Tag Info

5

I assume you're talking about the numerical instabilities that arise from having an infinite potential at $r=0$. Here are three common solutions: Use a soft-core potential that behaves like $1/r$ except very close to $r=0$ where it levels off to a finite value. For example, $1/\sqrt{\epsilon+r^2}$ instead of $1/r$ is common. Add hard sphere collision ...

3

We can detect it, but it takes very precise measurements. Stellar parallax, i.e. the relative displacement of close-by stars against the background of far away ones can be detected, but it's a very difficult measurement to make because the "motion" is very small (usually fractions of an arc second): Recently we have learned to build satellites that can ...

3

What you are asking for is not simple at all. Retrograde motion occurs when the line joining two planets rotates with respect to a fixed coordinate system (or the fixed stars) in the opposite direction as the planets. With both planets in motion in orbits that are not nicely aligned with each other, the times between retrograde motion will only be described ...

2

But you can see the motion of the earth in it's orbit; it's called aberration of starlight, first measured by James Bradley in 1729. Even earlier, parallax of stars had been detected, by 1680. But you have to take detailed observations at widely separated times for the effects to be even the least bit obvious.

2

Take the orbit equation (using $u=\frac1r$): $$u(\theta)=A\left(1+e\cos\theta\right)$$ Where $A$ is some constant that you can work out from the differential equation if desired, and $e$ is the eccentricity of the orbit. Now, what happens when $\theta$ goes through an angle of $2\pi$? Notice that we would not have returned to the original radius if ...

2

The correct formula is actually $$M = \frac{4\pi^2 a^3}{GP^2}$$ and is a form of Kepler's third law. $M$ in this formula is the central mass which must be much larger than the mass of the orbiting body in order to apply the law. In reality the formula that should be used is $$M_1 + M_2 = \frac{4\pi^2 a^3}{GP^2},$$ where $M_1+ M_2$ is the sum of the masses ...

2

You ask: could the gravitational force of an average asteroid be extrapolated to the rest of the asteroids in its family? But the question is what properties the members of a family have in common. You could argue that all the asteroids in a family will be made up from the same material and will have the same density. That means as long as you can ...

2

I am certain that mathematical analysis of tidal locking has been done many times but I have failed to find such an analysis where the mechanism for the transfer of angular momentum to spin angular momentum is included which is the question which has been asked. Perhaps someone is able to produce a reference or an analysis? Having experienced on a number ...

1

The most straightforward observation to show that the Earth moves is stellar parallax. If you take photographs of a groups of stars over a period of six months (half an orbit), some of the stars will seem to shift in position compared to the others. These stars are much closer to Earth and so seem to move more. This is similar to how, when you are riding in ...

1

The relation is indeed correct. Let me show you why. This type of problem is solved by equating forces. Other classical mechanics problems are by equating energies or, in rare cases, momentum. The gravitational potential indeed is $$V(r) = G \frac Mr \,.$$ This is not directly useful for this problem. Here you rather want to look at the gravitational ...

1

It's called the virial theorem: in a bound system, the average of the potential energy of a $\frac{1}{r}$-potential V is related to the average of the kinetic energy T like $\frac{\langle V \rangle}{2} = - \langle T\rangle$.

1

The final orbit has a smaller radius so we know the final PE must be less (more negative) and the final KE must be greater (moving faster in a smaller orbit) but the final total energy must be less (more negative), since $E_{total}=-\frac{1}{2}G\frac{Mm_{e}}{r}$. This means the satellite, or more correctly, the Earth-satellite system, has to lose energy. ...

1

At it's simplest level, satellite tracking is a 3D version of triangulation. Suppose one were able to have 3 observers at the vertices of a triangle on the ground all simultaneously and instantaneously measure the distance from themselves to the satellite. Then one could use the known coordinates of the 3 observers along with the measurements they each ...

1

The classic thought experiment (which I think is in Feynman's lectures) is to imagine a cannon on top of a hill shooting cannon balls. Ignore air friction for this. Typically the ball curves in an ellipse (we often say parabola, but this is an approximation for a uniform gravitational field, though the two are basically identical for this case) and hits ...

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