<p>The energy a rotating planet has is derived from the rotation the <a href="https://www.scientificamerican.com/article/why-and-how-do-planets-ro/" rel="noreferrer">initial interstellar cloud</a> of gas and dust had prior to it flattening into a protostellar disk.</p>
<p>The rotation of a planet is bound by the laws of <a href="https://en.wikipedia.org/wiki/Angular_momentum" rel="noreferrer">angular momentum</a>. Alone, without any other influences, the rate of rotation will remain constant for ever. However, some planets have moons &amp; these will affect the planet's rate of rotation. Earth's rotation rate is slowing due to the Moon, particularly through the <a href="http://curious.astro.cornell.edu/physics/37-our-solar-system/the-moon/the-moon-and-the-earth/147-the-moon-slows-the-earth-s-rotation-but-how-fast-was-it-spinning-billions-of-years-ago-intermediate" rel="noreferrer">action of tides</a>. During the era of the dinosaurs a day of Earth was went from <strike>19</strike> <a href="https://www.abc.net.au/news/science/2020-01-27/hours-in-dinosaurs-day-physics-astrophysics/11886448" rel="noreferrer">21</a> hours (620 Ma ago) to 23 hours (65 Ma ago), now its 24 hours. I'll let others who are more knowledgeable about this than me comment on how a star might affect the rotation of planets orbiting it.</p>

CC BY-SA 4.0
2022-08-23T19:18:37.120Z
Also the star itself does inflict tides on the planets. On Earth 1/3 of the tide is caused by the Sun. As such, moons are not a prerequisite for loss of spin rotation. Solar tides will gradually lead to a longer year as the spin's angular momentum is thus transferred into orbital angular momentum. See this question: https://astronomy.stackexchange.com/questions/18283/effects-of-the-solar-tide-on-planets
planetmaker
CC BY-SA 4.0
2022-08-24T02:15:17.647Z
The moon is tidally locked to Earth. Most of the Jovian moons are tidally locked. If a moon orbiting a planet can lose its angular momentum then a planet orbiting a star can as well, although the process seems to be a lot slower.
Adam Chalcraft
CC BY-SA 4.0
2022-08-24T04:38:00.373Z
@AdamChalcraft - Also see Mercury's 3:2 spin-orbit resonance (which is the result of the same effect on an eccentric orbit instead of a circular one).
TLW
CC BY-SA 4.0
2022-08-24T04:43:55.800Z
Be aware that time to tidal lock scales, to a first approximation, with the _sixth_ (6th) power of the orbital radius (SMA).
TLW
CC BY-SA 4.0
2022-08-24T08:25:17.297Z
`Alone, without any other influences` - which means it's not going to happen; there is no place within the universe where gravity does not reach... just maybe verrrry slowly in case of a planet without a sun, deep in intergalactic emptiness, if such a thing should ever happen for whatever reason.
AnoE
CC BY-SA 4.0
2022-08-24T14:01:51.000Z
@planetmaker A planet without a star is called a *rogue planet* - it just needs to be big enough to be of "planetary mass", which is big enough to become roughly spherical, but not so big it turns into a star.
Nuclear Hoagie
CC BY-SA 4.0
2022-08-24T21:45:26.640Z
@Fred "During the era of the dinosaurs a day on Earth was about 19 hours, now it's 24 hours." Do you have a source for this? I wish there were a nice chart of that over time. My quirky calculation shows it was 23 hours long 250 million years ago.
Ramon Snir
CC BY-SA 4.0
2022-08-24T23:55:02.467Z
@RamonSnir: That 19 hours was something I came across some years ago. I've edited the answer according to the latest info I found online - 21 hours 620 Ma ago & 23 hours 65 Ma ago).
Fred
CC BY-SA 4.0
2022-08-25T14:32:23.723Z
Not mentioned, the moon is slowly drifting away from us.  Some of the Earth's lost rotation is due to the transfer of energy to the Moon's widening orbital path.
Reginald Blue
CC BY-SA 4.0
2022-08-26T10:54:28.117Z
Also, I seem to have read somewhere that it would take longer for Earth to become tidally locked to either the Sun or the Moon than it would take for the Sun to go supernova.
Nzall
CC BY-SA 4.0
2022-08-26T16:31:09.407Z
@Nzall No, because the Sun will never go supernova. It's way too small. (Unless in the far distant future, when it's a white dwarf, and the Milky Way has merged with the Andromeda galaxy, it happens to collide with another white dwarf).
PM 2Ring
CC BY-SA 4.0
2022-08-26T19:27:44.980Z
@PM2Ring Sorry, with supernova I meant a generic "star-shattering kaboom". I essentially was saying that I read somewhere that it would take Earth 10 billion years to tidally lock while the sun will turn into a Red Giant in 5 billion years.
Nzall
CC BY-SA 4.0
2022-08-26T20:00:26.263Z
@Nzall Sure, the Sun will gradually get hotter and then spend several billion years going through its red giant phases, emitting a fair bit more energy than it does now as well as spewing out a lot of gas & dust. (The Earth will be unliveable in a billion year or so, long before the red giant phases start). Eventually, the Sun *will* shed a large chunk of matter, creating a pretty [planetary nebula](https://en.wikipedia.org/wiki/Planetary_nebula) when it turns into a white dwarf & stops doing fusion. But I wouldn't exactly call that "star-shattering".
PM 2Ring
CC BY-SA 4.0
2022-08-23T18:24:26.703Z
2022-08-24T23:52:58.657Z
2022-08-24T23:52:58.657Z
Fred
<p>In addition to the interaction with other bodies (moons, other planets, stars) as described in another answer, you also have gravitational waves produced by spinning body that isn't perfectly symmetrical. Which the earth isn't. Those waves carry some energy and angular momentum with them, which means the planet loses it.</p>
<p>Mind, that contribution is tiny. It's relevant only for extremly dense and massive bodies, so it wouldn't matter to a planet. But it is an effect.</p>

CC BY-SA 4.0
2022-08-24T16:56:48.917Z
Even the gravitational radiation due to Earth's orbital motion around the Sun is tiny. According to https://physics.stackexchange.com/a/412990/123208 it's ~196 watts. It'd be interesting to compare the gravitational radiation due to the Earth's rotation to that of the orbit of the Earth & Moon about their barycentre.
PM 2Ring
CC BY-SA 4.0
2022-08-24T22:56:38.533Z
So on the order of... a race between losing energy to gravitational waves and theoretical proton decay?
Michael
CC BY-SA 4.0
2022-08-25T13:28:35.660Z
It's on the time scale of "way too long to matter", yes. For the earth specifically, it would've long since been swalloed by the sun as it expands into a red giant.
BurnNote
CC BY-SA 4.0
2022-08-24T11:38:20.857Z
2022-08-24T11:38:20.857Z
BurnNote
<p>In addition to tidal/gravitational effects, a planet might lose energy also due to electromagnetic emission, provided that the planet has a magnetic field and if the magnetic field dipole moment is <strong>not</strong> aligned to the rotation axis (this is indeed the case for the Earth, the &quot;magnetic axis&quot; is tilted by about 11°).</p>
<p>The component of the dipole orthogonal to the rotation axis, due to rotation, will radiate electromagnetic waves. The radiated power by such a dipole is (in far field condition, assuming an ideal dipole) proportional to <span class="math-container">$m^2 \omega^4$</span>, where <span class="math-container">$m$</span> is the dipole moment and <span class="math-container">$\omega$</span> the angular velocity. A good reference on this is, for example, D. J. Griffiths <em>Introduction to Electrodynamics</em>, Chapter 11 &quot;Radiation&quot;.</p>
<p>There is an exercise in the same book in which one is asked to compute such emitted power in the case of the Earth, just to find out that such power is really tiny - of the order of few <span class="math-container">$\mu\text{W}$</span>.</p>

CC BY-SA 4.0
2022-08-25T12:14:58.043Z
2022-08-25T12:14:58.043Z
Alessio Perinelli
<p>Well, kenetic energy is of at least three types.  Translational energy, rotational energy, and vibrational energy.  As energy is neither created nor distroyed, the planet really does not lose energy, rather transfers it to other bodies.  If it is in some sort of debrie field, it might impart its rotational energy into other masses which would show up as translational energy.
Think of playing billards.  If a ball hits another ball off center, you will see that the hit ball has both rotational energy and translational energy.  Eventually, drag will dissipate into heat energy (which is just kinetic [translational and vibrational] energy of the molecules).</p>

CC BY-SA 4.0
2022-08-25T16:27:34.670Z
2022-08-25T16:27:34.670Z
Walter
<p>When a planet rotates/spins, is it true that the planet loses energy? If this is true, then where does this energy lost by the planet go?
My confusion may be in the basic understanding of the mechanism. When I imagine a planet rotating, what I think is either of the two things may happen -</p>
<ol>
<li>The planet already has energy within the system, which is lost when the planet rotates.
OR</li>
<li>The planet rotates because it receives energy from the outside.</li>
</ol>
<p>So which one of the above actually happen?</p>
<p>This is what I think may happen. If I am wrong, please correct me.</p>

CC BY-SA 4.0
2022-08-24T13:08:42.697Z
This might be an entirely different question, but doesn't rotation cause stress or strain on a physical body, some of which is converted to heat?
Barry Carter
CC BY-SA 4.0
2022-08-25T15:23:14.307Z
@BarryCarter How so?
Jason Goemaat
CC BY-SA 4.0
2022-08-25T16:22:57.287Z
@JasonGoemaat I thought things like conservation of angular momentum only applied in theory, not in practice. When a ball is rotating, the atoms in the ball are going at many different velocities. The electric force keeps them together, but the ball must deform slightly, which changes it shape, and the process repeats. I think that's how stress and strain work. Even if you use a spring without breaking it, every use weakens in slightly (I think)
Barry Carter
CC BY-SA 4.0
2022-08-26T14:04:31.707Z
@BarryCarter: Angular momentum is absolutely conserved. Has to be, by Noether's theorem, since the universe is rotation-invariant. There's no preferential direction.
MSalters
<rotation><planetary-science>
Do planets lose energy while rotating?
apk
# Do planets lose energy while rotating?

(Asked by: username_0 on 2022-08-23)

When a planet rotates/spins, is it true that the planet loses energy? If this is true, then where does this energy lost by the planet go?
My confusion may be in the basic understanding of the mechanism. When I imagine a planet rotating, what I think is either of the two things may happen -

1. The planet already has energy within the system, which is lost when the planet rotates.
   OR
2. The planet rotates because it receives energy from the outside.

So which one of the above actually happen?

This is what I think may happen. If I am wrong, please correct me.

**Tags:** rotation, planetary-science

> **Comments:**
> **username_1**: This might be an entirely different question, but doesn't rotation cause stress or strain on a physical body, some of which is converted to heat?  
> **username_2**: @username_1 How so?  
> **username_1**: @username_2 I thought things like conservation of angular momentum only applied in theory, not in practice. When a ball is rotating, the atoms in the ball are going at many different velocities. The electric force keeps them together, but the ball must deform slightly, which changes it shape, and the process repeats. I think that's how stress and strain work. Even if you use a spring without breaking it, every use weakens in slightly (I think)  
> **username_3**: @username_1: Angular momentum is absolutely conserved. Has to be, by Noether's theorem, since the universe is rotation-invariant. There's no preferential direction.  

---

## 4 answers

### Answer by username_4 on 2022-08-23. Score: 32

The energy a rotating planet has is derived from the rotation the [initial interstellar cloud](https://www.scientificamerican.com/article/why-and-how-do-planets-ro/) of gas and dust had prior to it flattening into a protostellar disk.

The rotation of a planet is bound by the laws of [angular momentum](https://en.wikipedia.org/wiki/Angular_momentum). Alone, without any other influences, the rate of rotation will remain constant for ever. However, some planets have moons & these will affect the planet's rate of rotation. Earth's rotation rate is slowing due to the Moon, particularly through the [action of tides](http://curious.astro.cornell.edu/physics/37-our-solar-system/the-moon/the-moon-and-the-earth/147-the-moon-slows-the-earth-s-rotation-but-how-fast-was-it-spinning-billions-of-years-ago-intermediate). During the era of the dinosaurs a day of Earth was went from 19 [21](https://www.abc.net.au/news/science/2020-01-27/hours-in-dinosaurs-day-physics-astrophysics/11886448) hours (620 Ma ago) to 23 hours (65 Ma ago), now its 24 hours. I'll let others who are more knowledgeable about this than me comment on how a star might affect the rotation of planets orbiting it.

> **Comments:**
> **username_5**: Also the star itself does inflict tides on the planets. On Earth 1/3 of the tide is caused by the Sun. As such, moons are not a prerequisite for loss of spin rotation. Solar tides will gradually lead to a longer year as the spin's angular momentum is thus transferred into orbital angular momentum. See this question: https://astronomy.stackexchange.com/questions/18283/effects-of-the-solar-tide-on-planets  
> **username_6**: The moon is tidally locked to Earth. Most of the Jovian moons are tidally locked. If a moon orbiting a planet can lose its angular momentum then a planet orbiting a star can as well, although the process seems to be a lot slower.  
> **username_7**: @username_6 - Also see Mercury's 3:2 spin-orbit resonance (which is the result of the same effect on an eccentric orbit instead of a circular one).  
> **username_7**: Be aware that time to tidal lock scales, to a first approximation, with the _sixth_ (6th) power of the orbital radius (SMA).  
> **username_8**: `Alone, without any other influences` - which means it's not going to happen; there is no place within the universe where gravity does not reach... just maybe verrrry slowly in case of a planet without a sun, deep in intergalactic emptiness, if such a thing should ever happen for whatever reason.  
> **username_9**: @username_5 A planet without a star is called a *rogue planet* - it just needs to be big enough to be of "planetary mass", which is big enough to become roughly spherical, but not so big it turns into a star.  
> **username_10**: @username_4 "During the era of the dinosaurs a day on Earth was about 19 hours, now it's 24 hours." Do you have a source for this? I wish there were a nice chart of that over time. My quirky calculation shows it was 23 hours long 250 million years ago.  
> **username_4**: @username_10: That 19 hours was something I came across some years ago. I've edited the answer according to the latest info I found online - 21 hours 620 Ma ago & 23 hours 65 Ma ago).  
> **username_11**: Not mentioned, the moon is slowly drifting away from us.  Some of the Earth's lost rotation is due to the transfer of energy to the Moon's widening orbital path.  
> **username_12**: Also, I seem to have read somewhere that it would take longer for Earth to become tidally locked to either the Sun or the Moon than it would take for the Sun to go supernova.  
> **username_13**: @username_12 No, because the Sun will never go supernova. It's way too small. (Unless in the far distant future, when it's a white dwarf, and the Milky Way has merged with the Andromeda galaxy, it happens to collide with another white dwarf).  
> **username_12**: @username_13 Sorry, with supernova I meant a generic "star-shattering kaboom". I essentially was saying that I read somewhere that it would take Earth 10 billion years to tidally lock while the sun will turn into a Red Giant in 5 billion years.  
> **username_13**: @username_12 Sure, the Sun will gradually get hotter and then spend several billion years going through its red giant phases, emitting a fair bit more energy than it does now as well as spewing out a lot of gas & dust. (The Earth will be unliveable in a billion year or so, long before the red giant phases start). Eventually, the Sun *will* shed a large chunk of matter, creating a pretty [planetary nebula](https://en.wikipedia.org/wiki/Planetary_nebula) when it turns into a white dwarf & stops doing fusion. But I wouldn't exactly call that "star-shattering".  

---

### Answer by username_14 on 2022-08-24. Score: 12

In addition to the interaction with other bodies (moons, other planets, stars) as described in another answer, you also have gravitational waves produced by spinning body that isn't perfectly symmetrical. Which the earth isn't. Those waves carry some energy and angular momentum with them, which means the planet loses it.

Mind, that contribution is tiny. It's relevant only for extremly dense and massive bodies, so it wouldn't matter to a planet. But it is an effect.

> **Comments:**
> **username_13**: Even the gravitational radiation due to Earth's orbital motion around the Sun is tiny. According to https://physics.stackexchange.com/a/412990/123208 it's ~196 watts. It'd be interesting to compare the gravitational radiation due to the Earth's rotation to that of the orbit of the Earth & Moon about their barycentre.  
> **username_15**: So on the order of... a race between losing energy to gravitational waves and theoretical proton decay?  
> **username_14**: It's on the time scale of "way too long to matter", yes. For the earth specifically, it would've long since been swalloed by the sun as it expands into a red giant.  

---

### Answer by username_16 on 2022-08-25. Score: 6

In addition to tidal/gravitational effects, a planet might lose energy also due to electromagnetic emission, provided that the planet has a magnetic field and if the magnetic field dipole moment is **not** aligned to the rotation axis (this is indeed the case for the Earth, the "magnetic axis" is tilted by about 11°).

The component of the dipole orthogonal to the rotation axis, due to rotation, will radiate electromagnetic waves. The radiated power by such a dipole is (in far field condition, assuming an ideal dipole) proportional to $m^2 \omega^4$, where $m$ is the dipole moment and $\omega$ the angular velocity. A good reference on this is, for example, D. J. Griffiths *Introduction to Electrodynamics*, Chapter 11 "Radiation".

There is an exercise in the same book in which one is asked to compute such emitted power in the case of the Earth, just to find out that such power is really tiny - of the order of few $\mu\text{W}$.


---

### Answer by username_17 on 2022-08-25. Score: 1

Well, kenetic energy is of at least three types. Translational energy, rotational energy, and vibrational energy. As energy is neither created nor distroyed, the planet really does not lose energy, rather transfers it to other bodies. If it is in some sort of debrie field, it might impart its rotational energy into other masses which would show up as translational energy.
Think of playing billards. If a ball hits another ball off center, you will see that the hit ball has both rotational energy and translational energy. Eventually, drag will dissipate into heat energy (which is just kinetic [translational and vibrational] energy of the molecules).


---

