06 '16 at 19:35

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Keep the units with the numbers, even if you convert units. $G$ has units too; since you used the $\mathrm{m}^3 \mathrm{kg}^{-1} \mathrm{s}^{-2}$ value and put $R$ in meters and $M_{\mathrm{center}}$ in kilograms, your $T$ should be in seconds. Sanity check: by Kepler's third law $T^2 \propto R^3$, a 2.3 AU orbit around the Sun (half the mass of your central star) should take 3.5 years.

Mike G
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