New answers tagged

-2

Gravitation was thought to be a force acting at a distance: a massive body, it was thought, exerted a pull on another body and, in doing so, exerted an equal and opposite pull on itself in obeyance to third law of motion and consequent conservation of momentum: Even accepting this obsolete explanation of gravity, the description was not correct, was ...


0

There is something that escapes me: how can angular momentum or KE be affected by what happens on Earth? How does the Moon know there are tides on Earth? its pull , which is their only relation, is always the same, isn't it? Yes. But the pull is not towards the centre of the Earth. The Earth is not a spherically symmetrical body. It is a body with a tidal ...


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The Earth's orbit remains pretty much the same from year to year. Each year takes the same amount of time - about 365.24 x 24 hours. The trouble for calendar makers is that this is not a whole number of days. If we just rounded it off to the nearest day then, over time, New Year's Day would arrive a bit earlier each year compared to the seasons. For ...


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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 ...


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Is Earth's orbit changing? Yes. But not large enough to alter our calendar in near future Well, there are some long-period oscillations, but those are very small, and don't imply that we're systematically moving towards or away from the Sun. There is an effect which is making us move very slowly away from the Sun. That is the tidal interaction between the ...


0

Have a read of Wikipedia Leap year A common year has 365 days and a leap year 366 days, with the extra, or intercalary,day designated as February 29. A leap year occurs every four years to help synchronize the calendar year with the solar year, or the length of time it takes the earth to complete its orbit about the sun, which is about 365¼ days. So ...


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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 ...


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You are orbiting at the surface of the earth, so $r$ is fixed. Since $r$ is fixed you are not free to change $a$; it is fixed by Newton's Law of gravitation $$ a = \frac{F}{m} = G\frac{m_\mathrm{earth}}{r_\mathrm{earth}^2} = 9.8\;\;\mathrm{m/s}^2$$ Since $a$ is fixed and $r$ is fixed, $v$ is determined.


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Assuming no air resistance, the only force acting on the satellite is the force of gravity, $\bf{F_g}$. The stated circular orbit also means that centripetal force, $\bf{F_c}$ is involved, which causes the satellite to follow the circular path. Due to this, the centripetal acceleration can be directly equated to g, which allows you to immediately solve for ...


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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 ...


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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 ...


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Assuming point particles and Newtonian gravity they will collide iff, in an inertial frame based in the centre of mass of the system: angular momentum is zero; total energy measured is less than zero. Where the zero of potential energy is when the particles are infinitely far apart. (1) means that the velocities are radial, (2) means there is a time ...


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If the particles collide, it means they couldn't escape each other given their initial conditions of mass, locations, and velocities. For example, two masses of any value, positioned at any location, will definitely collide if their initial velocities are zero with regard to some reference grid you've pre-selected. Their initial kinetic energies, zero in ...


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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

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 ...


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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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There's an exhaustive discussion here.


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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 ...


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By increasing its radius. Angular momentum is the cross product of radius and velocity: $$\mathbf{L} = \mathbf{r} \times \mathbf{v}$$ It's a trade-off between radius and speed if the angular momentum is constant (which is for a satellite)


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


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The charged accelerometer will register a non vanishing acceleration. The reason why the setup you are proposing (interaction via electric charge) gives a physically distinct result from the setup in the answer you linked (interaction via gravity), even though they can be described by the same mathematical force, is because the Equivalence Principle applies ...


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 ...


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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 ...


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The orbit (in polar coordinates) of a body under a inverse-square force, $-K/r^2$, is given by $$r(\phi)=\frac{L^2/mK}{1+\sqrt{1+\frac{2L^2E}{mK^2}}\cos\phi},$$ where $E$ and $L$ are the energy and the angular momentum of the particle. The equation above is just the polar representation of a conic section of conic parameter $L^2/mK$ and eccentricity $$\...


-3

My answer is more metaphysics than physics. The reason we do not "feel" the acceleration is that the change is within the tolerances of our bodies. That said, I am sure there have been people born who are more attuned to these forces. But for the most part, for most of use, there are so many forces acting on our senses that the acceleration of the earth is ...


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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 ...


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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, ...


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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 ...


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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 ...


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Air friction (simply a form of friction) as we observe in our everyday life opposes the motion/state of the body (in this case motion of satellite ). It's true that air friction is responsible for decreasing the speed of the satellite, thereby decreasing the kinetic energy and ultimately the total energy of satellite. But as we know that any form of system ...


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. ...


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

I am unsure how this is the precession of the ellipse and how the precessing ellipse can be proved to give a nearly circular orbit. Will showing that the angular velocity $w_0$ being nearly constant work? Consider the figure below showing the effective potential (continuous line) of a particle of mass $m$ and angular momentum $L$ under the ...


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. ...


-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 ...


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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 ...


1

Since the derivation of the Keppler's first law given in the other answers involves a non-trivial integration I think it is worth to see an easier way. Let $\vec p$ and $\vec L$ are the momentum and angular momentum of the planet, respectively, $m$ its mass, $K$ comes from the gravitational force $\vec F=-K\hat r/r^2$, and $\hat r$ is the radial unit vector ...


-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 ...


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.


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 ...


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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 ...


0

What you just did was to find a condition for attractive power-law forces to have stable orbits where stable means they remain bounded when perturbed around the circular orbit. You got the correct result. The Bertrand's Theorem though says something different: the only forces whose bounded orbits imply closed orbits are the Hooke's law and the attractive ...


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 ...


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 ...



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