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No you can't and the behaviour of bodies with mass and of light is completely different near a compact, massive object if you use Newtonian physics rather than General Relativity.
In no particular order; features that GR predicts (and which in some cases have now been observationally confirmed) but which Newtonian physics cannot:
An event horizon. In ...
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Imagine "gravity" spreading out in a sphere, like light from a bulb.
For each doubling of the distance, the sphere has four times the area. The surface area of the sphere is proportional to the square of the radius. If the same gravity is stretched over that sphere, the force of gravity would be inversely proportional to the square of the radius.
...
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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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I am not an expert in physics and the explanation of the others is excellent. However, I noticed a flaw in your reasoning which they did not address.
You have written:
Considering the Newton's Law of Gravity equation $F = GM/r^2$, if the radius of an object becomes super small, then it can technically have immense gravity.
Hence I deduce that you read the $...
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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 ...
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Both expressions are incorrect. The first should be
$$\frac{GM_{\text{moon}}}{(R_{\text{moon}}-r_{\text{planet}})^2} - \frac{GM_{\text{moon}}}{{R_{\text{moon}}}^2}\tag{1b}$$
or
$$\frac{GM_{\text{moon}}}{{R_{\text{moon}}}^2} - \frac{GM_{\text{moon}}}{(R_{\text{moon}}+r_{\text{planet}})^2}\tag{1a}$$
where $R_{\text{moon}}$ is the distance between the center of ...
9
While admiring @ProfRob's answer I'll add some additional perspective/background that may serve as a helpful stepping-stone since not every Astronomy SE reader is prepared to embrace General Relativity in all its glory.
Can Newton's gravity equation explain why black holes are so strong?
The simple equation1 $F = GMm/r^2$ doesn't explain anything but it ...
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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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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
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 ...
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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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Planetoids can have moons and the minimum size is "pretty small". For example 2003 SS84is a small Near-Earth asteroid, with a diameter of 120m and a moon of about 60m in diameter, which orbits at a distance of 270m ever 24 hours. It probably didn't form by "attracting the moon" but the moon probably formed as a result of impact splitting ...
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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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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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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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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 ...
5
The ratio $T/\Omega$ tells you about the acceleration of the system - or more specifically, the second derivative of its moment of inertia - it does not tell you about the velocity.
If the system collapses because it has $T/\Omega<0.5$, then when it reaches $T/\Omega=0.5$ it stops accelerating. That doesn't mean it stops collapsing. It overshoots in the ...
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If Einstein's GR equations are expanded in terms of familiar coordinates (Cartesian, spherical,...), the dominant or leading terms of the expansion (for the acceleration) can be written as the single Newtonian term GM/r^2. The next terms of the expansion can be considered as GR corrections to this leading term.
Before the publication of GR, 19th-century ...
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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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The validation is the same as the validation of any astronomical theory: it fits the data. Newton's law of gravitation was formed empirically, by observing the motion of the planets.
In particular, he took Kepler's laws, which were observational, and $F=ma$ from his physics and determined that the elliptical orbits we see must be caused by something with an ...
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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 ...
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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 ...
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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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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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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 ...
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