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Real-life physics in Bionicle

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I don't know if you've covered this or not, but Bara Magna gravity.  The Toa and Matoran lived on Aqua Magna for about 1000 years, which is a moon/fragment of Bara Magna.  So theoretically it would be smaller, right?  Therefore the gravity would also be lighter.  So how do they all land on Bara Magna and cope with the (theoretically) much higher gravity?

Edited by evil_jaga_genius

Avatar by Nicholas Anderson (NickonAquaMagna)

My blog: The Jaga's Nest

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I don't know if you've covered this or not, but Bara Magna gravity.  The Toa and Matoran lived on Aqua Magna for about 1000 years, which is a moon/fragment of Bara Magna.  So theoretically it would be smaller, right?  Therefore the gravity would also be lighter.  So how do they all land on Bara Magna and cope with the (theoretically) much higher gravity?

We've been over this quite a bit, just look back a few pages. I suggested the use of aerojel-like mineral to decrease the gravity of the mega-planet.

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What about Vakama absorbing the fire into his body wouldn't he melt?


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I don't know if you've covered this or not, but Bara Magna gravity.  The Toa and Matoran lived on Aqua Magna for about 1000 years, which is a moon/fragment of Bara Magna.  So theoretically it would be smaller, right?  Therefore the gravity would also be lighter.  So how do they all land on Bara Magna and cope with the (theoretically) much higher gravity?

You might have seen these by now, but here are the calculations I did for this: http://www.bzpower.com/board/topic/16353-real-life-physics-in-bionicle/?p=841140

 

RahiSpeak also did some three posts before mine.

 

The only incorrect part of my post is the comment about average density being the same as Earth's as long as SM has the same surface gravity. ToV pointed out that with my numbers, SM's average density is only about 50 kg/m^3, which is impossible for a rocky planet (or a gas giant of that size, for that matter). But with the masses and volumes that I calculated, the gravity will be the same for SM, AM and BotaM, as long as we can find a solution for the density problem (i.e. aerogel, or as I proposed, energized protodermis being able to oscillate between having positive and negative mass. I explained this a few posts later.). This is because, as I've said a few times, surface gravity is inversely proportional to the radius of a planet - the square of the radius actually, meaning that size is an exponentially greater factor in determining surface gravity than mass is.

 

-Letagi

Edited by Letagi

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This is because, as I've said a few times, surface gravity is inversely proportional to the radius of a planet - the square of the radius actually, meaning that size is an exponentially greater factor in determining surface gravity than mass is.

You may have said it a few times, but that doesn't necessarily make it true. ;)

Surface gravity is actually proportional to the radius of the planet.

If it were inversely proportional to the square, like you say, then a planet of twice the size would have a quarter of the surface gravity, which makes no sense.

 

You can see from the equations:

g=Gm/r^2 (where g is surface [acceleration due to] gravity, G is the gravitational constant, m is the mass of the planet and r the radius) and

m=VD=(4/3)(pi)(r^3)D (where V is volume of the planet and D is average its density)

 

Then by substituting: g=[G(4/3)(pi)(r^3)D]/[r^2] => g=(4/3)(pi)rD

ie g is proportional to r (sorry if the above is hard to read)

(@Letagi I think you just read the r^2 in the gravitational force equation and forgot that mass also depends on radius, which must be taken into account when working out the proportionality. :))

 

 

On the subject of the aerogel hypothesis; I pretty much disproved that as being something that could solve the problem. From my calculations, even with negligible density aerogel, you would have to fill 99% of Spherus Magna with aerogel to make the planet have the right mass to give reasonable gravity, given that the surface has been observed to be rock (realistic-density aerogel would require an even larger percentage)! There are other problems, not least that this aerogel would be exposed in the Shattering, but I shan't re-type them here. So some sort of solution with Energised Protodermis providing antigravity / absorbing gravity etc. is necessary.

(For the record, my calculations basically amounted to: average density of SM for desired gravity=50 kgm^-3, density of rock=5000 kgm^-3, 50/5000*100%=1%.)

Edited by Xelphene

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How about the masks of Growth and Diminishment? Weren't you supposed to die if you became a drastically different size? Simple spatial manipulation?

 

My theory is this:  Matter is mostly made of empty space.  Atoms are surrounded by clouds of electrons which repel each other and in between is your empty space.  So I think the masks would manipulate that space, making it bigger or smaller and as a result the size of the wearer would change.  I'm not sure if the wearer would die from changing shape, or how the masks would actually work but that could be what happens.

 

(BTW, thanks for the answers on gravity.)


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My blog: The Jaga's Nest

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Well, it could be a aerogel-like mineral that looks like stone. And nowhere does it specify the weight of the stone on the surface, so it's still plausible. In fact, it would have been very unlikely for Telluris to survive that rock slide if they were earth-like rocks...


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How about the masks of Growth and Diminishment? Weren't you supposed to die if you became a drastically different size? Simple spatial manipulation?

 

 

My theory is this:  Matter is mostly made of empty space.  Atoms are surrounded by clouds of electrons which repel each other and in between is your empty space.  So I think the masks would manipulate that space, making it bigger or smaller and as a result the size of the wearer would change.  I'm not sure if the wearer would die from changing shape, or how the masks would actually work but that could be what happens.

 

(BTW, thanks for the answers on gravity.)

I may not have a theory on this, but I can tell you right now that that cannot work. Movies like "Honey, I Shrunk the Kids" make it seem plausible that all one needs to do to make something grow or shrink is change the amount of empty space within each of the atoms. The physics, however, is pretty clear that this can't happen. Electrons are held where they are by electro-magnetic forces and quantum mechanics. To move them apart or push them together you would have to change physics; if you changed physics then it would sort of cease to be "real-life physics in BIONICLE".

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This is because, as I've said a few times, surface gravity is inversely proportional to the radius of a planet - the square of the radius actually, meaning that size is an exponentially greater factor in determining surface gravity than mass is.

You may have said it a few times, but that doesn't necessarily make it true. ;)

Surface gravity is actually proportional to the radius of the planet.

If it were inversely proportional to the square, like you say, then a planet of twice the size would have a quarter of the surface gravity, which makes no sense.

 

You can see from the equations:

g=Gm/r^2 (where g is surface [acceleration due to] gravity, G is the gravitational constant, m is the mass of the planet and r the radius) and

m=VD=(4/3)(pi)(r^3)D (where V is volume of the planet and D is average its density)

 

Then by substituting: g=[G(4/3)(pi)(r^3)D]/[r^2] => g=(4/3)(pi)rD

ie g is proportional to r (sorry if the above is hard to read)

(@Letagi I think you just read the r^2 in the gravitational force equation and forgot that mass also depends on radius, which must be taken into account when working out the proportionality. :))

 

 

On the subject of the aerogel hypothesis; I pretty much disproved that as being something that could solve the problem. From my calculations, even with negligible density aerogel, you would have to fill 99% of Spherus Magna with aerogel to make the planet have the right mass to give reasonable gravity, given that the surface has been observed to be rock (realistic-density aerogel would require an even larger percentage)! There are other problems, not least that this aerogel would be exposed in the Shattering, but I shan't re-type them here. So some sort of solution with Energised Protodermis providing antigravity / absorbing gravity etc. is necessary.

(For the record, my calculations basically amounted to: average density of SM for desired gravity=50 kgm^-3, density of rock=5000 kgm^-3, 50/5000*100%=1%.)

 

I was going to write a big long reply to this, but instead, I'll ask you a very simple question: what happens to the density term in your rearranged equation as radius increases? Work with it a bit more and you'll see that these are actually the exact same proportionalities as those that I've described before. Remember that density is mass divided by volume, and that volume is proportional to the cube of the radius. The radii cancel and we're left with radius squared in the denominator.

 

I wouldn't have said it multiple times if I weren't absolutely certain about it. ;)

 

I totally agree with you on the aerogel thing though. Aside from the problems you mentioned, there's very little chance it could be produced in stellar reactions. I prefer my idea about EP having negative mass under certain circumstances and thus producing antigravity.

 

-Letagi

Edited by Letagi

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@Letagi

Nothing happens to my density term - it is a constant - that is the point of rearranging to a form that involves it.

 

By ignoring that the mass is also a function of radius, you are basically assuming that the mass is constant as the planet gets larger, which means that the density rapidly decreases. In real life, which is what I believe we are discussing here, as a planet gets larger its mass will increase. Its density is constant, because the material a planet is made of does not get lighter just because the planet is bigger.

 

So g=(4/3)(pi)rD is the appropriate equation, and therefore surface gravitational force is proportional to the radius of the planet (with the natural assumption that the density of the planet is constant).


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You're right that mass is a function of radius. As a planet gets bigger, more mass is added. I never ignored this. But you're not realizing that the density term in that equation is not always a constant. ToV hit the nail on the head, more succinctly than I could have probably, but I'll add a bit more detail. This is something I discussed a few pages back but it turned out not to be the correct idea at the time.

 

Pressure inside a planet is enormous. The pressure at Earth's inner core is so great that the iron is solid, despite being at well over its melting point. Most of Saturn and Jupiter are not made of gas, but rather metallic hydrogen, due to the pressure at those depths.

 

If we want to increase the radius of a planet, we have to add mass, as you say. Depending on the material, this can go two different ways: either we have to add mass faster than radius, or the radius can increase faster than mass; however, these two possibilities result from different mechanisms. I'll give you an example of each.

 

Jupiter and Saturn are very close to each other in radius; Saturn's radius is about 85% that of Jupiter. However, Jupiter has more than three times Saturn's mass. This means, as we add more and more mass, that the radius increases relatively slowly. Very slowly, in fact. The effect of this is that Jupiter's surface gravity is 2.5 times greater than that of Saturn, despite having very similar radii. We can add mass to either planet all we want, and as you say, radius will increase correspondingly (or vice-versa); but due to pressure, the average density still changes.

 

An example of the converse is Earth. Due to its rotation, Earth is flattened at the poles and elongated at the equator. This means that there's more mass distributed throughout Earth horizontally and less vertically. Since average density stays the same, then according to your formula, gravity should be stronger at the equator. The opposite is true: gravity is stronger at the poles. There's less mass under the poles, creating less gravity, but there's also less radius. And when we look at Newton's equations for gravity, we see radius squared in the denominator. Therefore, and as you see from this example, radius has a greater and opposite effect on gravity than mass.

 

To sum up: the only way in which surface gravity can increase along with radius is if the density changes, as in the Jupiter and Saturn example. If average density stays the same, then radius is a more significant determining factor than mass with regards to surface gravity, as in the example of Earth's flattening, and serves to decrease gravity as radius increases. So yes, a planet with constant average density and increasing radius (if such a thing were possible, which it isn't; refer to Saturn and Jupiter) would indeed have decreasing surface gravity. Your equation doesn't work if density stays constant. If you need me to pull out more formulas I can; we just covered this a few weeks ago in my university geophysics course.

 

-Letagi

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That's your problem, right at the start. When you say "surface gravity is inversely proportional to the square of the radius", you are looking at the gravity equation and assuming that the mass of the planet stays constant as you increase its size. That is what I am disagreeing with. If you accept that mass increases, you need to substitute in the expression for mass in terms of radius to be able to see what is actually going on.

 

I agree with what ToV said too. It makes sense that density increases slightly as the planet has more mass added. (The amount density increases by can be ignored to a certain extent due to its small size; the density of rock does not vary that much under pressure.) The thing is, that is exactly the opposite of what your use of the equations would suggest. By holding the mass constant, you are saying that as the planet's size increases, its density massively decreases (due to D=m/V).

 

I'm not going to go into analysing Jupiter and Saturn, because their compositions vary so can be expected to have different densities. And I'm not totally sure what you mean about the rates of radius increase compared to mass, seeing as mass is always proportional to the radius cubed, with the small corrections mentioned above for pressure increasing density (which could only force mass to increase more rapidly in relation to radius). But at no point can the radius "increase faster than mass" like you claim above.

 

Regarding your example on Earth, it is an interesting effect but unrelated to the matter at hand. The poles have greater gravity than the equator because, as you say, the effective radius is less, but also because (and this is the crucial point) the mass attracting the objects on the surface in each case is the same (ie the mass of the whole Earth) because it is the same planet. That is a completely different scenario to planets of different radii, and therefore different masses.

 

Finally, regarding your summary; most of it is incorrect (not wanting to be rude, but there isn't a nice way of putting it, sorry). Surface gravity DOES increase in proportion to radius if you keep the density constant. As shown by the equation g=(4/3)(pi)GrD, which I derived from the gravity equation and the equation for the mass of a sphere.

(Side note: Oops, I think I forgot the G constant when I wrote it above; my points still stand though.)

Of course the equation works if you keep density constant (why wouldn't it?), and shows that surface gravity increases as radius increases. (I'm not sure why you think it predicts a decrease - maybe you misread - but that is exactly the thing you said earlier that I disagree with.)

 

tl;dr I hope someone else with physics knowledge comes along and helps resolve this, because these posts are getting loooooong. :P


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Hmm. If I increase mass, that object's gravity would increase. I'm inclined to say that wouldn't affect the density at all. If density increases with gravity, one would have a lot of planets imploding on their own gravity. Interplanetary collisions don't cause things to implode as far as I know. 

 

Gravity is the force of a mass of an object that acts on other objects, not on itself. 

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With regards to the Earth example, it is very relevant. More so than the Jupiter and Saturn example. When we calculate gravity at the poles and at the equator, we use different values for radius. The reason is that the surface gravity at any point on a planet is related to the distance to the centre of mass. That distance (radius) is smaller at the poles and larger at the equator.

 

We also have to take into account the different masses. Since we have two different radii, we also effectively have two different masses; in other words, there's more mass distributed along the equator, and less along the polar axis. We don't calculate these masses specifically, but rather we use a formula that finds gravity directly.

 

This is, the more I think about it, actually a perfect example, because it's one in which both the mass and radius change, but average density stays the same (or the difference is negligible). It doesn't matter that these two measurements are on the same planet.

 

There are three factors that determine the surface gravity at any point on a planet: radius, mass distribution and rotation. Rotation is a factor because it causes centrifugal force, but this has a very small impact. We'll just calculate gravity from radius and gravity from mass distribution.

 

We can calculate each one independently. Let's do radius first. We're not taking mass into account just yet; radius is the only variable we have to worry about here. re is equatorial radius, 6378 km. rp is polar radius, 6357 km.

 

First, equatorial surface gravity:

 

g= GM/re2

 

g= [(6.67*10-11 m3/kgs2)(5.972*1024 kg)]/[6.378*10m]2

 

g= 9.79211720754 m/s2

 

Now, polar surface gravity:

 

g= GM/rp2

 

g= [(6.67*10-11 m3/kgs2)(5.972*1024 kg)]/[6.357*10m]2

 

g= 9.85691950813 m/s2

 

Finding the difference,

 

gp - ge = 9.85691950813 m/s2 - 9.79211720754 m/s2 = 0.06480230059 m/s2

 

Gravity is greater at the poles when we only look at radius and ignore mass distribution.

 

Now, moving on to mass distribution. This is a little more complex. The formula is as follows:

 

g = [GM/r2] - (3/2)(GM/r2)(J2)(3cos2θ - 1)

 

Where θ is the colatitude (not to be confused with latitude) and J2 is the dynamic form factor for Earth, a constant measured to be 0.001082626.

 

As a side note, you may notice that the formula does appear to take radius into account, since we use the two different radii below. But it doesn't use those radii values to calculate gravity; rather, it uses them to calculate mass distribution, from which the formula then calculates gravity.

 

There are some pretty tricky derivations using calculus to get this formula that I haven't memorized, don't fully understand, would take an hour to copy down here, and are outside the scope of this topic anyways. I can provide them if you're really curious and/or suspicious that I'm making stuff up. But here are the calculations:

 

For equatorial surface gravity:

 

ge = [GM/re2] - (3/2)(GM/re2)(0.001082626)(3cos290 - 1)

 

ge9.79844119644 m/s2

 

For polar surface gravity:

 

gp = [GM/rp2] - (3/2)(GM/rp2)(0.001082626)(3cos20 - 1)

 

gp9.74441180446 m/s2

 

Finding the difference,

 

gp - ge = 9.74441180446 m/s2 - 9.79844119644 m/s2 = -0.05402939197 m/s2

 

Gravity is greater at the equator when we only take into account mass distribution. Let's add the two numbers we've found and see what happens.

 

0.06480230059 m/s2 + (-0.05402939197 m/s2) = 0.01077290861 m/s2

 

Gravity is larger at the poles by a small margin. The reason for this, if it's not clear from the above math, is that radius has a larger and inverse effect on surface gravity than mass does. It's not that your rearranged equation doesn't work; I just tried it using Mercury's data, and g matched up perfectly with the value that Google gives me. But when we talk about changing masses and radii, things get more complicated.

 

The point I was trying to make with Jupiter and Saturn is that, in those cases, it is fundamentally impossible to change radius and mass without increasing average density due to the effect of pressure. This is what I meant by mass increasing faster than radius. You add mass, but the radius doesn't increase as much as you expect it to due to the exponentially increasing compressional effect, and the density thus increases. I think - and I admit I have no math to back this up, but it seems reasonable - that the opposite would be true for rocky planets. Solid materials should have enough resistance to pressure that radius can increase at an appropriately high rate relative to the addition of mass, causing a decrease in average density. This is the fundamental reason why your rearranged equation doesn't work in this case: the density term by definition cannot remain constant. In a case where mass increases faster than radius (exaggerated compressional effect; gas giants), it increases. In a case where radius increases faster than mass (negligible compressional effect; rocky planets), it decreases.

 

But the polar versus equatorial Earth example is even better, because it's a rare case where density does stay constant. However, the only reason density stays constant is that we're not adding any mass or distance to the radius; we're just taking measurements at different points on the surface where those values happen to differ already.

 

In response to fishers64: It's not that simple. Adding mass causes an additional compressional effect, which results in the radius not increasing as much as you would think, and the density correspondingly does increase. See the Jupiter and Saturn example. As for objects not exerting gravitational forces on themselves, they actually do. This is how objects with sufficiently high mass become spherical in space, and how things with sufficiently high mass and sufficiently low radius collapse into black holes.

 

-Letagi

Edited by Letagi

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Mmm, many numbers. But still irrelevant.

 

The complex mathematics of surface gravity due to a deviation from a spherical shape has no bearing on the question of whether a (roughly spherical) planet made of the same stuff as another (roughly spherical) planet, but with a bigger radius, has stronger surface gravity. (Hint: it does, both according to common sense and to the equations.)

 

All you need for the question at hand is g=(4/3)(pi)GDr, which is surface gravity on a spherical planet. Density can be considered a constant because we are looking at rocky planets, and the density of rock changes very little under pressure.

 

You also seem to misunderstand density. Density is a property of a particular material that is dictated by the mass of each atom in the material and the separation between the atoms. Yes, density increases noticeably in larger gas giants because the gas becomes compressed into a solid. But your suggestion that the density of rocky planets decreases as they get bigger is utterly wrong, because that would somehow require the atoms to get further apart in the material even though they are under more pressure from the added mass.


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I wouldn't have spent an hour doing the calculations and typing them out if I thought there was the slightest chance of them being irrelevant. The point of the calculations for surface gravity anomaly due to radius is to prove to you that as radius increases, surface gravity decreases. The point of the calculations for surface gravity anomaly due to mass distribution is to show that mass has an less significant impact on surface gravity than radius, and that the effects are opposite. Density doesn't play a role in these calculations. Maybe they're less pertinent than I thought at the time, but still very much relevant.

 

My understanding of density is quite sufficient. But you're right that I got ahead of myself when I said that the opposite effect would happen in rocky planets. Note to self: don't to physics in the middle of the night...

 

But here's the thing: we can't both be wrong, because the derivations you did are correct. We're working with essentially the same equation.

 

Here's something we should both be able to agree on. As we increase radius of a planet, density must necessarily increase as well due to pressure. There's an inverse-cube relationship between mass and radius with regards to density. If we try to increase radius by a factor of 3, mass must increase by a factor of 27 in order to keep average density constant. This is fine, because a radial increase of 3 times also means a volume increase of 27 times, leaving exactly enough room for the new mass. But, the additional mass means that the compressional effect increases as well, meaning that we get less than a factor of 3 in our final radius increase - or, we need to add more than 27 times the original mass to achieve the desired 3 times radius increase. Either way, the density goes up. This proves that you can't add mass and radius to a planet and expect to have the same average density. Density will always increase.

 

This is the reason why your rearranged equation makes it look like surface gravity is proportional to radius. In fact, the important variable in that equation is density. Density always increases, and so surface gravity increases as well. Another way of saying this is that you have to add more mass to attain a relatively small (and getting exponentially smaller) increase in radius.

 

We both made the same mistake, but drew different conclusions from it: looking at your equation and thinking that density can stay constant when radius and mass both increase. It can't. It's fundamentally impossible.

 

-Letagi

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How about the masks of Growth and Diminishment? Weren't you supposed to die if you became a drastically different size? Simple spatial manipulation?

Kinda. Obviously, you can't create something from nothing. So that means the mask of growth pulls in matter from a pocket dimension. Though, certain wavelengths of radiation can convert energy directly in matter, such as Gamma-rays. It explains Bruce Banner's exponential growth when he transforms into the Hulk.

 

As for the mask of dimishment, that's a toughy. I guess theoretically you can have a device that compresses the space in-between each individual atom of the user, without any ill effects--the limit a user can shrink down to is six inches based on the Bionicle Wiki, so I guess that makes sense.

Edited by The Bone-Snatcher

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I wouldn't have spent an hour doing the calculations and typing them out if I thought there was the slightest chance of them being irrelevant. The point of the calculations for surface gravity anomaly due to radius is to prove to you that as radius increases, surface gravity decreases. The point of the calculations for surface gravity anomaly due to mass distribution is to show that mass has an less significant impact on surface gravity than radius, and that the effects are opposite. Density doesn't play a role in these calculations. Maybe they're less pertinent than I thought at the time, but still very much relevant.

 

My understanding of density is quite sufficient. But you're right that I got ahead of myself when I said that the opposite effect would happen in rocky planets. Note to self: don't to physics in the middle of the night...

 

But here's the thing: we can't both be wrong, because the derivations you did are correct. We're working with essentially the same equation.

 

Here's something we should both be able to agree on. As we increase radius of a planet, density must necessarily increase as well due to pressure. There's an inverse-cube relationship between mass and radius with regards to density. If we try to increase radius by a factor of 3, mass must increase by a factor of 27 in order to keep average density constant. This is fine, because a radial increase of 3 times also means a volume increase of 27 times, leaving exactly enough room for the new mass. But, the additional mass means that the compressional effect increases as well, meaning that we get less than a factor of 3 in our final radius increase - or, we need to add more than 27 times the original mass to achieve the desired 3 times radius increase. Either way, the density goes up. This proves that you can't add mass and radius to a planet and expect to have the same average density. Density will always increase.

 

This is the reason why your rearranged equation makes it look like surface gravity is proportional to radius. In fact, the important variable in that equation is density. Density always increases, and so surface gravity increases as well. Another way of saying this is that you have to add more mass to attain a relatively small (and getting exponentially smaller) increase in radius.

 

We both made the same mistake, but drew different conclusions from it: looking at your equation and thinking that density can stay constant when radius and mass both increase. It can't. It's fundamentally impossible.

 

-Letagi

You're right that we can't both be wrong. But neither can we both be right, seeing as we disagree. :P

 

Sorry, your hour was wasted on those calculations, because they don't allow us to compare different sized planets. They only compare places on the same planet, with the same total mass.

 

But all this talk about density not staying constant can't save you, because you are arguing that density must always increase as radius increases (something I agree with, but I believe the effect can be ignored due to its small magnitude), which contradicts the conclusion you are trying to make.

 

As we know, g=(4/3)(pi)GrD.

If D increases as radius increases, as you say, then we could model that as D=Br^k, where k is a small positive number, and B is just a constant to make the rest work. (I would argue that D can just be treated constant, but for a planet made of more compressible stuff, or to illustrate what you are saying, I'll do this.)

Then substituting that in, g=(4/3)(pi)Gr*Br^k => [combining the constants for clarity using C=(4/3)(pi)GB] => g=Cr^(1+k)

So that tells us that surface gravity is proportional to radius to the power of (1+k), remembering that k is positive. So from that equation there is NO WAY of reaching your "gravity is inversely proportional to radius squared", because that would be g=Cr^(-2).

 

Even though I constructed that expression using an equation you agree is correct and your own views on density, it still disagrees with your inverse square idea. I don't think you will be able to find any way to debate that. ;)


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All that makes sense. But you still haven't addressed the fact that radius is squared in the denominator of Newton's gravity equation. Hence, an inverse square relationship. You don't even have to worry about density to see that.

 

Frankly, we may very well both be right. The Earth example proves the inverse square relationship, and your last post indicates that the rearranged equation works. Would you mind if I consult my geophysics prof on this one? :P Not because I'm determined to prove you wrong, but more because I'm genuinely curious about this.

 

-Letagi

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Well, the radius squared in the denominator only remains when mass is assumed to be constant, ie not a function of radius. That is the only time that you get your inverse square relationship. It works for your Earth example because the mass will always be constant if you only look at the one planet. (Most of the time, the "r" is used to mean separation from the centre of mass of a particular object in this form of the equation, rather than just radius. It basically says that gravitational force decreases with the square of the distance to the object.)

 

But when considering planets of different sizes, as we are for Spherus Magna, mass varies and is a function of radius. In that case, in order to work out the proportionality between gravity and radius, you must make sure all your terms are written in terms of radius. Hence you substitute in that mass is proportional to r^3, which is then divided by the r^2 in the denominator. That gives that g is directly proportional to r, which is the correct relationship for roughly spherical planets of differing sizes, with similar compositions.

 

Anyway, I'd be interested to see what your professor says.


:v: :m_o: :t: :u: :k: :m_o:

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