Tag Info

90

TL:DR Jupiter isn't dense enough for its gravity gradient over Earth's radius to produce a 1g tidal acceleration, even right at Jupiter's surface. thanks to PeterCordes Jupiter's gravity will pull on the Earth itself, as well as everything on it. It's not like a vacuum cleaner that selectively lifts small and light objects, the gravitational force will ...

73

Essentially gas does, it just happens to form a star first. Mass is not the only factor in creating a black hole. You also need for this mass to reach a high density. In the process of doing this, a star usually forms. The energy producing processes in the star interior produce a pressure that balances the gravitation attraction. This prevents a star from ...

58

You do, but it's too small to really notice First, it's not correct to say that we don't feel Earth's rotation because it's rotating at a constant speed. Think about driving a car, or riding in an airplane. Whether you're cruising down the road at 90 kph, or soaring through the air at 900 kph, you don't really "feel the speed". However, When you take a ...

54

As you said, the mass of the Moon is 1.2 percent that of the Earth. Now, if you mean the gravitational acceleration at the surface, it is calculated like this $G\frac{M}{R^2}$, where $M$ is the mass, and $R$ is the radius of the celestial body. The moon's mass is a hundred times smaller, but the radius is four times smaller, meaning its surface gravity will ...

52

1. Is material on Earth's surface not in free fall around Earth's center? No. Material on the Earth's surface -- or inside it -- is not in orbit, and so is not in free fall. You can temporarily put yourself into an orbit (and thus into free fall) by jumping up into the air, or jumping off a higher surface. When you do this, you are briefly in a very ...

48

Two rocks placed in space with no relative motion are going to be attracted by gravity, and hit. 3 rocks, placed in space with no carefully rigged symmetry, will likely miss each other, as the gravitational attraction of the additional rock changes their course. Those near misses are the beginning of rotation. Multiply that effect by trillions, and you have ...

48

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 ...

45

You really need a full-blown stellar evolution model to answer this precisely and I'm not sure anyone would ever have done this with an oxygen-dominated star. To zeroth order the answer will be the similar to a metal-rich star - i.e. about 0.075 times the mass of the Sun. Any less than this and the brown dwarf (for that is what we call a star that never gets ...

45

No need to guess. There's solid research done in this field. Even Wikipedia has some info: As two black holes approach each other, a ‘duckbill’ shape protrudes from each of the two event horizons towards the other one. This protrusion extends longer and narrower until it meets the protrusion from the other black hole. At this point in time the event ...

43

It's more for safety than anything else. Space is a very dangerous place for so many reasons. And making mistakes can very easily cause death. Being weightless does not mean you lose mass, so momentum is just as difficult as ever. But whereas on the ground you can easily use friction to stop, in space if you try to stop against the floor you will just move ...

42

It is not true that "objects float around" in the solar system. Perhaps you have seen video from the space station, and you can see things floating. This is not because there is no gravity, but because everything in the space station going at the same speed in the same direction. This makes it look as if things are floating. In fact the space station and ...

42

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. ...

37

The answer to this is surprising: We are. And many (if not all) other galaxies. And they move faster than light. See, the universe is expanding, at an accelerating rate. The fabric of spacetime itself stretches out, so that galaxies seem to move away from each other. The interesting thing is that relativity does not forbid these from moving away faster ...

37

It depends on what object it's acting on. There are many objects, including stars, that have magnetic fields where Lorentz forces on charged particles like electrons and protons are stronger than the gravitational force on them. Also remember that the strength of the Lorentz force depends on the speed of the particle moving through it, so a fast enough ...

36

The actual image isn't much. I was able to find it from Science, and this is all it is: It's just a ripple, seen at slightly different times from two different observatories. The shift fits perfectly by shifting it by the speed of light difference in their locations. Thus is the proof of gravity waves. It should be noted that the reason there are two ...

33

To help with James K's excellent answer, a visual representation might help. Let's look at a thought experiment - Newton's Cannonball. Let's say you have a cannon, high enough that it's being held above Earth's atmosphere. You fire it, and it falls to Earth a little ways away ("D" in the below diagram). You fire another one with more power so it's moving ...

32

The first question as stated has a rather trivial answer: "If the sun magically disappeared, instantly, along with all its influences, how long would it take its gravity to stop having an effect on us?" Since the Sun's gravity is among its influences, it would instantly stop having an effect on us. That's just part of the magical situation, and doesn't ...

31

An object on Mimas' surface would be much more attracted to Saturn than it is to Mimas. You are missing that Mimas as a whole accelerates gravitationally toward Saturn. What this means is that a point on the surface of the Mimas will feel the acceleration at that point toward Saturn minus the acceleration of Mimas as a whole toward Saturn. This is the tidal ...

31

The issue here is whether pairs of planets can become gravitationally bound to each other. In the two-body problem the trajectories or orbits are ellipses (bound orbits), parabolas and hyperbolas (unbound). For all practical purposes, an encounter looks like they start out at infinity with some finite speed, approach each other, and then maybe fly away or ...

31

The part of your intuition that is correct is that Jupiter pulls the Sun towards it. The problem is that "pulls towards" does not mean "brings closer"! The gravitational force results in an acceleration towards an attracting body, which is not a displacement or even the derivative of displacement, but the second derivative of displacement....

30

First of all: "How gravity really works" is a deep question, and any serious scientist would quickly concede that all we have is an incomplete working model. You certainly have heard about General Relativity; the first image on the page is your trampoline. Our working model, General Relativity, is working because it explains a lot of observations very ...

29

First of all, I think your question belies a misunderstanding of the nature of the LIGO observatories. The nature of the detectors is that they act like a microphone, as opposed to a camera. What that means is that they are sensitive to gravitational waves coming in from most directions, but don't have an ability to distinguish where the waves came from. ...

29

Going from Newton's theory to Einstein's theory is not simple. It's not like you can just add a term to Newton's gravity, like $\textbf{F}=-{GmM \over r^3}\textbf{r} + \textbf{f}(\textbf{r})$ and obtain the right formula. To say it with Feynman's words, you can't make imperfections on a perfect thing, you have to make another perfect thing. And the new ...

28

As HDE 226868 noted in his answer, the Sun is not going to go supernova. That's something only large stars experience at the end of their main sequence life. Our Sun is a dwarf star. It's not big enough to do that. It will instead expand to be a red giant when it burns out the hydrogen at the very core of the Sun. It will continue burning hydrogen as a red ...

28

Potentially a short (less than a second) burst of gravitational waves (GWs) would be detected. Much depends on asymmetries in the core collapse, since a spherically (or even axially) symmetric collapse would not produce GWs (e.g. Morozova et al. 2019). However, theoretical models suggest that the GWs start at low frequency (tens of Hz) and are associated ...

28

There's a few parts to this question so there's more than one answer. Earth gets knocked a little bit out of its orbit all the time by gravitational influence of other planets in our solar system. Jupiter and Venus are the primary two, but all the planets have some effect. These are called orbital perturbations and they tend to alternate, not add up. They'...

26

"Perfectly" is a funny word. Perfect circles are a mathematical abstraction. Real objects are not "perfect". So supposing a "perfectly spherical planet" is to suppose something that does not and could not exist. All real planets are made of atoms and anything made of little clumps of matter cannot be perfectly spherical. Even if you built a planet that ...

26

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 ...

25

The local dark matter density is actually quite tiny, on the order of $\rho\sim10^{-19}\text{ g/cm}^3$ (see e.g. Bovy & Tremaine (2012)). This means that there is roughly $0.001$-$0.01M_{\odot}$ of dark matter per cubic parsec - a staggeringly small amount. 1000 cubic parsecs would contain about one solar mass of dark matter - and that's a cube 10 ...

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