New answers tagged

1

To continue this as an addendum, I played around with some of this. It is clear that for a circular orbit that the cosmological constant will only slightly adjust the radius. There will be no change in the radius of the orbit with time. For an elliptical orbit this might be different over a very long period of time. I will continue with this dynamical ...


2

Your question has some bearing on what some people interpret erroneously as the source of the Pioneer anomaly. As some people point out there is some issue with what happens with a solar system in a galaxy. Really the influence of the cosmological constant is most likely to occur on that scale instead of a stellar system of planets. I will set this up some ...


1

There are two reasons. Inside the Sun, the force is no longer an inverse-square law. It actually grows linearly with $r$. The second reason is that Goldstein (as well as any other classical mechanics book) is interested in orbits with a non vanishing angular momentum with respect to the center of the Sun. An oscillation along a line passing through this ...


1

What you need is Kepler's equation, $$M = E - e \sin E$$ where $M$ is a quantity called the mean anomaly, e is the eccentricity of the orbit, and $E$ is called the eccentric anomaly, defined by this diagram where the sun is at $F$ and $C$is the center of the ellipse (the distance $e$ in the diagram should be $ae$). The quantity $M$ is simply $2\pi t/T$ ...


2

R. Rankin's answer gives you the general solution when the velocity along a curve is known. If you're interested in elliptical orbits due to gravitational interactions, you can use Kepler's laws. Kepler's second law says that the time required for an object on an elliptical orbit is proportional to the area swept out by a line connecting the orbiting body ...


2

Given your instantaneous velocity $\vec{v}(\vec{x})$ , and your start and end points, say $p_{1},p_{2}$ , just integrate it in space: $$\triangle t=\intop_{p_{2}}^{p_{1}}(\frac{d\vec{x}}{dt})^{-1}d\vec{x}$$


0

First, the authors positioned their reference frame at the center of mass, if you calculate the momentums that way, the total momentum will be zero, at least instantaneously. But that is not the point. In addition to the effects of other objects in our own solar system, mentioned in the other answer, remember that the Sun travels around the milky way ...


1

I don't know what numbers you are using, or how much precision you are expecting, but the problem could be your assumption that the earth - sun system is an isolated system. First there is earth's moon that (if not taken into account) might induce errors. Then there is the effect of other planets, especially jupiter. All of these must be considered if you ...


7

Most asteroids are in an elliptical orbit around the Sun in the inner Solar System, i.e. a region comprising Mercury, Venus, Earth, Mars and the Asteroid Belt. What can happen is that an asteroid's elliptical orbit intersects a planet's orbit and this might gives rise to a collision. Most of times, when an asteroid gets too close to a planet, it has too ...


0

If your test mass is being placed in an orbit at same angular speed as the lunar orbit i guess you'd need to super impose its centripetal force with the gravitational force. The new null orbit will probably be much lower because the null point is already near the orbit altitude.


-1

Both are unstable (ring and sphere) for the same reason: the potential everywhere inside either one is zero. This is true for gravitational as well as electrical and magnetic forces, which are all inverse square law / central force situations, and it is that pattern which causes the result. The proof requires calculus, but is considered an elementary ...


-4

The moon is dynamically and almost rigidly (but for librations) tied to the earth by means of an invisible solid rod, as it were. That is why we inhabitants of earth do not get to see the far side of the moon from our base. Solar eclipse occurs when the invisible rod places the moon between earth and sun casting a tiny shadow on earth. Lunar eclipse ...


44

The Moon's orbit must be concave toward the Sun. The Moon's orbit with respect to the Sun is always convex. This is easily proven by comparing the minimum possible gravitational acceleration of the Moon toward the Sun (5.7 mm/s2) and the maximum possible gravitational acceleration of the Moon toward the Earth (3.1 mm/s2). The acceleration vector, and ...


45

Incorrect Path I'm curious as to what does the moon's orbit around the sun looks like? One might think the orbit (in the sun's rest frame) follows the path of an epitrochoid. A (very) over exaggerated view of this motion (for unrealistic parameters, thus, not an accurate representation) can be seen in the following animation: Note that if you change ...


16

Before answering let me mention that there is a terrific free applet showing the orbits, including the velocity vectors of the system Sun/Earth/Moon: https://phet.colorado.edu/en/simulation/gravity-and-orbits It is in java so pretty easy to download and use. The moon's orbit must be concave toward the sun. The Moon' orbit around the Sun is a ...


10

The orbital speed of the earth around the sun is about 30 km/s, whereas the orbital speed of the moon around the earth is about 1 km/s. From this it follows that at no point of its path around the sun the moon will ever show a backwards motion. The path is similar to the trajectory of a point (moon) on the perimeter of a (somewhat sliding) wheel rolling ...


1

What you need to pull an object off the surface of another, gravitationally, is for the tidal acceleration $F_{\rm tidal}$ from the "external" body at the surface of the Earth (or whatever other body) to exceed the gravitational acceleration $g$ of Earth on the surface. This comes with the caveat that if the tidal forces are too strong, they will start to ...


2

Assuming that the gravity well acts as an attracting mass whose force follows Newton's law of gravitation, your well will suck the object from Earth's surface when the force from the well is larger than the force from the earth. That is, $GM_{earth}m/R_{earth}^2<GM_{well}m/r_{well}^2$, where $r_{well}$ is the distance between object and well. Earth ...



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