Spin up the Moon

How much thrust would be needed to make the Moon visible an added 1/2 degree per day, so that the full globe would be visible from Earth within the period of 1 year?

diameter: 3476 km
mass: 7.35e22 kg
rotational speed at the equator: 10.3mph/16.7 kph

 

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Not sure what you are asking here...how quickly do you want this 1/2 degree per day rotation?

 

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A full Moon as we call it is when it is in such a position that it's face pointing towards Earth is in full sunlight. It would need it's rotation around Earth slowed almost to a stop for this to happen all the time (assuming this is what you mean), at which point it would quickly spiral into the Earth.

 

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I think he means slowing the rotation of the moon about its axis (not its orbit around Earth), such that at one point in the year the nearside would be completely visible and another point, the far side.

 

What actually causes one side of the moon to always face the Earth? Presumably it didnt always start out that way, so how was its angular velocity achieved? If we did hypothetically stop the moon from spining, would it slowly revert back its original rotation somehow?

 

Hamster interpreted correctly: I don't think it would matter, either slowing or making it faster. I was thinking about speeding it up.

1/2 degree would be 3476km divided by 180, which equals 19.31km divided by 2 = 9.65km. Current rotational speed is 16.7km/h.

The idea, as Communist Hamster noted, would be to view both sides of the Moon over a period of a year.

My question is, how much thrust (from the equitorial surface of the Moon) would that take, and just for fun, what specifications are required to accurately do the calculations.

I think the calculations would be much more complex if I wanted to shift the axis of rotation (in the process) to something similar to Earth's. Then, there would be other mechanical principles involved. But that is not what I am asking.

I'll stick with speeding up the rotation on the current axis.

 

The rotational speed of the Moon keeps the same face towards us. It is a beautiful anomaly.

I cannot answer your other two questions.

 

You still haven't mentioned how quickly. That would determine the amount of thrust required. Do you want it within a year? by 2040? in two seconds?

 

Well he did say within 1 year in the OP.

 

The Moon rotates 360 degrees in 27.321582 days, has a mean radius of 1737.1 km, and masses 7.3477*10[sup]22[/sup] kg. Modeling the Moon as a uniform sphere yields a rotational kinetic energy of about 3.14*10[sup]17[/sup] joules. (Since the Moon has a small iron core, the actual energy will be a bit smaller.)

Stopping the Moon uniformly over a period of one year requires 9.95 gigawatts. Since the Flux Capacitor has a power output of 1.21 gigawatts, all it takes is eight plus a fraction Flux Capacitors running continuously over a year to do the job.

 

Your presumption is correct, Nick. Both the Earth and the moon used to have much faster rotations about their axis than they now have. Tidal interactions have slowed the rotations of both over billions of years. Just as the moon's gravity causes tides on the Earth, the Earth's gravity caused tides on the moon. Water in the form of oceans is not necessary for a body to be affected by tidal forces, tidal forces also lift and pull against the solid surface of the two bodies. The Earth is a much more massive body than the moon, and it exerts a stronger tidal force on the moon than vice-versa. The rotation of the moon has slowed more quickly than the rotation of the Earth because it has both less mass and it experiences a stronger tidal force than the Earth. The rotation of the Earth has also slowed from about a 12 hour day to our present 24 hour day. The rotation of the Earth continues to slow due to the tidal interaction with the moon. In the distant future, the Earth will also be 'tidally locked' with the moon, the same side of the Earth will always face the moon, assuming neither are destroyed and the moon doesn't drift too far from the Earth. Most people do not realize that the land masses of the Earth also rise and fall as the moon passes overhead, just to a lesser extent than the more fluid oceans.

 

Very well stated, 2inquisitive.

 

Wow interesting stuff 2inquisitive, thanks.

 

How quickly would only be determined by the amount of force applied at one time. 100 booster rockets fired at the same time would take twice as long a burn as 200 booster rockets fired at the same time.

The question is: How much force (thrust) would needed?

 

I am trying to understand this more fully, because it seems I only partially grasp the idea. Orbital rotation causes tidal activity, bulges, on both bodies. I can grasp the idea that there is some affect when axial rotation is changed, but it is a difficult concept. If the axial rotation of the Earth would speed up, then the bulge would consequently move faster over the planet, and I'm assuming the same affect would occur on the Moon. What affect would it have on the thrust needed to speed up the axial rotation of the Moon, and would that have any affect on the tidal forces applied to the Earth?

Thinking about it more deeply, which may be a mistake, the effect might have a consequence for the density of the mass, causing the mass to grow larger and the density to decrease, although that affect might be so insignificant that it would not be measurable. Still, the variables would seem to exist. I'm guessing the mass of the Moon is packed fairly tightly at this point (relative density, and I'm not sure how that is measured). This is not part of the original question, just an observation.

My biggest handicap, not being a professional in the fields of Astronomy or Physics, is the vocabulary. Could be there are terms other than orbital rotation and axial rotation, that are used to describe the phenomenas.

 

I'm trying to delete this follow up, but do not see the option to do so.

 

It's because the moon is not EXACTLY spherical, but has a slight "bump", earth has this "bump" too, so slowly the two "bumps" have begun to face eachother. At least I think so.

 

Xmo1,

Perhaps it is best to gain an understanding 'tidal activity'. I started typing out a reply and attempted to briefly summarize how tidal forces, and tidal acceleration, effect the oceans of the world, which is what I suspect you are thinking of when when you mention tidal activity. Tidal interactions between the Earth and the moon are much more complicated to explain briefy. The tides in our oceans is only one of the effects of the tidal interactions between the Earth and the moon. The Earth's rotation only causes the movement of a distortion (ocean tides), not the distortion itself. Here is a link to good ole wiki that should help your understanding:
http://en.wikipedia.org/wiki/Tidal_force

The effect on the thrust needed to speed up the moon's rotation would be very minimal. However, once you did speed up the rotation of the moon, the tidal forces from Earth would eventually slow the rotation back down. The moon would eventually be 'tidally locked' with the Earth again in the future. No, rotation has no effect on the intensity of the tidal force in mainstream physics. Reiku did link to an 'alternative' hypothesis with a force named 'spinity' which would presumably be affected by the increased rate of rotation. But remember, it is a paper by a lone physics student, not a part of mainstream theory.

It is accepted to state the Earth 'rotates' about its axis, and 'revolves around' (orbits) the sun.

 

That is an excellent link 2inquisitive. I have become attracted to Newtonian mathematics, and it is interesting to find so many great people of history have teethed on tidal behaviors. My guess is that the bulk of variables to answer the thrust equations are in there somewhere, and I will be able to answer my own question. I am in good company, and on a better track because of your post. Thank you.

 

In this case, wouldn't the computation be a simple F=MA, where we simply use Force as Thrust? My understanding is that the computation would be slightly more complex, but for my purposes the question could be answered using the simplest equation. I am trying to get an idea of the requirements. Something like: It would take 20,000 space shuttle rocket boosters firing for 20 minutes to get the .4km/h increase in rotation necessary for a .5 degree increase in the visible area of the lunar surface per day. (I'm making up these numbers)

Something like that.

Aeronautical thrust computations get pretty hairy. No need for that.

The related question people like to answer is, could it be done. And they usually answer no, or there would be no practical reason to do it, and a lot of reasons for not doing it.