-4
$\begingroup$

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.

$\endgroup$
  • 2
    $\begingroup$ 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. $\endgroup$ – Rob Jeffries May 6 '16 at 6:49
  • 1
    $\begingroup$ 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... $\endgroup$ – Andy May 6 '16 at 8:22
  • $\begingroup$ 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 3.44*10^11, and when cubed that comes out to 4.0707584*10^34. Substituting, that comes down too 2pisqrt(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. $\endgroup$ – Colin Stricker May 6 '16 at 13:47
  • $\begingroup$ 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. $\endgroup$ – Mike G May 6 '16 at 19:35
1
$\begingroup$

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.

$\endgroup$

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.