# Tag Info

2

I think we must be a bit careful what we are discussing here. General relativity states that when an observer is on a geodesic, i.e. in freefall, they are not experiencing gravity. That's Einstein's famous falling elevator gedankenexperiment. It's not that gravity is there but cannot be measured, or that it is there but canceled out by the equivalent ...

-2

Suppose the universe consists of only a hollow sphere. The diameter of the sphere doesn't matter, but we require that the shell of the sphere be uniform thickness. The sphere will have finite mass, therefore, if an observer at any distance from the sphere were to measure the resulting gravitational field, he would find a non-zero field, and hence, a force in ...

15

This answer just amplifies on the correct answer by Barbierium. Are there places in the Universe where there are no gravitational forces? The answer to this question is that there is no answer to the question. To define whether or not a gravitational force would act on a test particle at a certain point in space, we have to define some frame of reference. ...

-2

If I understand Einstein's theory (and if it actually is correct) then gravity is the shape of space. That shape is imposed on it by mass. Thus if there is mass in the universe, there must be gravity. Since (per the Big Bang theory) all of space & mass was once a single point (or something really small) which expanded at less than the speed of light, ...

-2

It depends on what your definition of the Universe. As long as there's mass, there's gravity. Gravity acts infinitely at causality speed, also known as speed of light. Some data suggests that there's a tiny, but everywhere present force that pulls very far away objects in opposite directions, essentially adding more space in between matter, but only where ...

24

In two dimensions I think I can infer in a lame, unconvincing and rigorless numerical way that there are likely to be zeros in gravity from a random distribution of objects there can be points of zero gravity. I create a space with 20 randomly distributed point sources, calculate and plot the force field on a 2000 x 2000 grid then choose the smallest grid ...

12

Gravity extends to infinity, so no, strictly theoretically speaking there is always some gravity present. In theory, even in this case we could have points in space where gravitational forces cancel out, but given the complexity of our universe, this just won't happen in practice. As a more relaxed viewpoint - there are special points around orbiting objects ...

1

Yes, this is pretty much exactly how Modified Newtonian Dynamics (MOND) works. It is observed that Newtonian mechanics works very well for large accelerations, but an ad hoc correction function to Newtonian gravity is proposed at small accelerations. This function can be tuned such that the rotation curves of galaxies can be explained without the need for ...

0

"Acceleration" means changing the velocity vector, so an object in a circular orbit is always "accelerating" (because the direction of the velocity changes), even if its speed is constant. – Peter Erwin

0

A black hole does not necessarily suck in everything around it. Something that is outside $r = 2GM/ c^2$ radial distance (Schwarzschild radius) from the black hole, behaves in a similar way as if the black hole were a normal gravitating object. Here is a Schwarzschild radius calculator online For a mass equal to that of the earth, the Schwarzschild radius is ...

2

I will try to generalize your question to 'why does any object not become a black hole?'. It is indeed true that the center of mass of an object pulls the mass around it, so why does it not collapse? We need to see, what force is balancing the force of gravity. If you press an object (let's say: Iron), as hard as you can, why does it not get destroyed ...

8

The gravitational force on a small mass on the outside of a planet is always the Newtonian $$F_{G}=-\frac{GM}{r^2},$$ so any planet, and particularly, any mass in the universe produces a gravitational field acting on everything else. So if, for example, the mass is $M=2\times 10^{27}\rm kg$ (i.e. one Jovian mass), then the gravity field outside the planet ...

5

Anything with mass has gravity, so yes, such a planet would have gravity. However, gases tend to disperse in their surrounding environment, so you’d need a very massive gas cloud to collapse into such a planet for gases not to disperse. This raises the question of the pressure at the centre of this planet; it would be high enough to turn the gas at least ...

1

S62 is, according to Wikipedia, the so far closest detected star orbiting the supermassive black hole at the center of our galaxy, Sagittarius A*: At closest approach its velocity is about 0.10c (10% of the speed of light), so ~30km/s Arxiv paper: S62 on a 9.9 yr Orbit around SgrA*

-2

I think there was an ahem ACTUAL EXPERIMENT DONE ahem where two electrons were fired in the exact opposite direction and the negative charge changed to positive with same said change happening at the same time to the other electron. This experiment if valid and true would mean something is travelling faster than the speed of light.

1

The key word here is "eventually." The strength of the gravitation felt between two objects is inversely proportional to the square of the distance between the two bodies. As a result of this, raising an object from a nonzero distance from object A to an infinite distance from object A requires a finite amount of energy. In a Keplerian/Newtonian ...

0

For an atom of mass $m$ in the sun to be at a constant radius $R$, the centrifugal force due to any non-radial thermal velocity component $v$ must be equal to the gravitational force, i.e. $$m\frac{v^2}{R} = \frac{Gm}{R^2}\int_0^R \varrho(r) 4\pi r^2 dr$$ Assuming an isothermal gas ($v=const.$), this requires obviously a radial density distribution $$\varrho(... 2 If you have a spherical ball of matter, then outside that ball of matter, the gravitational field is the same as if all them mass were concentrated at a point (as a black hole) But inside the ball of matter, some of the mass of the ball is behind you and acts in the opposite direction. This means that the gravitational field is at a maximum on the surface ... 3 Just for fun I'll plot the data cited in @planetmaker's answer and use the equation there to plot the gravity. We'll apply Newton's shell theorem to their equation:$$g(R) = \frac{G}{R^2}\int_0^R \varrho(r) 4\pi r^2 dr = \frac{Gm(R) }{R^2} $$where m(R) is the mass enclosed by a sphere of radius R. I plotted using this script. It prints out 31.1 ... 4 Because you are somewhat larger than your Schwarzchild radius. In order to turn into a black hole and start experiencing exciting things like Hawking radiation, you'd need to be compressed down into a ball about 10^{-25} meters in diameter, about one ten-billionth the size of a single proton. At that size, you'd have a hard time sucking anything in: your ... -4 The earth is made from MANY particles, these particles get something called mass, from a thing called the higgs feild, this mass is how much gravity something has, the earth has its mass, and mass warps space time, like when you push your finger on your bed, it creates a curve that goes down, it happens for space, and things get attracted to earth. The ... 4 If you consider a radially symmetric mass distribution, the gravity experienced at a distance R from the center is caused by the mass inside the sphere with radius R, thus the mass$$M(R) = \int_0^R \varrho(r) 4 \pi r^2 dr In a radial-symmetric case one can show that the contribution of all masses outside $R$ cancel each other. Thus the experienced ...

2

Yes. Insofar as such thing might be possible (perhaps in the style of the novel “The Mote in God’s Eye”), and assuming a star that has a spherically symmetrical mass distribution: Gravity inside a star acts like if all the stuff in the shell of matter farther from the star’s center than you wasn’t there at all. You feel attraction only from the stuff “...

4

Actually, it doesn't. If a black hole and a star have the same mass, their gravity is the same. As the formula for the attraction (gravity) between two objects is given as $F=\dfrac{GMm}{r^2}$, where $G$ is the gravitational constant, $M$ is the mass of the black hole/star, $m$ is the mass of the smaller object, and $r$ is the distance between the two. Even ...

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