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Lower mass Moon 
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#1
Sep2413, 06:15 AM

P: 4

Hi everybody,
I've registered to PF because after searching the web and this forum too, I couldn't find a clear answer to my dilemma. I've read this article but something seems wrong: http://www.scientificamerican.com/ar...halfmassmoon What if the mass of the Moon would be lower, while Earth's mass is constant and also Moon's tangential velocity is constant? Would the Distance between the centers of Moon and Earth change? Would the (mean) Radius of Moon's orbit change? According to the article above, "A less massive moon would also orbit closer to Earth than the real one". Well... how come and why? According to my calculations this is actually the other way. First of all, I know that there are some factors I'm not taking into account (see below). If we oversimplify, ignore the barycenter and consider Moon orbits the center of Earth then the answer is very simple. Orbit (distance between Moon's center and it's orbit rotation center) is the same as distance between centers of the two and the orbit (and distance) does not depend on Moon's mass, but only on it's velocity (and Earth's mass): GmM/D^2 = mv^2/R and we approximate R=D, we raise by D^2 and simplify by m GM=v^2 *D D=GM/v^2 and it's the same no matter what is the mass of the Moon. Now, considering the Moon orbits the barycenter, we have: D = distance between centers, R = radius of Moon's orbit / m = mass of Moon r = radius of Earth, orbit / M = mass of Earth r+R =D R = D * M / (m + M) = D / (1 + m/M) ; the larger the mass difference between the two, the smaller m/M, and the more D is closer to R we go back to GmM/D^2 = mv^2/R or GM/D^2 = v^2/R now we have: GM/D^2 = v^2/[D/(1+m/M)]=(v^2)*(1+ m/M)/D we multiply by D^2 GM=(v^2)*(1+m/M)*D D = GM / [(v^2)*(1+m/M)] = GM / [((M+m)/M) * V^2] = GM^2 / [(m+M)v^2] This would clearly show that if m becomes lower, D will be greater, in contradiction with the article mentioned above. This doesn't take into account that:  D is not constant (but I suppose its variations can be neglected)  orbits are not perfect circles  external influences and I don't know any other possible factors BUT... I find it hard to believe the missing factors not taken into account could possibly affect the result so much that the conclusion would change. So ... where did I go wrong? Please note that the article is signed or the information in it attributed to a University professor of physics and astronomy. Could the information be wrong? Also please note: R = GM^3/v^2(m+M)^2 so not only the distance will be larger, but also the barycenter moves towards Earth's center and the Moon's orbit Radius will be higher by a square factor. Thanks in advance for any comments and replies. 


#2
Sep2413, 07:17 AM

P: 709

Hi paulcretu, welcome to PF!
Perhaps you've heard that the Moon is drifting away from Earth in its orbit? This is caused by tidal bulges in Earth's crust(and of the oceans) getting displaced from the perfect EarthMoon alignment by our planet's rotation, and thus pulling on our satellite tangentially as well as radially. This produces torque and boosts the Moon's orbital energy, raising its orbit. The strength of the tides depends strongly on the distance between the two bodies, but also on mass of the body causing the deformation. Assuming we start with the same initial distance after the collision that spawned the Moon, the lower mass Moon would cause less deformation and get boosted with less force. So over the same period of time between its birth and now, it'd end up drifting away less than it has(and the Earth would have respectively shorter days as the flip side of the same process). The article also states in the last paragraph on the first page that the lower mass Moon would be easier to pull on(less inertia), but overall the result would be net decrease in torque. I don't know how to calculate that, but I'll go with the author's expertise. 


#3
Sep2413, 08:43 AM

P: 4

Maybe I've posted in the wrong section? Maybe some additional information is needed... I all came to me from the idea "what happens if we mine moon, make factories there and use Moon as a spaceport, and construct everything there". OK, this is maybe a childish dream (or not) but for the sake of the argument... Leaving velocity aside (we could launch simultaneously equal masses from opposite points to minimize impulse change) I was wondering: all right, assuming we can find all the required materials there to build factories/robots that in turn will build spaceships/solar panel satellites/whatever (maybe we send the missing materials from earth like plastics/polymers/rare metals/semiconductors)... that means every finite product leaving the Moon will decrease its mass... OK again... in a very long time. So I was just wondering  we decrease Moon's mass, but then, what happens with its orbit? That's how I started looking for some information on the subject and the only site I've found was that article. Nothing personal with the site or the author (I'm not even on the same continent! :) ). I liked physics in highschool and from time to time I like writing some rusty formulas. Now I'm still puzzled about the sentence "A less massive moon would also orbit closer to Earth than the real one" somewhere in the middle of the article  no explanation given. I was hoping if I test this I can get a confirmation. 


#4
Sep2413, 09:23 AM

P: 386

Lower mass Moon
Half mass Moon at 80 % distance would:
The difference would be shorter month (70 % present). And also, the outspiral of the moon would not slow down the Earth rotation as much. 


#5
Sep2413, 11:14 AM

P: 709

It would seem I wasn't clear enough in my response.
paulcretu, the article and you are talking about two different things, that's why you get seemingly conflicting answers. If you halve the mass of the Moon now, it'll drift away in it's orbit, just as you've shown. If the two bodies were just point masses that would be the end of it. However, since they're large and not perfectly rigid, the tidal interactions also play a role. Halving the mass of the moon will produce 1/4th of the tidal torque at the same orbital radius(tidal torque is dependent on the second power of the mass of the tiderasing body, so it doesn't cancel out with inertia: http://arxiv.org/pdf/1209.1615v3.pdf (eq.1 p.5) ). Over time the lower torque will result in less of an orbital raising. So the change in mass does not have only the oneoff, immediate effect, but also longterm effect on the evolution of the system. In your elaborated scenario with spaceships, the Moon would recede in response to lowering of mass. But, from that point on, the tidal torque would be much lower causing much slower recession in the future. 


#6
Sep2513, 02:00 AM

P: 4

Okay. Thank you all. So this:
D = GM^2 / [(m+M)v^2] ignores masses are not point masses. But I suppose it's valid for nonpoint masses too (with some corrections maybe)? And this: T_{z}=(3/2)GM_{1}^{2}(R^{5}/R^{6})k_{2}sin2εg takes into account the tidal torque. And as I understand lower torque leads to lowering the orbit. So raising mass (assuming M1 is the same as moon's mass, m) lowers the torque proportionally with M1 squared and this leads to a decrease in distance (but is this also proportional to the square of the mass ????). On the other hand also D in the above will decrease proportionally with m squared. So I'm still not convinced :) If the current drift is 4cm/year we have a speed. If we'd had an acceleration we could calculate a force. If speed is constant, there is (almost) no resultant force. But let's assume there IS a (resultant) force (a tiny one) and points (currently) towards raising the distance. If the Moon's mass would be half, the force would be .... 4 times lower? But that wouldn't change the vector's direction would it? So the Moon would STILL be drifting away but slower. So... if we neglect the barycenter  orbit is the same IF we consider barycenter  orbit is larger IF we consider tidal friction/tidal bulges, the effect would be LOWER but still towards raising the orbit. Never mind, I could move on...Maybe the two effects can cancel each other and my robotmanufactured spaceships can safely leave moon :) I'll let you know if my novel is ever finished :D On a side note, Bah.. I shouldn't have put a poll there, it makes this look like something personal against/with Prof. Comins which I swear I don't know. I just can't see how you can take into effect such "obscure" (for the nonastronomer) effect of the tidal forces but decide to overlook/ignore the more elementary physics formulas. This is science. You can't pick only the formulas you like. So to be honest I'm still not convinced :). Even if the masses are not point masses that wouldn't completed invalidate the formulas, right? This is funny. Maybe Prof. Comins is a very known authority in the field and nobody dares to argue against the article? Or maybe my dilemma is plain stupid (but still... what about my physics formulas? where am I wrong?) Maybe this question should have been in a different section of the forum? I mean not general astronomy, but Classical Physics? Because it seems to me the classical physics is overlooked in the article, not the astronomy part. 


#7
Sep2513, 02:55 AM

P: 386

A steady force can cause acceleration on a level orbit. But the small and steady force can also cause a steady climb on an inclined orbit. Basically the Moon is climbing an inclined orbit... and the incline of the orbit will adjust to whatever the dragging force is, so the speed is still nearly constant. Earth rotates faster than Moon orbits. Therefore the friction propels Moon ahead, forcing it to climb an inclined orbit. If Earth were to rotate slower than Moon orbits, or in opposite direction, then the friction would brake the Moon and cause it to descend on an inclined orbit. If Earth were to rotate at the same speed as Moon orbits, the drag would vanish and the orbit of Moon would be nearly closed. Compare Deimos. It is smaller than Moon. For that reason Deimos drifts away from Mars much slower than Moon drifts away from Earth. Which is the reason Deimos still is just 30 000 km away, not yet 380 000 km. If Moon were smaller, it would have been drifting outwards at a slower rate, and therefore would be further inside as of now. 


#8
Sep2513, 03:39 AM

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P: 15,167

Secondly, you aren't reading that equation correctly. That torque is the torque exerted by the Moon on the Earth, and that torque is proportional to the square of the Moon's mass. Double the Moon's mass and the torque increases by a factor of four. That's the torque on the Earth. What does that have to do with the tidal acceleration of the Moon? The answer lies in conservation of angular momentum. That the Moon exerts a torque on the Earth means that the Earth must exert a corresponding force on the Moon to conserve angular momentum. That force is, at least per this equation, proportional to the Moon's mass squared. That means the tidal acceleration of the Moon is proportional to the Moon's mass. If the Moon was smaller the acceleration would be smaller and thus the Moon's outward drift rate would be smaller. The most widely accepted hypothesis regarding the formation of the Moon is that it formed shortly after a Marssized object collided with the Earth about 4.5 billion years ago. Per this giant impact hypothesis, the Moon formed at a distance of about 4 to 6 Earth radii. Tidal interactions made the Moon migrate from this initially close distance to the current distance of about 60 Earth radii. If the Moon was smaller, that outward migration rate would be smaller and the Moon would be at a lesser distance than 60 Earth radii from the center of the Earth. With regard to the issue that the authors of the cited paper are (rather strongly) arguing against that simplistic formula, their main complaint is that it is overly simplistic. By way of analogy, "When people thought the earth was flat, they were wrong. When people thought the earth was spherical, they were wrong. But if you think that thinking the earth is spherical is just as wrong as thinking the earth is flat, then your view is wronger than both of them put together" (Isaac Asimov). You, paulcretu, are at the equivalent stage of arguing that the Earth is flat. 


#9
Sep2513, 05:10 AM

P: 4

Thank you, DH. I did not read the paper cited by Bandersnatch, but only the equation he suggested as explanation. Now I understand, that according to you the paper argues AGAINST this equation... That leaves us with no explanation for the statement in the OP article ("A less massive moon would also orbit closer to Earth than the real one").
I'm an not arguing that the Earth would be flat nor spherical. I am perfectly able to understand there's more to this than only gravitation and inertia/centrifugal force. Be it tidal torque, conservation of angular momentum or excuse me whatever. And I am not talking about Moon's formation/"birth". I was thinking at lowering Moon's mass as a thought experiment, starting from the current status quo, and not what if Moon was "formed" zillions of years ago with a lower mass. I was just trying to understand why hasn't been addressed (in the article cited by me and here on this thread) the fact that from the simple physics formulas, GmM/D^2 = mv^2/R implies D = GM^2 / [(m+M)v^2]. Is the physics wrong? I cam here as this is a physics forum but my bad I've posted in the astronomy section. When I was young I've learned to understand that physics models are a series of approximations, each one being more precise than the previous. So these formulas can't be THAT off. I suppose they are still taught in highschool physics classes (at least they did, in my country, when I was attending highschool but that was almost 20 years ago). I was just trying to understand if my calculations are wrong or not and if they are wrong, where? On the other hand, if they are not wrong, then do the other factors really compensate in the opposite direction? Earth is not spherical; the tidal forces create 'bulges'... the earth is not flat (opps I knew that!). Ok, but those are responsible for a deformation of ... how much? Less than 10%? So could we assume the formulas for the spherical shapes are 90% correct? So what's wrong with them then? If I am wrong, I stand corrected, no need to patronize me (too much :) . Neither physics nor astronomy are my field of profession. I'm not trying to prove anything and not trying to step on someone's toes. 


#10
Sep2513, 07:33 AM

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P: 15,167




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