<p>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.</p>

CC BY-SA 3.0
2016-05-06T17:16:16.273Z
2016-05-06T17:16:16.273Z
Mike G
<p>I was using this formula to calculate the orbital period of a satellite in days: </p>

<blockquote>
  <p>$T = \sqrt{(4\times \pi^2)\times \frac{R^3}{GM_{center}}}$</p>
</blockquote>

<p>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. </p>

<p>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. </p>

<p>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.</p>

<p>When substituted, I get:</p>

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

<p>I simplified this too: </p>

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

<p>Which further simplifies too:</p>

<blockquote>
  <p>T = sqrt(1.607094708736*10^33)/(2.656092*10^20)</p>
</blockquote>

<p>Something definitely seems wrong at this point. But, this comes out to:</p>

<blockquote>
  <p>$T = \sqrt{(6.0505988073304689747192491826337340724643574093066053\times 10^{12})}$ </p>
  
  <p>$T = 2.459796\times 10^{6}$</p>
</blockquote>

<p>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. </p>

<p>Somewhere, I am missing something. </p>

CC BY-SA 3.0
2016-05-06T06:49:11.967Z
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
CC BY-SA 3.0
2016-05-06T08:22:49.017Z
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](http://astronomy.stackexchange.com/questions/1201/what-was-the-length-of-year-1-million-years-back) and make sure you used the right mass as Rob suggested...
Andy
CC BY-SA 3.0
2016-05-06T13:47:13.907Z
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 = 2pi*sqrt(a^3/GM), the one that Andy gave in his link. 2.3AU comes out to 3.44*10^11, and when cubed that comes out to 4.0707584*10^34. Substituting, that comes down too 2pi*sqrt(4.070584*10^34/(6.674*10^-11)*(3.978*10^30). 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
CC BY-SA 3.0
2016-05-06T19:35:02.990Z
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](https://en.wikipedia.org/wiki/Significant_figures).
Mike G
<astrophysics><orbital-elements>
Calculating Orbital Period
Colin Stricker
# Calculating Orbital Period

(Asked by: username_0 on 2016-05-06)

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.

**Tags:** astrophysics, orbital-elements

> **Comments:**
> **username_1**: 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.  
> **username_2**: 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](http://astronomy.stackexchange.com/questions/1201/what-was-the-length-of-year-1-million-years-back) and make sure you used the right mass as Rob suggested...  
> **username_0**: 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 = 2pi*sqrt(a^3/GM), the one that username_2 gave in his link. 2.3AU comes out to 3.44*10^11, and when cubed that comes out to 4.0707584*10^34. Substituting, that comes down too 2pi*sqrt(4.070584*10^34/(6.674*10^-11)*(3.978*10^30). 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.  
> **username_3**: 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](https://en.wikipedia.org/wiki/Significant_figures).  

---

## 1 answers

### Answer by username_3 on 2016-05-06. Score: 1 | Accepted answer

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.


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