Tag Info

15

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 ...

9

(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 ...

6

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 ...

4

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 this question. Most small-scale N-body simulations (e.g., ...

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 ...

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 ...

3

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 ...

3

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 $\... 3 If you project the orbits onto a plane, for example the plane of the ecliptic, the projections will cross. But that's only because you're looking at a 3D problem in 2D. If you look at the orbits in 3D, you'll see that Pluto's orbit is highly inclined (17º) from the ecliptic, so it never actually passes through Neptune's orbit. Each time it seems to cross (in ... 3 I can attempt to address the second part of your initial question: "Is it a particle, a wave,...?" Einstein's theory of general relativity states that mass and energy bend space-time. Space-time, in turn, tells matter how to move (John Wheeler put this more elegantly). This concept is completely different from the theories of the other three fundamental ... 2 Adding to @Guillochon's answer, there are even a number of general relativistic tests in our solar system, the most famous being the precession of the perihelion of Mercury. In short, the location of the point of closest approach to the Sun (perihelion) for the planet Mercury is a changing quantity. Essentially, given one full revolution, it doesn't trace ... 2 "in a spherically symmetric distribution of matter, a particle feels no force at all from the material at greater radii, and the material at smaller radii gives exactly the force which one would get if all the material was concentrated at the central point". This may be poorly worded. The idea is not that the distance from the center of the body must be ... 2 Close to a black hole, but still outside the event horizon, orbiting arbitrarily fast, wouldn't result in a circular orbit, but in a trajectory towards the black hole. In this extreme scenario, centrifugal force could be interpreted as attractive. In more usual settings, centrifugal forces point outward. Since the centrifugal force (in usual settings)$m \...

2

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 ...

2

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 ...

2

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, ...

2

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 ...

2

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 ...

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 ...

2

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 ...

2

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 ...

1

Yes, and it is not uncommon for an orbit have an eccentricity close to one. The wikipedia site, linked in a comment above, notes C/1980 E1, which entered the inner solar system with an eccentricity close to one, but had a close encounter with jupiter and was accelerated. It left the inner solar system with a eccentricity of 1.05, and so is on a hyperbolic ...

1

Presumably you're supposed to start by calculating the minimum orbital period a comet can have for a 10,000 AU aphelion distance. Then divide that period (number of years) by five, assuming we see on average one of these every five years. That will give you an estimate for how many there are out there. (I would also tell the person setting the question ...

1

As far as our current understanding of gravity, no. The common analogy is a rubber sheet with marbles. The sheet can only be pulled downwards, so you cannot have a repelling force. To pull the sheet upwards would require something like inverse gravity, which is not yet known to current physics. Of course dark energy seems to look like this, but as the ...

1

This sounds very ambitious! I think your way in to this topic might be to have a look at the works of two authors - Matthew Bate and Mark Krumholz. The reason I suggest these two names is that they are representative of the two basic approaches to simulating the collapse of molecular clouds (which has little to do with n-body simulations). These approaches ...

1

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 ...

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