Given that the mean distance of an asteroid from the Sun is $450 \times 10^6$ km is it possible that the asteroid completes one rotation around the Sun in two years?

According to the third law of Kepler the farther a planet if from the Sun the longer is its orbital period. Because the distance from the Sun to earth is $149 \times 10^6$ km and it takes a year for earth to complete the rotation then for the asteroid it would take 3 years.

I tried to actually calculate how much time would take for the asteroid to complete one rotation combining Newton's law of gravitation and Kepler's third law and given $G=6.673 \times 10^{-11} Nm^2/kg^2$ and converting all distances to meters:

$$ T^2=\frac{4\pi^2r^3}{G\cdot M_{Sun}}=\frac{4\pi^2\cdot 450^3\cdot 10^{21}}{6.673\cdot 10^{-11}\cdot 1.989\cdot 10^{30}}=\frac{4\pi^2\cdot 450^3\cdot 10^2}{6.673\cdot 1.989}=2.99\cdot 450^3\cdot 10^2=27246375\cdot 10^3. $$

Then $T=\sqrt{27246375\cdot 10^3}=165064$ seconds $=45$ hours. Is there something wrong in my calculations or I'm not using the correct formula?

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    $\begingroup$ You are not cubing the distance properly. $\endgroup$ – antispinwards Nov 16 '19 at 14:53
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    $\begingroup$ Well you are still computing the cube of the distance incorrectly. Either convert km to m before computing the cube, or convert km³ to m³ afterwards. These have different conversion factors! $\endgroup$ – antispinwards Nov 16 '19 at 15:21
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    $\begingroup$ Also please make the distinction between 'radius' and 'orbital radius' clear. The 'radius' of an object is its physical size, which has nothing to do with orbital distance, for small masses at least. $\endgroup$ – AtmosphericPrisonEscape Nov 16 '19 at 15:23
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    $\begingroup$ It's a lot easier to do this calculation if you work in AU and years. Your asteroid has a mean orbital radius of roughly 3 AU, so its period is roughly $3\sqrt3$ years. $\endgroup$ – PM 2Ring Nov 16 '19 at 15:49
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    $\begingroup$ Concur with @PM2Ring; in au-yr units, $GM_\odot = 4\pi^2$. In SI units, it's easier to get the arithmetic right if you rewrite $450 \times 10^6$ km as $4.5\times10^{11}$ m. $\endgroup$ – Mike G Nov 16 '19 at 21:41

Here's how I would do it. I'd convert everything to a single, standard set of units as recommended in the comments, and also stick to one digit before the decimal in scientific notation:

$$ T^2=\frac{4\pi^2r^3}{G\cdot M_{Sun}}$$

Using all numbers in the same units:

$r \ 4.5 \times 10^{11} \ (m) $

$G = 6.674 \times 10^{-11} \ (m^3 \ kg^{-1} s^{-2}) $

$M = 1.989 \times 10^{30} \ (kg)$

$$ T^2=\frac{4 \cdot \pi^2 \cdot 9.11 \times 10^{34}}{6.674 \times 10^{-11} \cdot 1.989 \times 10^{30}} = \frac{3.60 \times 10^{36}}{1.33 \times 10^{20}} = 2.71 \times 10^{16} (s)$$

$T = 1.65 \times 10^8 \ (s) $

$\frac{1.65 \times 10^8}{365.25 \times 24 \times 3600} = 5.21 \ (years)$

Just fyi you can also call the product $GM$ the standard gravitational parameter and those numbers are more accurate than $G$ and $M$ are separately, as discussed in this answer.

The comments under your question also mention that it is sometimes easier to use AU and years as units and use Earth's orbit for scaling to the 3/2 power, but only for things orbiting our sun. With $r = 3$ AU:

$T^2 = 3^3$

$T = 3^{3/2} = 5.2 \ (years)$

| improve this answer | |

Judging by the fact that after cubing the distance you have ended up with an exponent of 1021 (21 = 6×3+3) at the first point you have substituted the numbers into the equation, you appear to be assuming that 1 km3 is 1000 m3. However, if you write the same linear quantity in metres and kilometres and cube them, you can see this is not the case.

$$\begin{align} \left(1\,\mathrm{km}\right)^3 &= 1\,\mathrm{km}^3 \\ \left(1000\,\mathrm{m}\right)^3 &= 1\,000\,000\,000\,\mathrm{m}^3 \end{align}$$

In summary, you forgot to take the cube of the unit conversion factor.

| improve this answer | |

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