# Tidal Heating Formula?

## Tidal Heating Formula?

I know there is a lot to this, but is there some sort of simplified (i.e. dumbed down) way to calculate the amount of heat generated by a gas giant on its moon? For example, in a world-building exercise, I generated a 69.6 earth mass gas giant with a moon that orbits at 504207 km in 5.08 days. The moon has a radius of 7224 km (very large, but it

-M-

__was__randomly generated), and a mass of 1.7 earth's. The base temp. came out to be 215°K...subsequent calculations of albedo, atmosphere and greenhouse gases bumped it up to 274°K. How much more would tidal heating elevate this? What would the formula look like, so I can run this on other worlds I've generated? Many thanks!-M-

**Sunchaser**- Planetesimal
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## Re: Tidal Heating Formula?

Wikipedia has an article with an equation for tidal heating.

https://en.wikipedia.org/wiki/Tidal_heating

https://en.wikipedia.org/wiki/Tidal_heating

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**Sirius_Alpha**- Admin
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## Re: Tidal Heating Formula?

Looks helpful, but how do I figure out the shear modulus and the dissipation factor? (Sorry, I'm an artist, not a physicist. But I want to make sure my space art is true-to-life.)

-M-

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**Sunchaser**- Planetesimal
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## Re: Tidal Heating Formula?

A conservative value of Q for terrestrial bodies is 100. Q is, as I understand it, dominated by the response factor of the atmosphere - for the solar system gaseous planets Q is >10

I am less clear on mu, but I take it that it will mostly depend on the

^{5}, and there is effectively no intemediate value taken. For thin (Earth-like, rather than Neptune-like) atmospheres, a terrestrial value for Q will suffice.I am less clear on mu, but I take it that it will mostly depend on the

*surface*composition of a planet, as that is where most of the tidal response occurs. An earth-like composition gives a silicate-dominated surface, and by looking through some chemistry papers (ew) it seems like 70 - 100 GigaPascals is a reasonable representation of mu for silicates.**Shellface**- Neptune-Mass
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## Re: Tidal Heating Formula?

If the atmosphere is more dense, and Q for Earth would be 100, then would Q be expressed as a product of the two numbers multiplied? Is it similar for determining mu as well?

Very helpful...

-M-

Very helpful...

-M-

**Sunchaser**- Planetesimal
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## Re: Tidal Heating Formula?

For Q, I'm afraid it's not that simple (tides are weird). I would advise sticking to a value of 100 for thin atmospheres, and then switching to a value of about 10If the atmosphere is more dense, and Q for Earth would be 100, then would Q be expressed as a product of the two numbers multiplied? Is it similar for determining mu as well?

^{5}- 10

^{6}for thick atmospheres (see Quinn et al.). Again, there doesn't seem to be much of a smooth transition between the two values.

Mu I am still not clear on, but I am not really sure how much the concept of shear makes sense to gases, since they obviously do not have a defined shape. Mu seems to generally decrease for weaker materials, so I suspect that it may become small for gases, if it is valid at all.

Since the mu * Q term is in the denominator, and Q is >10

^{5}for gaseous planets, unless mu is very, very small for gases, tidal heating has to be much smaller for gaseous planets than terrestrial planets. I can't recall ever seeing tidal heating of giant planets discussed in literature, so it may be negligibly small?

This sounds about right, seeing as if the tidal deformation acts at the surface, then for a gaseous planet the atmosphere can simply be deformed (similar to how the Earth's water is pulled by the Moon) without producing much heat, while for rocky planets the deformation will cause a lot of internal motion, and thus a lot of heat.

…So I imagine one can safely ignore tidal heating for bodies with atmospheres thick enough to take the brunt of the tides.

Hope that helps, somehow or other.

**Shellface**- Neptune-Mass
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## Re: Tidal Heating Formula?

Actually, it kinda does! Many thanks!

-M-

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**Sunchaser**- Planetesimal
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## Re: Tidal Heating Formula?

Although, I do have one other question about this: would the final number derived from the equation be added in to the satellite's base temp?

-M-

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**Sunchaser**- Planetesimal
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## Re: Tidal Heating Formula?

Well, the output is heat, not temperature.

I'm not wise enough to actually explain the difference, and I'm quite bad at thermodynamics anyway, so I'll just give the equation:

Q = c*m*ΔT, where Q is heat energy (q in the tidal heating equation) in Joules, m is the mass of the object being heated (kg), ΔT is the change in temperature (K), and c is the specific heat capacity of the object.

c is significantly dependent on the material's chemistry, though for the most common terrestrial elements, c is usually within a factor of a few. I couldn't find a value for Earth's c, but using some simplifications I get an approximate value of ~750 Jkg

From this I get a temperature change that is negligible. However, it seems to

I'm not wise enough to actually explain the difference, and I'm quite bad at thermodynamics anyway, so I'll just give the equation:

Q = c*m*ΔT, where Q is heat energy (q in the tidal heating equation) in Joules, m is the mass of the object being heated (kg), ΔT is the change in temperature (K), and c is the specific heat capacity of the object.

c is significantly dependent on the material's chemistry, though for the most common terrestrial elements, c is usually within a factor of a few. I couldn't find a value for Earth's c, but using some simplifications I get an approximate value of ~750 Jkg

^{-1}K^{-1}, which should generally be valid within a hundred or two for terrestrial planets. Presumably this should apply to your case.From this I get a temperature change that is negligible. However, it seems to

*always*be negligible, even for cases where it should not be (e.g Io). I suspect that the tidal heating equation is incorrect somehow, particularly as the units on either side of the equality do not come out the same. I will try to investigate the issue.**Shellface**- Neptune-Mass
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## Re: Tidal Heating Formula?

(Apologies for the delay, exams and such)

Solving for the units of the equation given in the wikipedia article, I get J = m

Using some google-fu, I came across this page, which gives the following equation for tidal heating

H = 21 k

--- ---- ---------- e

2 Q

Where G is the gravitational constant, M is the (host) planet mass, R is the planet radius, n is again the mean motion (1/period) of the moon, and a is the semi-major axis of the moon, all in SI units. e is eccentricity, and Q

The units of this equation solve correctly (kg⋅m

You haven't given values for R or e, but solving for the rest of the terms gives

q = 7.36... * 10

For, say, R = 6 Rearth (38226000 m), q = 6.01... * 10

If you can give your values for R and e, it will be possible to derive a value for ΔT.

Solving for the units of the equation given in the wikipedia article, I get J = m

^{2}, which is obviously not true (J = kg⋅m^{2}⋅s^{-2}). Thus, the equation seems to be incorrect.Using some google-fu, I came across this page, which gives the following equation for tidal heating

*rate*(i.e heat/time, in J/s)H = 21 k

_{2,0 }G M^{2}R^{5}n--- ---- ---------- e

^{2}2 Q

_{0}a^{6}Where G is the gravitational constant, M is the (host) planet mass, R is the planet radius, n is again the mean motion (1/period) of the moon, and a is the semi-major axis of the moon, all in SI units. e is eccentricity, and Q

_{0}is the same as Q from before. k_{2,0}refers to the Love number of the moon (and planet?) in some fashion; I have never understood their calculation, but I know that they are on order of unity, so I will take the very broad assumption of k = 1.The units of this equation solve correctly (kg⋅m

^{2}⋅s^{-3}= kg⋅m^{2}⋅s^{-3}), so it seems to be workable. One can derive that this can be made into an equation for heat (q) by removing the n term, as that removes the normalisation to orbital period.You haven't given values for R or e, but solving for the rest of the terms gives

q = 7.36... * 10

^{-11}* R^{5}* e^{2}JFor, say, R = 6 Rearth (38226000 m), q = 6.01... * 10

^{27}* e^{2}J, which is quite reasonable (note that e^{2}is probably <0.01).If you can give your values for R and e, it will be possible to derive a value for ΔT.

Last edited by Shellface on 25th May 2016, 11:29 am; edited 1 time in total

**Shellface**- Neptune-Mass
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## Re: Tidal Heating Formula?

Sorry...I was trying to get past the brain cramp all that gave me...(typical me, I bite off more than I can chew.)

The satellite in question has a radius of 1.1326 earth's, but I couldn't figure what e was for...if eccentricity, I gave it .02

Thanks again for your help!

-M-

The satellite in question has a radius of 1.1326 earth's, but I couldn't figure what e was for...if eccentricity, I gave it .02

Thanks again for your help!

-M-

**Sunchaser**- Planetesimal
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## Re: Tidal Heating Formula?

Haha, sorry, I got a bit carried away.

You're right that e is for eccentricity (I forgot to describe that), but the missing R is for the planet.

You're right that e is for eccentricity (I forgot to describe that), but the missing R is for the planet.

**Shellface**- Neptune-Mass
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## Re: Tidal Heating Formula?

R is 45837 (7.2 earth's), mass is 69.6 earths

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-M-

**Sunchaser**- Planetesimal
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## Re: Tidal Heating Formula?

Right, so the heat equation fills out like so:

q = 21 1 6.674 * 10

--- --- -------------------------------------------------------- 0.02

2 100 (504207 * 1000)

q = 5.96... * 10

This is quite a lot of energy. However, upon converting this to temperature…

q = c * m * ΔT

5.96... * 10

ΔT = 0.00078... Kelvin

…One finds that this causes a negligible temperature increase on the moon. This is because the moon is massive relative to the amount of energy it gains, so the energy-per-unit-mass imparted on the moon due to tides is rather small.

For comparison, I calculate ΔT = 1.9 K for Io (assuming, probably inaccurately, the same values for Q and c). Thus, even in extreme cases, tidal heating has a small impact on lunar temperature, though it may produce large amounts of energy.

Hopefully that's a satisfactory answer!

q = 21 1 6.674 * 10

^{-11}* (69.6 * 5.97 * 10^{24})^{2}* (45837 * 1000)^{5}--- --- -------------------------------------------------------- 0.02

^{2}2 100 (504207 * 1000)

^{6}q = 5.96... * 10

^{24}JoulesThis is quite a lot of energy. However, upon converting this to temperature…

q = c * m * ΔT

5.96... * 10

^{24}= 750 * (1.7 * 5.97 * 10^{24}) * ΔTΔT = 0.00078... Kelvin

…One finds that this causes a negligible temperature increase on the moon. This is because the moon is massive relative to the amount of energy it gains, so the energy-per-unit-mass imparted on the moon due to tides is rather small.

For comparison, I calculate ΔT = 1.9 K for Io (assuming, probably inaccurately, the same values for Q and c). Thus, even in extreme cases, tidal heating has a small impact on lunar temperature, though it may produce large amounts of energy.

Hopefully that's a satisfactory answer!

**Shellface**- Neptune-Mass
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## Re: Tidal Heating Formula?

Very interesting....I was thoroughly under the impression that almost every moon of decent size would be rendered too warm due to tidal heating. I guess now it would come down to its atmosphere/albedo/hydrosphere combo to adjust any temperature!

Thank you very much!

-M-

Thank you very much!

-M-

**Sunchaser**- Planetesimal
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## Re: Tidal Heating Formula?

The way I would approach this is to balance the energy being radiated by the planet with the energy being deposited by tidal and solar heating.

E.g. for Io, the power from tidal heating

Where

If the object is also being heated by a nearby star, you need to add the power being received from the star to the power received from tidal heating. For example, the day side of Io receives about 5.2x10

˝

(Note that the tidal power is halved and the right-hand side now begins with 2π rather than 4π since this is only considering half of the surface area)

Solving this I get a temperature of 148 K, as opposed to 145 K when I set the tidal heating power to zero.

Alternatively, assuming that Io efficiently transfers the heat from the Sun evenly across its entire surface:

Which gives me 127 K for the tidal+solar heating case, and 122 K for the case with solar heating alone.

E.g. for Io, the power from tidal heating

*P*_{tidal}is about 10^{14}W. Assuming this is evenly distributed across the surface of Io and radiated as thermal (black-body) radiation:*P*_{tidal}= 4π*R*^{2}⋅*σT*^{4}Where

*σ*is the Stefan-Boltzmann constant,*R*is the radius of Io and*T*is the blackbody temperature. Rearranging the equation and solving for temperature, I get*T*=81 K.If the object is also being heated by a nearby star, you need to add the power being received from the star to the power received from tidal heating. For example, the day side of Io receives about 5.2x10

^{14}W from the Sun (solar flux at Jupiter's orbit multiplied by the cross-sectional area of Io). So setting the model where the whole of the day side of Io is the same temperature, and none of the solar heating is transferred to Io's night side:˝

*P*_{tidal}+*P*_{solar}= 2π*R*^{2}⋅*σT*^{4}(Note that the tidal power is halved and the right-hand side now begins with 2π rather than 4π since this is only considering half of the surface area)

Solving this I get a temperature of 148 K, as opposed to 145 K when I set the tidal heating power to zero.

Alternatively, assuming that Io efficiently transfers the heat from the Sun evenly across its entire surface:

*P*_{tidal}+*P*_{solar}= 4π*R*^{2}⋅*σT*^{4}Which gives me 127 K for the tidal+solar heating case, and 122 K for the case with solar heating alone.

**Lazarus**- dG star
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## Re: Tidal Heating Formula?

Incidentally if I go to the paper Peale, Cassen & Reynolds (1979) where the activity on Io was predicted, they give the tidal heating formula as:

Google-fu suggests that

So, for a 1.7 Earth mass satellite with a radius of 7224 km in a 5.08 day orbit with eccentricity 0.02, using 50 GPa as the shear modulus and a

The isolated temperature would then be 134 K, which can be combined with the base temperature by adding fourth powers then taking the fourth root. So from a base temperature of 215 K, the final temperature is 222 K.

Of course, given the huge amount of tidal heating in this moon, it may be more appropriate to use the Io-like shear modulus of 2×10

For a final case, if I use Io's orbital eccentricity of 0.0041 (note that the free eccentricity is 0.00001, the higher value is maintained by the resonance with the other satellites) and the Earth-like shear modulus, the power drops to 5.1×10

*P*_{tidal}= (36π*ρ*^{2}*n*^{5}*R*^{7}*e*^{2}) / (19*μQ*)Google-fu suggests that

*μ*for Io is quite low, Google Books preview of*Io After Galileo: A New View of Jupiter's Volcanic Moon*by Lopes and Spencer gives an upper limit of 2×10^{8}Pa, while Earth seems to be at around 5×10^{10}Pa (typical solids are in the range 10^{10}– 10^{11}Pa, which suggests liquid plays a large role in the physics of Io's interior). Using this upper limit for*μ*and a*Q*value of 100 (seems to be the usual choice for terrestrial-type planets) I get about 5×10^{14}W for Io, which seems to be roughly in line with the observed value, which is nice.So, for a 1.7 Earth mass satellite with a radius of 7224 km in a 5.08 day orbit with eccentricity 0.02, using 50 GPa as the shear modulus and a

*Q*factor of 100, I obtain a tidal power of 1.2×10^{16}W. This translates into a tidal flux of 18 W/m^{2}(versus 2.4 W/m^{2}for Io), so this sounds like it would be an extremely volcanic world.The isolated temperature would then be 134 K, which can be combined with the base temperature by adding fourth powers then taking the fourth root. So from a base temperature of 215 K, the final temperature is 222 K.

Of course, given the huge amount of tidal heating in this moon, it may be more appropriate to use the Io-like shear modulus of 2×10

^{8}Pa. This increases the tidal heat power to 3.0×10^{18}W, giving a flux of 4600 W/m^{2}and a tidal base temperature of 535 K, which combines with the base 215 K to give 538 K.For a final case, if I use Io's orbital eccentricity of 0.0041 (note that the free eccentricity is 0.00001, the higher value is maintained by the resonance with the other satellites) and the Earth-like shear modulus, the power drops to 5.1×10

^{14}W, the tidal flux is then 0.8 W/m^{2}and the tidal base temperature is 60 K, adding in the solar heating base temperature of 215 K gives a 0.3 K increase.**Lazarus**- dG star
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## Re: Tidal Heating Formula?

Hypervolcanic does not sound good...

I changed some of the values: planetary and satellite masses are the same (69.59792 EM and 1.7 EM respectively)

Orbital distance is 797564 km with a period of 9.78 days (assuming I did the math correctly)

And for what it's worth; the pair orbits an F0V star with a mass of 1.78 solar and luminosity of 7.48 solar at a distance of 3.8653 AU.

Thanks!

-M-

I changed some of the values: planetary and satellite masses are the same (69.59792 EM and 1.7 EM respectively)

Orbital distance is 797564 km with a period of 9.78 days (assuming I did the math correctly)

And for what it's worth; the pair orbits an F0V star with a mass of 1.78 solar and luminosity of 7.48 solar at a distance of 3.8653 AU.

Thanks!

-M-

**Sunchaser**- Planetesimal
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