<p>The formula for orbital period is given on <a href="https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body" rel="nofollow noreferrer">Wikipedia</a>:</p>
<p><span class="math-container">$$T=2\pi \sqrt\frac{a^3}\mu$$</span></p>
<p>where:</p>
<ul>
<li><span class="math-container">$T$</span> is the orbital period in seconds</li>
<li><span class="math-container">$a$</span> is the orbit's semi-major axis in meters</li>
<li><span class="math-container">$\mu = GM$</span> is the standard gravitational parameter
<ul>
<li><span class="math-container">$G$</span> is the <a href="https://en.wikipedia.org/wiki/Gravitational_constant" rel="nofollow noreferrer">gravitational constant</a></li>
<li><span class="math-container">$M$</span> is the <a href="https://en.wikipedia.org/wiki/Sun" rel="nofollow noreferrer">mass of the more massive body</a> in kilograms</li>
</ul>
</li>
</ul>
<p>So <span class="math-container">$T = 2 \pi \sqrt \frac { (172 \cdot 10^9) ^ 3 } { 6.674 \cdot 10^{-11} \cdot 1.9884 \cdot 10^{30} }$</span>. Can you take it from here?</p>
<p>As for density, the <a href="https://en.wikipedia.org/wiki/Sphere#Enclosed_volume" rel="nofollow noreferrer">volume of a sphere</a> is given by the formula <span class="math-container">$V = \frac43\pi r^3$</span>; density (<span class="math-container">$\rho = \frac MV$</span>, with <span class="math-container">$M$</span> the mass) is usually given in grams per cubic centimeter, so it makes sense to convert to those units. That gives the following calculation:</p>
<p><span class="math-container">$$\frac { 6.15 \cdot 10^{27} } { \frac43\pi (6.743 \cdot 10^8)^3 }$$</span></p>

CC BY-SA 4.0
2020-07-29T20:34:20.217Z
2020-07-29T20:34:20.217Z
Glorfindel
<h2>Orbital Period</h2>
<p>According to Kepler's Third Law, the orbital period <span class="math-container">$T$</span> is defined as</p>
<p><span class="math-container">$$T=2\pi\sqrt \frac{a^3}{\mu}$$</span></p>
<p><span class="math-container">$T$</span> is, as said before, the orbital period (i.e. the time for an object - in this case, the planet - to complete an orbit around the massive, central object - in this case, the star) measured in seconds.<br />
<span class="math-container">$a$</span> is the object's semi-major axis (the longest diameter of an ellipse - in this case, the greatest distance between star and planet).<br />
<span class="math-container">$\mu=GM$</span> with <span class="math-container">$G$</span> being the Gravitational Constant and <span class="math-container">$M$</span> the mass of the massive object (the star).<br />
(from <a href="https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body" rel="nofollow noreferrer">Wikipedia - Orbital Period</a>)</p>
<p>By inserting the values, we get <span class="math-container">$$T=2\pi\sqrt \frac{(1,15AU)^3}{GM}$$</span></p>
<p>Note that in this case, the mass of the planet is not relevant. What we need instead is the mass of the star, which you have not given. Since you assumed a sun-like star, we can insert the sun's standard gravitational parameter for <span class="math-container">$G\times M$</span>:</p>
<p><span class="math-container">$$T=2\pi\sqrt \frac{(1.72\times10^{11}m)^3}{1,33\times10^{20} \frac {m^3}{s^2}}\approx38863930\,s$$</span></p>
<p>which is around 449.81 days.</p>
<hr />
<h2>Density</h2>
<p>For the density, we know that <span class="math-container">$$\rho=\frac{m}{V}$$</span></p>
<p>Approximating the planet's shape to a sphere with a Volume <span class="math-container">$V=\frac{4}{3}\pi r^3$</span>, we get <span class="math-container">$V\approx1.284\times10^{21}m^3$</span></p>
<p>Thus, the density is <span class="math-container">$$\rho\approx 4789\frac{kg}{m^3}$$</span></p>
<hr />
<h2>Notes and Disclaimer</h2>
<p>I am no physics expert myself, I am only a student. I do not take any responsibility regarding the correctness of my calulations.</p>
<p>I encourage you to only take the formulas (either from this post or just look them up) and do the math yourself. Since you participate in a competition (I do not know the rules of the competition, i.e. whether it is allowed to ask others for help), it is anyway better to actually do the work yourself. Please only consider this post as a reference to check whether your calculations seem to be correct (assuming that mine are, which I hope) and don't simply copy it.</p>
<p>I do not take any responsibility if you use this post other than described above, which includes potentially being ruled out from the competition.</p>
<p>I hope this helps.</p>

CC BY-SA 4.0
2020-07-29T21:09:18.680Z
2020-07-30T08:02:06.033Z
2020-07-30T08:02:06.033Z
jng224
<p>I'm a student studying in 9th grade, I am making this fictional habitable planet named &quot;Darwin B&quot; for a planet making competition. It orbits a sun-like star at a distance of 1.15 AU or 172 million kilometres in a nearly circular orbit. Its rotational period is 19 hours, 38 minutes. Its mass is <span class="math-container">$6.15×10^{24}kg$</span> and its radius is about 6,743 kilometres. I have to calculate its orbital period and density but I'm weak in maths and don't know how to. Please help.</p>

<orbit><planet><mass><density>
How to calculate the orbital period and the density of a planetary body?
Ansh Saxena
# How to calculate the orbital period and the density of a planetary body?

(Asked by: username_0 on 2020-07-29)

I'm a student studying in 9th grade, I am making this fictional habitable planet named "Darwin B" for a planet making competition. It orbits a sun-like star at a distance of 1.15 AU or 172 million kilometres in a nearly circular orbit. Its rotational period is 19 hours, 38 minutes. Its mass is $6.15×10^{24}kg$ and its radius is about 6,743 kilometres. I have to calculate its orbital period and density but I'm weak in maths and don't know how to. Please help.

**Tags:** orbit, planet, mass, density


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## 2 answers

### Answer by username_1 on 2020-07-29. Score: 1

The formula for orbital period is given on [Wikipedia](https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body):

$$T=2\pi \sqrt\frac{a^3}\mu$$

where:

* $T$ is the orbital period in seconds
* $a$ is the orbit's semi-major axis in meters
* $\mu = GM$ is the standard gravitational parameter
  + $G$ is the [gravitational constant](https://en.wikipedia.org/wiki/Gravitational_constant)
  + $M$ is the [mass of the more massive body](https://en.wikipedia.org/wiki/Sun) in kilograms

So $T = 2 \pi \sqrt \frac { (172 \cdot 10^9) ^ 3 } { 6.674 \cdot 10^{-11} \cdot 1.9884 \cdot 10^{30} }$. Can you take it from here?

As for density, the [volume of a sphere](https://en.wikipedia.org/wiki/Sphere#Enclosed_volume) is given by the formula $V = \frac43\pi r^3$; density ($\rho = \frac MV$, with $M$ the mass) is usually given in grams per cubic centimeter, so it makes sense to convert to those units. That gives the following calculation:

$$\frac { 6.15 \cdot 10^{27} } { \frac43\pi (6.743 \cdot 10^8)^3 }$$


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### Answer by username_2 on 2020-07-29. Score: 0

## Orbital Period

According to Kepler's Third Law, the orbital period $T$ is defined as

$$T=2\pi\sqrt \frac{a^3}{\mu}$$

$T$ is, as said before, the orbital period (i.e. the time for an object - in this case, the planet - to complete an orbit around the massive, central object - in this case, the star) measured in seconds.  
$a$ is the object's semi-major axis (the longest diameter of an ellipse - in this case, the greatest distance between star and planet).  
$\mu=GM$ with $G$ being the Gravitational Constant and $M$ the mass of the massive object (the star).  
(from [Wikipedia - Orbital Period](https://en.wikipedia.org/wiki/Orbital_period#Small_body_orbiting_a_central_body))

By inserting the values, we get $$T=2\pi\sqrt \frac{(1,15AU)^3}{GM}$$

Note that in this case, the mass of the planet is not relevant. What we need instead is the mass of the star, which you have not given. Since you assumed a sun-like star, we can insert the sun's standard gravitational parameter for $G\times M$:

$$T=2\pi\sqrt \frac{(1.72\times10^{11}m)^3}{1,33\times10^{20} \frac {m^3}{s^2}}\approx38863930\,s$$

which is around 449.81 days.

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## Density

For the density, we know that $$\rho=\frac{m}{V}$$

Approximating the planet's shape to a sphere with a Volume $V=\frac{4}{3}\pi r^3$, we get $V\approx1.284\times10^{21}m^3$

Thus, the density is $$\rho\approx 4789\frac{kg}{m^3}$$

---

## Notes and Disclaimer

I am no physics expert myself, I am only a student. I do not take any responsibility regarding the correctness of my calulations.

I encourage you to only take the formulas (either from this post or just look them up) and do the math yourself. Since you participate in a competition (I do not know the rules of the competition, i.e. whether it is allowed to ask others for help), it is anyway better to actually do the work yourself. Please only consider this post as a reference to check whether your calculations seem to be correct (assuming that mine are, which I hope) and don't simply copy it.

I do not take any responsibility if you use this post other than described above, which includes potentially being ruled out from the competition.

I hope this helps.


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