# Tag Info

4

First, let us look at the current rate at which the moon slows down. I have a few different sources, and they don't all give me the same answer. First, there is this claim that Earth slows down at a rate of about 0.005 seconds per year per year. A year has approximately $365.25 \cdot 24 \cdot 3600 = 3.15\cdot 10^7 \mathrm{sec}$, so 0.005 seconds change ...

3

Assuming that you have some sort of slow, continuous drag on the object, then velocity at any point is equal to the velocity of a circular orbit at that point. The object doesn't have a single speed from the start to the end of your graphic, but a slowly increasing one. As the drag occurs, the slowing of the object and the loss of altitude happen ...

2

The other answers here are in the spirit of what you can do, but allow me to elaborate a little more. To understand if the trajectory of the movement under a potential $V$ is stable or not you have to understand what this stability means. The most simple example is the harmonic oscillation- $V=-{1 \over 2}kx^2$.In Newtonian mechanics, for a point of ...

2

In the case of radial freefall is from rest at some initial $r=R$, the motion will be periodic if you treat the gravitating body as a point-mass and ignore collisions. Since the radius is strictly positive, it makes sense to substitute $$r = R\cos^2\left(\frac{\eta}{2}\right) = \frac{R}{2}\left(1+\cos\eta\right)\text{.}$$ while conservation of specific ...

2

All the stars would be attracting each other and hence they would stick to each to attain equilibrium. Why doesn't this happen? This is an old question. Even Newton himself had thought about this question. His idea was that in long distances or separations (say, inter-galactic distances) the force of gravity might appear to be repulsive. That's why not ...

2

The Moon rotates around the Earth slower than the rotation of the Earth itself. That's why, from a fix point on the Earth, the Moon appears to be moving. The Moon creates the tide on Earth. So the tide "follows" the Moon. However as the Earth rotates faster than the Moon it will tend to carry the tide with itself "forward". The Moon pulls the tide toward ...

2

Even if the original dust cloud only had a relatively small angular velocity (which it might have had for all sorts of reasons), the process of collapsing would have amplified it. That is, the collapse process preserves the angular momentum, but it translates to a much larger rotational speed in the newly-collapsed system. Think of what happens to a spinning ...

1

Angular momentum is a conserved quantity (in a closed system) and this is true also for the angular momentum that is carried by the electromagnetic (EM) field. This conservation is a manifestation of rotational symmetry and the azimuthal part of the EM field emitted must be single valued. In other words, when rotating the EM field in the azimuthal ($\phi$) ...

1

The answers are yes and no. Special relativity does make ellipses precess, but it only accounts for 7" out of 43" per century of Mercury's anomalous precession. I wonder if Einstein and/or Sommerfeld knew that. To first order, incorporating special relativity results in a small inverse cube correction to the gravitational force, which is well known to ...

1

All the stars would be attracting each other and hence the would stick to each to attain equilibrium. Why doesn't this happen? You are forgetting angular momentum. Consider a binary star pair. Ignoring the expansion of spacetime, and in the absence of some mechanism that removes angular momentum from the system, those stars will orbit one another ...

1

In a classical context, LRL vector is conserved only for potentials behaving like $\frac{k}{r}$, indeed we can see the general construction of LRL vector : \begin{eqnarray} \frac{d\vec{p}}{dt}\times \vec{L} &=& -\partial_r v(r) \frac{\vec{r}}{r} \times \vec{L}, \nonumber\\ \mu r^3 \frac{d\hat{r}}{dt} &=& ...

1

The foci are simply points that define the ellipse by the relation $c^2 = a^2 - b^2$, where $c$ equals the length of each one of the foci to the center and $a$ is the length of a focus to the end of the ellipse. For a circle, $a$ = $b$. Given any two foci, a point on the ellipse is a point that is equal I he sum of he lengths of the foci.

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A circle is a degenerate ellipse, and you can also think of a circle as having two foci (on top of one another) as the eccentricity approaches 1. The foci of conic sections in general originate from the approach in which the curves are defined - using a focus (point) and directrix (line). This approach leads to rational parametric expressions for the conic ...

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