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The Hubble expansion has no bearing whatsoever on the length of the year. This is because the whole Milky Way galaxy (and in fact most galaxies, if not all, and even local groups) has decoupled from the Hubble flow long ago. In fact, it could only form after it decoupled. Note that M31, our sister galaxy, is in fact falling onto the Milky Way rather than ...
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You can use Gauss's law for gravitation to work out the gravity as a function of (interior) radius.
$$ \oint \vec{g} \cdot d\vec{A} = -4\pi G \int \rho\ dV\ .$$
What this means is that the flux of gravitational field $\vec{g}$ out of a closed surface is proportional to the mass enclosed within that surface.
Let us assume that the density is a function of (...
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(Disclaimer: As I already pointed out in a comment to the question above, I never did a calculation with $H_0$ before and I might be utterly, horrible wrong with my interpretation.)
If you completely ignore the slowly changing orbit of earth and only take expansion of space into account and assume the Hubble-parameter to be pretty constant in the timeframe ...
8
The full equation for the time for an object to drop is
$$t = \frac{ \arccos \Big( \sqrt{ \frac{x}{r} }\Big) + \sqrt{ \frac{x}{r} \ ( 1 - \frac{x}{r} ) } }{ \sqrt{ 2 \mu } } \, r^{3/2},$$
where $x$ is the radius of the sun, $r$ is distance of the object, and $\mu=GM=1.327\times10^{20}$. (in SI units, so you will need to convert your distance to metres, ...
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I like to classify solutions of the problem of the time evolution of the
complete initial state of a set of objects at some epoch time, where the
objects are subject to Newtonian gravitation into two main groups.
One approach is to use orbital elements of some sort.
The other is to use a numerical initial value problem solver,
aka a numerical integrator. The ...
7
Stellar clusters around supermassive black holes are systems in which relativity likely plays a role. Currently, only bright stars can be seen in our own galactic center because there is a ton of neutral gas between us and the galactic center that obscures it. As a result, we only have a few "test particles" out of the many stars that actually orbit the ...
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Running the math for a 5 meter long pendulum and 1 kg mass, I get an amplitude of 0,017 mm.
You are off by quite a bit. There is essentially no horizontal deflection when the Moon is at the horizon. The maximum horizontal deflection occurs when the Moon is about 45 degrees above or below the horizon.
The tidal acceleration at some point on the surface of ...
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If you're asking whether it's sufficient to use a retarded (time-delayed) positions to calculate gravitational forces, then no, that would be much worse than Newtonian gravity. For example, that would predict that the Earth should spiral into the Sun on the order of about 400 years. See also answers to
Besides retarded gravitation, anything else to worry ...
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Leckner's paper deals with the effect of induced polarization on the spheres. Electrons are redistributed, making the force different from what one would expect. The gravitational counterpart is tidal distortion: since the gravitational field is non-radial when you have two heavy masses close to each other, matter will move to make the surface an ...
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I can attempt to address the second part of your initial question (*"Is it a particle, a wave,...?") Einstein's theory of general relativity states that mass and energy bend space-time. Space-time, in turn, tells matter how to move (John Wheeler put this more elegantly).
This concept is completely different from the theories of the other three fundamental ...
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The short answer because Jupiter is a gas giant, so it's kind of got a very large atmosphere and atmosphere's aren't very dense. Also, if you look at your chart, the gravitation inside the Earth increases until you get to the outer core. This is likely to be much more pronounced on gaseous bodies like gas giants and stars.
Longer answer:
In layman's ...
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A photon is an entity defined in the context of a relativistic field theory, and so it doesn't really make sense to talk about the Newtonian bending of a photon. Necessarily, we need to substitute an analogous question that's sensible in the Newtonian framework. To do so, we can imagine a classical corpuscle of light--appropriately enough, a theory of light ...
4
The magnitude of the force of gravity between two bodies is proportional to the product of their masses:
$$F=G\frac{m_1m_2}{r^2}$$
This doesn't change depending on which body you're applying the force to, i.e. if you interchange the masses. The magnitude is the same.
What does change is the direction of the force. Force is a vector quantity, denoted as $\...
4
Apart from the field-theoretical standpoint presented by Stan, one can repel objects in a sense, when taking orbital mechanics into account.
The slingshot maneuver extracts angular momentum and energy of an orbiting mass by the use of gravity. The trick here is that the probe's velocity just gets redirected in the planet's reference frame. But this results ...
answered Dec 15 '15 at 15:21
AtmosphericPrisonEscape
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3
Ignoring details such as the oblateness of the Earth, atmospheric drag, third body influences such as the Moon and the Sun, relativity, ..., the period of a satellite of negligible mass (even the International Space Station qualifies as a "satellite of negligible mass") is $T=2\pi\sqrt{\frac {a^3}{\mu_\mathrm{Earth}}}$. Neither Newton's gravitational ...
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In general relativity, gravity neither pushes nor pulls. To explain why ball travels in an arc you note the start and end points of the throw in 4d space time (3 space co-ordinates and 1 time coordinate) You then find the shortest path between these two 4d points in the curved spacetime surrounding the Earth. This shortest path is the path in spacetime that ...
3
I'm guessing that this misunderstanding is a result of the oft-used rubber sheet analogy. The rubber sheet analogy says that, according to general relativity, mass curves space-time like a heavy bowling ball on a near-taut blanket (or rubber sheet) curves the blanket/sheet. This resulting curve makes other bits of matter/energy move in different ways. I'm ...
3
If you project the orbits onto a plane, for example the plane of the ecliptic, the projections will cross. But that's only because you're looking at a 3D problem in 2D. If you look at the orbits in 3D, you'll see that Pluto's orbit is highly inclined (17º) from the ecliptic, so it never actually passes through Neptune's orbit. Each time it seems to cross (in ...
3
"in a spherically symmetric distribution of matter, a particle feels no force at all from the material at greater radii, and the material at smaller radii gives exactly the force which one would get if all the material was concentrated at the central point".
This may be poorly worded. The idea is not that the distance from the center of the body must be ...
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James K gave a good answer to this, but I just want to add that if the Sun was unmoving relative to the center of the Milky-way, it would fall towards the center along with your object. The sun orbits within the Milky way with tangential velocity about 230 km/s.
It's rather pointless, but we can compare mass and distance and find the sweet spot where the ...
2
The original result is Newton's shell theorem. Since we can break up a spherically symmetric distribution into spherically symmetric concentric shells, it is sufficient to consider the corresponding statement for one such shell: for each shell taken individually, there is no force on a particle inside, and a force on a particle outside as if all of the mass ...
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Adding to @Guillochon's answer, there are even a number of general relativistic tests in our solar system, the most famous being the precession of the perihelion of Mercury.
In short, the location of the point of closest approach to the Sun (perihelion) for the planet Mercury is a changing quantity. Essentially, given one full revolution, it doesn't trace ...
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Newtonian treatments of the bending of light go back to Laplace who, in 1798, wrote about light escaping from massive bodies, ie: black holes! See Appendix A of Hawking and Ellis "Large Scale Structure of Space-Time" where there is a nice translation of Laplace's paper.
Newtonian treatments cannot properly deal with all aspects of light bending. Notably, ...
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The Newtonian gravitational acceleration of a test object (a small object with negligible mass) toward a point mass is given by
$$\vec g =\frac {GM}{||\vec r||^3} \vec r \tag{1}$$
where $\vec g$ is the acceleration vector, $G$ is the Newtonian gravitational constant, $M$ is the mass of the point mass, and $\vec r$ is the displacement vector directed from ...
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Close to a black hole, but still outside the event horizon, orbiting arbitrarily fast, wouldn't result in a circular orbit, but in a trajectory towards the black hole.
In this extreme scenario, centrifugal force could be interpreted as attractive.
In more usual settings, centrifugal forces point outward.
Since the centrifugal force (in usual settings) $m \...
2
$\mu$ is a quantity that can be easily observed (semi-) directly. It can be easily derived from orbital period of orbiting bodies or acceleration of falling bodies, even if their mass is significantly smaller than the mass of the body you're measuring $\mu$ for.
Now the only way to really obtain G is to divide $\mu$ by mass of your celestial body. And ...
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I'm not sure how accepted this is in the physics community, but a book by Andrew Thomas called Hidden in Plain Sight posits a theory that gravity may be repulsive within black holes (i.e. on scales smaller than the Schwartzchild radius for a given mass).
It's an interesting theory, and it avoids the need for a singularity, but to my knowledge it's ...
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I suggest reading the paper on the 1919 expedition to get a clearer picture of why they did it at that time and why it hadn't been done before. From reading chapter 2 I think the main reason was the astrophotography equipment required for the experiment and the alignment of bright enough stars close to the sun to observe the effect.
Of course before ...
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When a body spins around another body due to gravity and maintains a consistent orbit, we can know clearly two things about the body:
1. The speed of the orbit
2. The average radius of the orbit
Think about it. The Moon spins around the Earth due to gravity of the Earth. Right? And it's orbit always stays the same(when averaged). For it to stay in this ...
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A NEO only impacts Earth if both bodies are in the orbit intersection zone at the same time, closely enough for gravity $(F \propto 1/r^2)$ to bring them in contact.
In a simulation, the timestep should also be short enough to detect a collision.
An object at a moderate relative speed of 10 km/s crosses Earth's diameter in about 20 minutes; a longer timestep ...
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