<p>It's almost completely unconstrained. There's a limit to how fast a planet can spin before it starts to break up, which is $p = 2 \pi / \sqrt{m G / r^{3}}$ (where $m$ is the planet's mass and $r$ is its radius). For something like the Earth (and Venus is pretty similar), this corresponds to about 1.4 hours. So the day has to be longer than that.</p>

<p>But there's no limit to how <em>slowly</em> the planet can rotate. If it's rotating slowly enough so that one rotation period = one year, then as seen from the planet's surface the star would never seem to move in its sky and the solar day (time from one noon to the next) would be infinite. This is what you get if the planet is 1:1 tidally locked to the star. </p>

CC BY-SA 3.0
2016-09-09T14:45:40.563Z
2016-09-09T14:51:26.583Z
2016-09-09T14:51:26.583Z
Peter Erwin
<p>You can use Kepler's third law to get the length of a year
$$\frac{P^2}{a^3} = \frac{4 \pi^2}{G(M+m)},$$
where $P$ is the period of rotation, $a$ is the distance between the planet and the star, $M$ is the mass of the star, $m$ is the mass of the planet and $G$ is the universal gravity constant.
As the mass of Venus is roughly the same as the Earth's ($M_V = 0.815M_\oplus$), you will get something close but a bit larger than 365 days.</p>

<p>If you want the length of a day on this planet, you need more information as the rotation rate of the planet which gives the length of a day cannot be constrained easily. For example on Earth, one day is 24 hours today, and the length of a day circa 620 million years ago has been estimated from rhythmites (alternating layers in sandstone) as having been about 21.9 hours. The origin of the evolution of the length of a day is the tidal interactions between Earth and the Moon (as well as to a smaller extent with the Sun) , an effect known as tidal acceleration (or in this case, deceleration).</p>

<p>This is also the reason why we only see one side of the moon: it is tidally locked to the Earth so that the Moon revolves around the Earth in exactly one moon day. In the very far future, the Earth will revolve around the Moon in exactly one 'future' day, so that one day will last 27.7 present days!</p>

<p>Finally, the initial rotation rate of the planet at its formation is largely unconstrained so that even though you give us a planet's mass and age and the properties of its orbit and its star, it will still be impossible to answer your question entirely.</p>

CC BY-SA 3.0
2016-09-09T14:22:02.090Z
2016-09-09T14:29:22.937Z
2016-09-09T14:29:22.937Z
cphyc
<p>Take the following planet:</p>

<ul>
<li>Same distance from the main star as Earth</li>
<li>Main star is the same size as the sun</li>
<li>Planet is the size of Venus</li>
</ul>

<p>How would I calculate the length of the day on this planet? If you need more information, please ask.</p>

CC BY-SA 3.0
2016-09-09T14:54:49.707Z
The piece of information you need is pretty much "the length of the day". It is independent of your variables. (Though if the planet is really close to the star, it would tend to be tidally locked.)
SE - stop firing the good guys
<planet>
How would I calculate the length of the day of this planet?
John Doe
# How would I calculate the length of the day of this planet?

(Asked by: username_0 on 2016-09-09)

Take the following planet:

* Same distance from the main star as Earth
* Main star is the same size as the sun
* Planet is the size of Venus

How would I calculate the length of the day on this planet? If you need more information, please ask.

**Tags:** planet
> **Comments:**
> **username_1**: The piece of information you need is pretty much "the length of the day". It is independent of your variables. (Though if the planet is really close to the star, it would tend to be tidally locked.)  

---

## 2 answers

### Answer by username_2 on 2016-09-09. Score: 4

It's almost completely unconstrained. There's a limit to how fast a planet can spin before it starts to break up, which is $p = 2 \pi / \sqrt{m G / r^{3}}$ (where $m$ is the planet's mass and $r$ is its radius). For something like the Earth (and Venus is pretty similar), this corresponds to about 1.4 hours. So the day has to be longer than that.

But there's no limit to how *slowly* the planet can rotate. If it's rotating slowly enough so that one rotation period = one year, then as seen from the planet's surface the star would never seem to move in its sky and the solar day (time from one noon to the next) would be infinite. This is what you get if the planet is 1:1 tidally locked to the star.


---

### Answer by username_3 on 2016-09-09. Score: 2

You can use Kepler's third law to get the length of a year
$$\frac{P^2}{a^3} = \frac{4 \pi^2}{G(M+m)},$$
where $P$ is the period of rotation, $a$ is the distance between the planet and the star, $M$ is the mass of the star, $m$ is the mass of the planet and $G$ is the universal gravity constant.
As the mass of Venus is roughly the same as the Earth's ($M\_V = 0.815M\_\oplus$), you will get something close but a bit larger than 365 days.

If you want the length of a day on this planet, you need more information as the rotation rate of the planet which gives the length of a day cannot be constrained easily. For example on Earth, one day is 24 hours today, and the length of a day circa 620 million years ago has been estimated from rhythmites (alternating layers in sandstone) as having been about 21.9 hours. The origin of the evolution of the length of a day is the tidal interactions between Earth and the Moon (as well as to a smaller extent with the Sun) , an effect known as tidal acceleration (or in this case, deceleration).

This is also the reason why we only see one side of the moon: it is tidally locked to the Earth so that the Moon revolves around the Earth in exactly one moon day. In the very far future, the Earth will revolve around the Moon in exactly one 'future' day, so that one day will last 27.7 present days!

Finally, the initial rotation rate of the planet at its formation is largely unconstrained so that even though you give us a planet's mass and age and the properties of its orbit and its star, it will still be impossible to answer your question entirely.


---

