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61

We don't feel any acceleration because the Earth and all of us humans on it is in free fall around the Sun. We don't feel the centripetal acceleration any more than the astronauts on the ISS feel the acceleration of the ISS towards the Earth. This happens because of the way general relativity describes motion in gravitational field. The motion of a freely ...


49

John Rennie's answer is right from the viewpoint of General Relativity -- but since the question is tagged with Newtonian mechanics, it deserves a Newtonian answer too. In the Newtonian framework, I think the best answer to "why don't we experience this force" is that we can't feel forces that apply to our body at all. What we actually experience with our ...


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 ...


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 ...


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 ...


12

According to the Equivalence Principle a free falling system cannot locally detect a gravitational field. However Earth is a large enough system such that non-local effects turn out to be appreciable. Solar tides are - although small - detectable. So in principle one can experience the Sun's gravitational field even though we are in free fall. What I claim ...


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 ...


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 ...


6

There is indeed a nett force on the body owing to the electrostatic attraction / repulsion. Therefore, there is nonzero four acceleration, and the body will have a different orbit from the ones defined by the spacetime geodesics for the metric describing the massive body's neighborhood. From the standpoint of an observer stationary with respect to the ...


4

John Rennie has answered the question in terms on general relativity, but it can also be answered with Newtonian physics. Your question is very similar to this one: Why does the moon stay with the Earth? and I can refer you to my answer there. In short, the Sun isn't only pulling on the Earth itself, it's pulling on everything on it as well, including us, ...


4

Even if the orbit were a perfect circle, there's some acceleration towards the sun. If there weren't acceleration then the earth would move in a straight line (instead of a circle); but it doesn't move in a straight line therefore there's acceleration. In a sense, the earth doesn't feel the acceleration because it doesn't try to resist it: if you stand on ...


3

All you need to do is calculate the perigee distance $r_p$ that is the distance of closest approach. Then if $r_p < R_A$ your satellite will crash and burn. Once again we start from the vis-viva equation: $$ v^2 = GM\left(\frac{2}{r} - \frac{1}{a} \right) \tag{1} $$ The parameter $a$ is the semi-major axis of the ellipse, and it is related to the ...


3

No. It's due to inaccuracy in the calendar. In a nutshell, Earth's orbit actually takes 365 and 1/4 days. So every four years, we add an extra day so the calendar doesn't get all out of whack. Of course, it gets more interesting than that. When I said it took 1/4 of a day longer than the calendar allows for, it actually takes a few minutes longer (11, to ...


2

Let me call radial stability the stability of $r$ around $r_0$, where $r_0$ is the radius of the circular orbit. The difference between this one and the Lyapunov stability is that the latter looks not only to $r$ but also to the polar angle $\theta$ (for a central force) and their conjugated momenta. So in this sense I would say Lyapunov is stronger. ...


2

This cannot work in a 2-body scenario. In a 3-body scenario, yes, and this can even be shown from real world examples. Lagrangian orbits are reasonably well understood, where a 3rd body orbits a location defined by the masses and orbits of the first 2 bodies. the L4 and L5 locations follow or lead the 2nd body, but have stable orbits around them that do ...


2

The answer is simply that not every space-time has a corresponding effective potential in the sense that we have a coordinate $x$ such that $\dot{x}=\sqrt{2(E-V_{eff})}$. But this is true even in Newtonian mechanics, consider a problem with a Lagrangian $$L = \frac{m}{2}(\dot{r}^2 + r^2 \dot{\varphi}^2) - V(\varphi)$$ Obviously, $p_r\equiv m \dot{r}$ is ...


2

If you add mass to the Earth, or the "Earth system", it makes not the slightest bit of difference to the orbit unless you also change the Earth's angular momentum around the Sun. That is because the basic dynamical equation controlling the orbit is that the inward gravitational force due to the the Sun is equal to the mass times centripetal acceleration. ...


2

Juno probe's speed is in relation to what frame of reference? The escape speed of Jupiter is ~59.5 km/s and 150,000 kph ~ 67.1 km/s, so this speed must be in reference to the sun otherwise the spacecraft would not stay in orbit. Jupiter's orbital speed about the sun is ~13.1 km/s, which subtracted from the 67.1 km/s would result in ~54 km/s, thus more ...


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: $$\triangle t=\intop_{p_{2}}^{p_{1}}(\frac{d\vec{x}}{dt})^{-1}d\vec{x}$$


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 ...


2

A circular orbit occurs, if the test body has the "right" velocity: $$ v_\text{circ} = \sqrt\frac{GM}r $$ If the velocity is less, the body will enter a non circular elliptical orbit or fall onto the central body (if it has a non zero extension). If the velocity is higher, the body will enter a non circular elliptical orbit or escape in a hyperbolic orbit.


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 ...


1

If you think of it in terms of conservation of momentum and collisions, the simplest version works just the same as tossing a handball at a on-coming freight train. The interaction is elastic, and the ball returns with the same speed it had going in in the center of momentum frame, but the center of momentum frame is moving in the ground frame, so the ball ...


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 $2\pi t/T$ ...


1

You definitely don't need to use General Relativity to answer this question. It depends upon what you mean by "feel". If "feel" means "detectable by sophisticated instruments" then, yes, it can be "felt". But your body is not a very sophisticated detection instrument. According to what I've read elsewhere, the Earth speeds up by $1000$ $m/s$ as it moves ...


1

Without seeing the particular example you have in mind it's impossible to be definitive, but without any context, for a probe in the Solar System, I'd assume heliocentric coordinates. Unless it's an orbital velocity around some other body, or an escape velocity from some other body, or of course if it's specifically stated to be in some other frame.


1

This turns out to have a really boring answer. We can find the two circular orbits by finding the maxima and minima of the effective potential, and we get: $$ r = \frac{L^2}{2M}\left(1 \pm \sqrt{1 - \frac{12M^2}{L^2}}\right) \tag{1} $$ where the $+$ gives the outer stable orbit and the $-$ gives the inner unstable orbit. Note that both orbits exist only ...


1

Well, the "object" has two kinds of energy at all times: kinetic (movement), and gravitational potential. This calculation you want to do is easiest if there is only one significant body, like the sun. In this case, at the beginning your KE (kinetic energy) is 0.5mv2, and your gravitational energy is Ug = -Gm1m2/r. Now, an object can never completely leave ...


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 ...



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