<p>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:</p>

<p><span class="math-container">$$ T^2=\frac{4\pi^2r^3}{G\cdot M_{Sun}}$$</span></p>

<p>Using all numbers in the same units:</p>

<p><span class="math-container">$r \ 4.5 \times 10^{11} \ (m) $</span></p>

<p><span class="math-container">$G = 6.674 \times 10^{-11} \ (m^3 \ kg^{-1}  s^{-2}) $</span></p>

<p><span class="math-container">$M = 1.989 \times 10^{30} \ (kg)$</span></p>

<p><span class="math-container">$$ T^2=\frac{4 \cdot \pi^2 \cdot 9.11 \times 10^{34}}{6.674 \times 10^{-11} \cdot 1.989 \times 10^{30}} = \frac{3.60 \times 10^{36}}{1.33 \times 10^{20}} = 2.71 \times 10^{16} (s)$$</span></p>

<p><span class="math-container">$T = 1.65 \times 10^8 \ (s) $</span></p>

<p><span class="math-container">$\frac{1.65 \times 10^8}{365.25 \times 24 \times 3600} = 5.21 \ (years)$</span></p>

<hr>

<p>Just fyi you can also call the product <span class="math-container">$GM$</span> the <a href="https://en.wikipedia.org/wiki/Standard_gravitational_parameter" rel="noreferrer">standard gravitational parameter</a> and those numbers are more accurate than <span class="math-container">$G$</span> and <span class="math-container">$M$</span> are separately, as discussed in <a href="https://space.stackexchange.com/a/39930/12102">this answer</a>.</p>

<p>The comments under your question also mention that it is <em>sometimes easier</em> to use AU and years as units and use Earth's orbit for scaling to the 3/2 power, <strong>but only for things orbiting our sun</strong>.  With <span class="math-container">$r = 3$</span> AU:</p>

<p><span class="math-container">$T^2 = 3^3$</span></p>

<p><span class="math-container">$T = 3^{3/2} = 5.2 \ (years)$</span></p>

CC BY-SA 4.0
2019-11-16T23:21:12.070Z
2019-11-16T23:31:51.520Z
2019-11-16T23:31:51.520Z
uhoh
<p>Given that the mean distance of an asteroid from the Sun is <span class="math-container">$450 \times 10^6$</span> km is it possible that the asteroid completes one rotation around the Sun in two years?</p>

<hr>

<p>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 <span class="math-container">$149 \times 10^6$</span> km and it takes a year for earth to complete the rotation then for the asteroid it would take 3 years.</p>

<p>I tried to actually calculate how much time would take for the asteroid to complete one rotation <a href="https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law_of_Kepler" rel="nofollow noreferrer">combining</a> Newton's law of gravitation and Kepler's third law and given <span class="math-container">$G=6.673 \times 10^{-11} Nm^2/kg^2$</span> and converting all distances to meters:</p>

<p><span class="math-container">$$
T^2=\frac{4\pi^2r^3}{G\cdot M_{Sun}}=\frac{4\pi^2\cdot 450^3\cdot 10^{21}}{6.673\cdot 10^{-11}\cdot 1.989\cdot 10^{30}}=\frac{4\pi^2\cdot 450^3\cdot 10^2}{6.673\cdot 1.989}=2.99\cdot 450^3\cdot 10^2=27246375\cdot 10^3.
$$</span></p>

<p>Then <span class="math-container">$T=\sqrt{27246375\cdot 10^3}=165064$</span> seconds <span class="math-container">$=45$</span> hours. Is there something wrong in my calculations or I'm not using the correct formula?</p>

CC BY-SA 4.0
2019-11-16T14:59:58.137Z
@antispinwards you're right I didn't cube $450$. But even cubing it returns the result of $45$ hours
Yos
CC BY-SA 4.0
2019-11-16T15:23:20.683Z
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.
AtmosphericPrisonEscape
CC BY-SA 4.0
2019-11-16T15:49:26.227Z
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.
PM 2Ring
CC BY-SA 4.0
2019-11-16T18:42:06.427Z
Your assumption that 3 times as far, so it should take 3 years is incorrect as well.   It travels a path 3 times the distance but it also requires lower tangential velocity to follow that path and the lower velocity works out to the square root of 3, so 3*square root of 3. (as @PM2Ring points out above).
userLTK
CC BY-SA 4.0
2019-11-16T21:41:50.367Z
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.
Mike G
<orbit><gravity><distances>
How to determine the orbital period of an asteroid using Kepler laws given its orbit's radius?
Yos
# How to determine the orbital period of an asteroid using Kepler laws given its orbit's radius?

(Asked by: username_0 on 2019-11-16)

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?

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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](https://en.wikipedia.org/wiki/Kepler%27s_laws_of_planetary_motion#Third_law_of_Kepler) 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?

**Tags:** orbit, gravity, distances

> **Comments:**
> **username_0**: @antispinwards you're right I didn't cube $450$. But even cubing it returns the result of $45$ hours  
> **username_1**: 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.  
> **username_2**: 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.  
> **username_3**: Your assumption that 3 times as far, so it should take 3 years is incorrect as well.   It travels a path 3 times the distance but it also requires lower tangential velocity to follow that path and the lower velocity works out to the square root of 3, so 3*square root of 3. (as @PM2Ring points out above).  
> **username_4**: 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.  

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## 1 answers

### Answer by username_5 on 2019-11-16. Score: 5 | Accepted answer

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](https://en.wikipedia.org/wiki/Standard_gravitational_parameter) and those numbers are more accurate than $G$ and $M$ are separately, as discussed in [this answer](https://space.stackexchange.com/a/39930/12102).

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:

$T^2 = 3^3$

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


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