Hot answers tagged

18

There is no significance in the fact that the same face of the Moon is always oriented towards the Earth. It simply means that the rotation of the Moon around its own axis has the same period as the orbit of the Moon around the Earth. (This wasn't always so. It can be reconstructed that the Moon previously had a faster rotation rate than it has now. ...


11

Foucault's pendulum would, indeed, work on the Moon - though because the rotation is much slower (a period of about 2360 ks versus about 86 ks), the pendulum will also precess that much slower. (It will also swing with a longer period, due to the reduced gravity, but that's a different matter.) Regarding your question about whether the rotation is "real&...


4

Does the moon's elliptical orbit around Earth also affect Earth's spin? What effect does distance play? The moon's orbit affect's earth's spin. The eccentricity is not especially relevant, but the distance is. The closer the moon, the stronger the tidal forces, and (to a first-order), the greater the coupling between the two. That would suggest slowing ...


3

I think I understand your question. It's not so much about Foucault's Pendulum as it is about frames of reference. I believe this is what you are thinking: If you ignore the rotation of the moon and just think of it as an orbiting point, then an obvious way to use it to generate a coordinate system that follows it is to have one axis tangent to the orbit, ...


3

Yes, it would work. Of course, you should take into account that gravity at the Moon's surface is only 1/6 of the value on Earth and that a Moon (sidereal) day is 27 Earth days.


2

If we ignore other bodies, the orbital plane will remain fixed. It will not turn along with the Moon as the Moon orbits. That would require a force acting on the satellite which is not in the orbital plane. Orbits do precess in real-world examples, but that is under the influence of other bodies (or of uneven features of the body orbited), and it is usually ...


1

Let's make some assumptions: We know both the Earth and Moon radius, they are perfect, uniform spheres and their densities are the same. The acceleration at any point due to a mass is: \begin{equation} a = \frac{GM}{r^2} \end{equation} The mass of a sphere is given by: \begin{equation} M = \frac{4}{3} \pi r^3 \rho \end{equation} Substituting this into the ...


1

For a mass $m$ at the surface of the moon: $$F=G\frac{M_{moon}m}{R_{moon}^2}$$ So that: $$F=g_{moon}m$$ where: $$g_{moon}=G\frac{M_{moon}}{R_{moon}^2}$$


Only top voted, non community-wiki answers of a minimum length are eligible