World Builder's Handbook Feedback

[Geir]

Thanks for the clarification. I went with a DM+8 for G & H atmospheres to reflect that they start at an order of magnitude more oppressive than even extremely dense insidious atmospheres.

I noticed today on p.105, a note on the Tidal Lock Status chart says to reroll an axial tilt of more than 3 degrees as (2D-2)/10 degrees for both 3:2 and 1:1 tidal locks; however, on p.107 in the summary section on the Effects of Tidal Lock, point 3 says to roll 1D on the Axial Tilt Table for 1:1 locks only. (I did a search of the forum, but it doesn't look like this specific thing has been addressed to date.)

So, do we:
1. Ignore the chart note on p.105 for both 3:2 and 1:1 locks, in favor of the rule only for 1:1 locks on p.107,
2. Apply the chart note for 3:2 locks and the p.107 rule for 1:1 locks, or
3. Ignore the rule on p.107 in favor of the chart note on p.105?

And, just in case I haven't said it a dozen times already, the process of creating systems and worlds with this book has been spectacular! Thank you!

Thanks. Okay, now you're making me think... basically your option 3.
On page 107, The summary points with the p. 105 chart notes, so page 107 #4 is consistent, but p. 107 #3 should be the same text as the first sentence of the chart note:
"On either a 3:2 or 1:1 tidal lock, If a world’s axial tilt is more than 3°, reroll the world’s axial tilt as (2D-2) ÷ 10 degrees"

(Pedantically, it has to be that way, because if you followed p. 107 #3 and rolled 1D, then you could get a 1 and a table whose lowest column is 2-4)

 

[Geir]

And, just to pile on, is it a safe assumption (at the very least for the sake of simplicity), that when a terrestrial world and a moon are in a mutual 1:1 tidal lock, the moon's orbital period (and, thus, its rotational period) is changed to be equal to the parent world's rotational period?

That becomes slightly problematic, because it can push the moon either too low (Roche) or too far (Hill). The statement on page 107 is:
"the length of the sidereal day of the locked body is equal to the period of the moon" so the length of the planet's day is changed to match the moon's orbital period, which is computed on page 77.

Yes, in reality it would be a little of both, but this is a point-in-time generation system, not an evolving system model - which would be more 'fun' but I don't see how it could be done without a computer and multiple iterations and lots more math.

 

[Levi Dash Card]

That becomes slightly problematic, because it can push the moon either too low (Roche) or too far (Hill). The statement on page 107 is:
"the length of the sidereal day of the locked body is equal to the period of the moon" so the length of the planet's day is changed to match the moon's orbital period, which is computed on page 77.

Yes, in reality it would be a little of both, but this is a point-in-time generation system, not an evolving system model - which would be more 'fun' but I don't see how it could be done without a computer and multiple iterations and lots more math.

Got it. Thanks! (I'm sure I'll be back...)

 

[GabrielGABFonseca]

The WBH has come and gone, so this isn't really an errata submission per see, but I did notice something today I hadn't before...

On handily building a few worlds with my copy of the book (the internet was out for most of today), I noticed that the Gas Giant sizing rules on page 55 have it so that their diameter and mass are uncoupled. Previously I didn't give it much thought, but today it dawned on me that due to the ranges involved and their independence to one another, you can get some shockingly dense "Gas Giants". Small Gas Giants can get up to 4.375 times as dense as Earth, Medium Gas Giants up to ~1.574, and Large Gas Giants can have densities a whopping 7.8125 times as dense as Earth — that's nearly twice as dense as Osmium!

So I just wanted to ask why this system of gas giant size/mass determination was chosen over a density-based one like that used for terrestrial planets. Looking at the density table, the Exotic Ice densities column seems like it would be a good fit for Gas Giants, even.

Say, for instance: to determine gas giant density, you roll 2D3+DMs; for Small Gas Giants and Large Gas Giants, the DM is +6; for Medium Gas Giants the DM is +3; If the Gas Giant's HZCO deviation is smaller than -1, DM-2.
This reflects the fact that gas giants larger than Jupiter start to become denser due to their own gravity compressing the gases, and that small gas giants (i.e.: ice giants) are made of generally denser gases than the true giants. The HZCO Deviation also accounts for their atmospheric envelopes being "puffed up" due to proximity to the star.
You'd then either determine the mass or diameter of the Gas Giant; personally, I prefer the mass method. This would give them the diameter ranges: SG 3.11–5.5⊕; MG 5.5–14.15⊕; LG 10.2–26.7⊕, assuming non-'puffy' giants.

I do want to emphasise that this is not intended as a "grrrr, ur doin it wrong!!1!" type of post on my part, I'm just genuinely curious. I love the WBH entirely too much to be angry at such a small thing!

 

[Geir]

The WBH has come and gone, so this isn't really an errata submission per see, but I did notice something today I hadn't before...

On handily building a few worlds with my copy of the book (the internet was out for most of today), I noticed that the Gas Giant sizing rules on page 55 have it so that their diameter and mass are uncoupled. Previously I didn't give it much thought, but today it dawned on me that due to the ranges involved and their independence to one another, you can get some shockingly dense "Gas Giants". Small Gas Giants can get up to 4.375 times as dense as Earth, Medium Gas Giants up to ~1.574, and Large Gas Giants can have densities a whopping 7.8125 times as dense as Earth — that's nearly twice as dense as Osmium!

So I just wanted to ask why this system of gas giant size/mass determination was chosen over a density-based one like that used for terrestrial planets. Looking at the density table, the Exotic Ice densities column seems like it would be a good fit for Gas Giants, even.

Say, for instance: to determine gas giant density, you roll 2D3+DMs; for Small Gas Giants and Large Gas Giants, the DM is +6; for Medium Gas Giants the DM is +3; If the Gas Giant's HZCO deviation is smaller than -1, DM-2.
This reflects the fact that gas giants larger than Jupiter start to become denser due to their own gravity compressing the gases, and that small gas giants (i.e.: ice giants) are made of generally denser gases than the true giants. The HZCO Deviation also accounts for their atmospheric envelopes being "puffed up" due to proximity to the star.
You'd then either determine the mass or diameter of the Gas Giant; personally, I prefer the mass method. This would give them the diameter ranges: SG 3.11–5.5⊕; MG 5.5–14.15⊕; LG 10.2–26.7⊕, assuming non-'puffy' giants.

I do want to emphasise that this is not intended as a "grrrr, ur doin it wrong!!1!" type of post on my part, I'm just genuinely curious. I love the WBH entirely too much to be angry at such a small thing!

I see what you're getting at.

There does seem to be a lot a variety in gas giant densities, but I didn't focus on anything other than randomness there. Medium gas giants don't seem too horribly out of bounds, but Small and Large have a statistical have a risk of getting a bit out of whack. Large giants, I suppose could be a 'stripped core'. But using the diameter, or a derivation of it, as the multiplier constant in the mass equation is a good 'fix' to look at for any future version.

And tossing in a DM for HZCO deviation could be a decent modifier - but I could argue for both 'puffy' and 'stripped core' results for hot giants, so maybe not...

I've added these thoughts to my little list of 'future errata'.

 

[GabrielGABFonseca]

And tossing in a DM for HZCO deviation could be a decent modifier - but I could argue for both 'puffy' and 'stripped core' results for hot giants, so maybe not...

I've added these thoughts to my little list of 'future errata'.

Personally, I'd argue that stripped cores of gas giants would be classified as terrestrials, in the form of Cthonian Planets. One could even argue that the very-high-density planets naturally generated in the system's hot zone could, through referee a posteriori interpretation, be said to be stripped gas giant cores.

All that being said, my emphasis has always been how planets move (i.e.: celestial mechanics), not as much what they're made of, so I'm a bit out of my depth here.

And given you're compiling notes for the future and we're talking gas giants, I'd love to swap notes with you about a method I've been tinkering with to generate dynamically stable moon systems for the gas giants at some point; using orbital periods and then converting them to Planetary Diameters/km, using a variation of the concept of Dermott's Law (which is as much of a law as the Titius-Bode """law""", but is a very useful little tool nonetheless).

 

[Levi Dash Card]

So, I was looking at hierarchies of tidal locks, and, after reading a little on the relative strengths of tidal pulls (in our case, Sol vs. Luna), I've devised a way to quantify relative pulls for determining the order of checking for the world's tidal lock. Here's what I wrote up for my own notes:

"A moon will almost always have a stronger tidal effect on a world than its star. (Luna has about twice the strength of Sol on Earth.) The relationship between a moon’s pull vs. a star’s pull is:

a/A = (m*0.000003*(R*150,000,000)^3/M*(r*D)^3), where

a is the moon’s tidal acceleration, A is the star’s tidal acceleration, m is the moon’s mass in planetary masses, M is the star’s mass in solar masses, r is the moon’s orbital distance in planetary diameters, R is the world’s orbital distance in AU, and D is the world’s planetary diameter in km.

For two moons of the same world, their relationship can be expressed as:

a/A = (m*R^3/M*r^3), where

a is the smaller moon’s tidal acceleration, A is the larger moon’s tidal acceleration, m is the smaller moon’s mass in planetary masses, M is the larger moon’s mass in planetary masses, r is the smaller moon’s orbital distance in planetary diameters, and R is the larger world’s orbital distance in planetary diameters.

In both cases, if the result is greater than 1, the smaller body has the greater tidal pull, and should be checked first. If the result is 1, then the pulls are equal, but the smaller body should still be checked first. If the result is less than 1, then the larger body has the greater tidal pull, and should be checked first.

A world can only be locked to one body. As such, use a hierarchy of tidal locks, comparing the moons to establish the order from greater to lesser effect, then comparing moons to the star to determine its place within the hierarchy (which I expect will always be last)."

 

[Geir]

So, I was looking at hierarchies of tidal locks, and, after reading a little on the relative strengths of tidal pulls (in our case, Sol vs. Luna), I've devised a way to quantify relative pulls for determining the order of checking for the world's tidal lock. Here's what I wrote up for my own notes:

"A moon will almost always have a stronger tidal effect on a world than its star. (Luna has about twice the strength of Sol on Earth.)

For "significant moons" you are probably right. At first blush I was going to challenge this assertion (with Phobos and Deimos for instance) but I think that for the Galileans at Jupiter and probably all of Saturn's "dwarf planet-sized" moons, you are probably correct. Didn't check the math, but given that Luna is Size 2 and far away, yeah.

The relationship between a moon’s pull vs. a star’s pull is:

a/A = (m*0.000003*(R*150,000,000)^3/M*(r*D)^3), where

a is the moon’s tidal acceleration, A is the star’s tidal acceleration, m is the moon’s mass in planetary masses, M is the star’s mass in solar masses, r is the moon’s orbital distance in planetary diameters, R is the world’s orbital distance in AU, and D is the world’s planetary diameter in km.

For two moons of the same world, their relationship can be expressed as:

a/A = (m*R^3/M*r^3), where

a is the smaller moon’s tidal acceleration, A is the larger moon’s tidal acceleration, m is the smaller moon’s mass in planetary masses, M is the larger moon’s mass in planetary masses, r is the smaller moon’s orbital distance in planetary diameters, and R is the larger world’s orbital distance in planetary diameters.

In both cases, if the result is greater than 1, the smaller body has the greater tidal pull, and should be checked first. If the result is 1, then the pulls are equal, but the smaller body should still be checked first. If the result is less than 1, then the larger body has the greater tidal pull, and should be checked first.

A world can only be locked to one body. As such, use a hierarchy of tidal locks, comparing the moons to establish the order from greater to lesser effect, then comparing moons to the star to determine its place within the hierarchy (which I expect will always be last)."

Makes sense.
The fun part is an unknown (or at least unknown to me) interaction factor - if the lessor tidal pull is still significant, it might disrupt or delay the lock. Obviously this doesn't prevent Jupiter's moons from being locked to Jupiter, but it might be a factor in Venus's odd retrograde rotation, which seems to correspond awfully closes to 5 rotations per closest approach - the Wikipedia article on Venus says this is discounted, but I'm not a big fan of coincidences and I've yet to find a concrete (or at least simple enough to be usable) set of conditions to force a lock.

 

[Levi Dash Card]

For the record, I only populate more detailed information (i.e., gravity, mass) for moons of Size 2 and larger. It's possible a Size 1 or S moon close to the parent world will have enough of a measurable effect on the world's rotation, but I have over 200 systems to build for my new pocket empire sector and need to save time somewhere (and those pesky "essentially a giant rock(s) in space" (as Spock put it) just aren't going to be that important with regards to creating the sector's astropolitical starscape).

 

[Levi Dash Card]

I'm trying to decide between two worlds as the Mainworld for one of my systems. One is SAH = 696, with an H.7.8 taint, RR 7, and Hab 7. The other is SAH = CD9, with an H.6.8 taint, RR C, and Hab 5. Neither world has native sophonts. I'm split between the two (696 has a deadlier High O2 taint, but is more habitable; CD9 has a non-lethal High O2 taint, but is less habitable, and both would make interesting mainworlds), while trying to keep to dice and objective decision-making for my sector.

The Final Mainworld Determination criteria considers Habitability, the presence of native sophonts, Resource Rating, and refueling. So, to that end, I think I have a decent formula for weighing these:

Habitability (H) - Habitability Rating x 2.5 (range of 0-25)
Native Sophonts Present (N) - Resource Rating x 1.5 (or 0, if extinct or no evidence of their existence) (0 or range of 3-18)
Resources (R) - Resource Rating (range of 2-12)
Better Refueling (F) - Size^2 x Hydrography / 400 (treat Hydrography as 0 if hydrogen isn't present in Fluid Oceans) (range of 0-5.625)

The sum of these four factors is the FMD Rating, for a theoretical range of 2-60.625. (I haven't calculated all the interactions that would affect the maximum rating, though I suspect a E98 temperate world, with sophonts present, very comfortable gravity, and maximized resources, would give you the best possible rating at 58.92. But, a variety of worlds would give you an H, N, R sum of 55.)

Overall habitability has the greatest individual weight, but the presence of native sophonts is an effectively-binary factor that, due to its significant weight, will likely make their homeworld the mainworld. The idea behind the refueling rating is that gas giants are a shared resource, so a world's water/fluid resources would be the primary determinant of advantage in this category.

In the above dilemma, world 696 has an FMD Rating of 25.04, and CD9 has a rating of 27.74. (Interestingly, their H, N, R sums are equal at 24.5, so F ends up being the determining factor. Bigger world, lots of ocean.)

I think it works well enough for my purposes, but I'm interested in hearing thoughts/suggestions for refinement.

 

[Geir]

Wouldn't E98 give you a probably high gravity super-earth with a tainted dense atmosphere?

I'm not sure the Size matters much for refueling, since even 1% of Earth's surface is 5 million square kilometers, enough to refuel any fleet... so maybe subtract something for no water, but size squared seems skewed- maybe just add half of the Hydrographics (rounding down) and treat as zero if it's not water (though both methane and ammonia actual have more hydrogen than water; even hydrochloric, hydrofluoric, and sulfuric acid have hydrogen - might cause a bit of a problem with the scoops and refineries getting eaten away though...)

Instead of scoring, I would filter it as sophonts automatically win (depends a bit on the setting - could be if primitives live on a horrible planet and there's a nice Earth-analog nearby, the nice world gets colonies and the primitives get a 'reservation world'), then add up habitability and resources ratings to see which looks best. In the above example the CD9 would get 17 points and 696 would only have 14 - but the setting could matter here as well - if it were a 2300AD setting with no ant-gravity for either comfort or surface-orbit interface, then I would penalize the bigger world.

But that's just opinion. The setting details are important to determination. Plus, it might be fun not to let the nicer world win: a crashed colony ship and people forced to live where they end up is a common enough trope. Or well, the world used to be called Ceti Alpha V... (technically still was, but you get the idea - bad things happen after the settlement).

 

[Levi Dash Card]

Wouldn't E98 give you a probably high gravity super-earth with a tainted dense atmosphere?

I'm not sure the Size matters much for refueling, since even 1% of Earth's surface is 5 million square kilometers, enough to refuel any fleet... so maybe subtract something for no water, but size squared seems skewed- maybe just add half of the Hydrographics (rounding down) and treat as zero if it's not water (though both methane and ammonia actual have more hydrogen than water; even hydrochloric, hydrofluoric, and sulfuric acid have hydrogen - might cause a bit of a problem with the scoops and refineries getting eaten away though...)

Instead of scoring, I would filter it as sophonts automatically win (depends a bit on the setting - could be if primitives live on a horrible planet and there's a nice Earth-analog nearby, the nice world gets colonies and the primitives get a 'reservation world'), then add up habitability and resources ratings to see which looks best. In the above example the CD9 would get 17 points and 696 would only have 14 - but the setting could matter here as well - if it were a 2300AD setting with no ant-gravity for either comfort or surface-orbit interface, then I would penalize the bigger world.

But that's just opinion. The setting details are important to determination. Plus, it might be fun not to let the nicer world win: a crashed colony ship and people forced to live where they end up is a common enough trope. Or well, the world used to be called Ceti Alpha V... (technically still was, but you get the idea - bad things happen after the settlement).

With low rolls on both composition and density, even an E world could have a gravity in the .7-.9 range (for a +1 Hab); the size also gives a +1 Hab, both of which would compensate for the -2 Hab from the tainted atmosphere. Hydro in a no penalty zone and a temperate climate would give you a 10 Hab. Extremely rare happenstance, but it fits.

I had considered sophonts being an automatic win vs. no sophonts, but I decided on the weighted "bonus" instead.

I'll have to consider the Hydro only vs. Size & Hydro. My thought process is that surface area is a function of the square of the radius (represented for our purposes by Size/2), so that a Size 8 world (12,800km dia.) with only 2 Hydro will still have about 4 times the raw surface area of water/fluids than a Size 2 world (3,200km dia.) with 8 Hydro.

 

[Levi Dash Card]

Regarding Primordial Systems rules (p.226), is it safe to say that any terrestrial world or moon in a system that is less than 1 million years old will always be Size 0 or S? Is there a bias toward 0 or S in that regard? (My current binary system - O3 V primary with a B4 V close companion - has an age of appx. 118,000 years.)

 

[Geir]

Given the age of the system, and since for both Protostars and Primordial systems I wrote the maximum size as the age in millions of years, I'd say Size 0 would be about it. I'd just make the system a giant planetoid belt starting where things got cool enough to not be vaporized rock, with possibly a few gas giant cores carving out blank spots.

 

[GabrielGABFonseca]

This is comically down the line from the original release, but...

Someone on the discord asked about the Tidal Heating Factor formula for moons, and if they'd done their computation right. In trying to help them, I suggested plugging in the values for Earth's moon and seeing if they matched what was given in the book as a way to test if they were using the right figures.

From the WBH, Page 126:

For a moon in orbit around a planet, use standard days for the value of period and millions of kilometres for distance. For Io, this formula results in a value of about 303,000. Plugging in the same numbers in for Enceladus results in 33,000. For Luna it is 52, a relatively inconsequential result. To address seismic and heating factors, divide by 3,000 to treat a result equivalent to Io’s condition as a tidal heating factor of 101 (Enceladus would be 11) and ignore numbers less than 1.

The formula given in the book is:
(M² * S⁵ * e²)/(3000 * a⁵ * P * m)
Where:

• M is the Primary's Mass (in M⨁)
• S is the World (i.e.: Moon's) size code
• e is the Moon's eccentricity
• a is the Moon's semimajor axis in millions of km
• P is the Moon's orbital period in days
• m is the World (i.e.: Moon's) mass (in M⨁)

For our Moon, those values are: M = 1; S = 2; e = 0.0549; a = 0.384399; P = 27.321661; m = 0.0123
Which when plugged into the equation give us:
(1² * 2⁵ * 0.0549²)/(3000 * 0.384399⁵ * 27.321661 * 0.0123) = (1 * 32 * 0.00301401)/(3000 * 0.00839288448 * 27.321661 * 0.0123) = 0.09644832/8.46144839481 = 0.01139855914

Which is a fair bit off of the given value of 52 for Luna.
Assuming that the formula given in the book as-presented is already incorporating the "divide by 3000" step, we still find a value of ≈ 34.195 and not the 52 given in the text.

I must assume I am doing something wrong, though what exactly I know not.

 

[mando501]

Hi. It's me, I'm the problem, it's me.

As long as we're bringing this here, a couple of other things tripped me up while working through this:

• The moon's orbital distance was given as 384.399 km on the example form, rather than 384,399. Since all other delimited numbers in the book use commas as separators and period as the decimal place, I'm assuming this was a simple typo
• The moon's orbital eccentricity was given as 0.549 on the example form. After seeing GAB's example above, I checked and realized that was also likely an error and should be 0.0549

 

[Geir]

Hi. It's me, I'm the problem, it's me.

As long as we're bringing this here, a couple of other things tripped me up while working through this:

• The moon's orbital distance was given as 384.399 km on the example form, rather than 384,399. Since all other delimited numbers in the book use commas as separators and period as the decimal place, I'm assuming this was a simple typo
• The moon's orbital eccentricity was given as 0.549 on the example form. After seeing GAB's example above, I checked and realized that was also likely an error and should be 0.0549

Typos for sure, and in this case they are certainly mine (actually its even worse, since the book has the distance as 389.399 - so doubly wrong.) I'll add it to my future errata document...

 

[Geir]

This is comically down the line from the original release, but...

Someone on the discord asked about the Tidal Heating Factor formula for moons, and if they'd done their computation right. In trying to help them, I suggested plugging in the values for Earth's moon and seeing if they matched what was given in the book as a way to test if they were using the right figures.

From the WBH, Page 126:


The formula given in the book is:
(M² * S⁵ * e²)/(3000 * a⁵ * P * m)
Where:

• M is the Primary's Mass (in M⨁)
• S is the World (i.e.: Moon's) size code
• e is the Moon's eccentricity
• a is the Moon's semimajor axis in millions of km
• P is the Moon's orbital period in days
• m is the World (i.e.: Moon's) mass (in M⨁)

For our Moon, those values are: M = 1; S = 2; e = 0.0549; a = 0.384399; P = 27.321661; m = 0.0123
Which when plugged into the equation give us:
(1² * 2⁵ * 0.0549²)/(3000 * 0.384399⁵ * 27.321661 * 0.0123) = (1 * 32 * 0.00301401)/(3000 * 0.00839288448 * 27.321661 * 0.0123) = 0.09644832/8.46144839481 = 0.01139855914

Which is a fair bit off of the given value of 52 for Luna.
Assuming that the formula given in the book as-presented is already incorporating the "divide by 3000" step, we still find a value of ≈ 34.195 and not the 52 given in the text.

I must assume I am doing something wrong, though what exactly I know not.

Or it's me... my spreadsheet has a^6, not a^5, but that was lost in the text. The correct formula should be a^6. Another oops on me... but at least in my calculation I got the value for a correct and the value for e as well.

 

[mando501]

Okay, updating my spreadsheet I have the following:

M = 1
S = 2
e = 0.0549
a = 0.384399
P = 27.321661
m = 0.0123

THF = (M^2 * S^5 * e^2) / (3000 * a^6 * P * m)
    = (1^2 * 2^5 * 0.0549^2) / (3000 * 0.384399^6 * 27.321661 * 0.0123)
    = (1 * 32 * 0.00301401) / (3000 * 0.00301401 * 27.321661 * 0.0123)
    = (0.096448) / (3.252572)
    ~= 0.02965

Still a bit off - are we missing something else? Have I done a dumb?

If I leave off the 3000 divisor, it comes up to 88 and change, which is closer.

 

[Geir]

Okay, updating my spreadsheet I have the following:

M = 1
S = 2
e = 0.0549
a = 0.384399
P = 27.321661
m = 0.0123

THF = (M^2 * S^5 * e^2) / (3000 * a^6 * P * m)
    = (1^2 * 2^5 * 0.0549^2) / (3000 * 0.384399^6 * 27.321661 * 0.0123)
    = (1 * 32 * 0.00301401) / (3000 * 0.00301401 * 27.321661 * 0.0123)
    = (0.096448) / (3.252572)
    ~= 0.02965

Still a bit off - are we missing something else? Have I done a dumb?

If I leave off the 3000 divisor, it comes up to 88 and change, which is closer.

I was wrong about being wrong…

Trying to reconstruct what I did 2+ years ago (with no notes, just spreadsheet scribbles)



Assumption One:

The factors I included for tidal energy change in the following manner: ~M^2*R^5* N* E^2/A^6

M=mass of planet

R = radius of world (or Size in this unitless case)

N = mean angular speed in orbit

E= eccentricity

A = orbital distance



From that, I considered N to be proportional to the circumference of the orbit / orbital period, so that’s proportional to 2*pi*A / P, where A is the orbital distance and P is the period… since the fudge factor to make the temperature work out is going to be a constant added later, we can ignore the 2*pi and we can reduce the power of A in the divisor from 6 to 5.

And this is the total amount of tidal energy.

So, Assumption Two:
The energy is going to be applied to the entire body and the actual temperature increase is therefore going to be proportional to the mass of the body heated – so 1/ Mass = the actual temperature increase on the body from tidal heating. (If you follow my assumption, all that energy goes into the object and it is applied in direct proportion to the mass – this is my weakest assumption, but there it is - I have a hindsight sneaky suspicion that it might be more correct to assume 1/ Mass^(1/4) ) .

So this gives the final formula as written:

~M^2*R^5* E^2/(A^5 * P * Mass of moon)

For Luna the values I plugged in were:

M =1

R = 2.17175 (here is where you’re going to get a difference for sure, since I took the radius of the moon divided by 1600 times 2 rather than just 2.)

E = 0.055 (not sure why I wasn’t so precise here, but at least I wasn’t off a full decimal point)

A= 0.3844

P= 27.322

M = 0.0123

Which should get you 51.81315 which rounds to 52, and then divide by 3000 to get 0.017271.



Is this correct? Don’t know. A real astrophysicist would probably complain a lot about my assumptions (especially the second one), but if you look at the Wikipedia article https://en.wikipedia.org/wiki/Tidal_heating (not my only source, but I never found anything more useful) it talks about things like “the imaginary portion of the second-order Love number)” and nope, I just wanted to get a number that gave a plausible temperature that varied in a plausible way based on values we already had.