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Hot answers tagged gravity

5

It's probably easier to measure gravitational masses for anti-hydrogen or anti-ions, because those particles can be cooled and trapped using only electromagnetic fields. The problem is that antiparticles are generally "hot" when they are created. The processes that create antiparticles have an energy threshold beneath which they don't work at all (this is ...

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As you may know, the friction is proportional to the normal force of an object or in this case the force of attraction between magnet and refrigerator. If your force is strong enough then the friction will be sufficient and the magnet will not slip (on earth the force of friction must exceed the mass of your magnet multiplied by 9.81 m/s). If we assume ...

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In general, air pressure in the Earth's atmosphere is hydrostatic pressure, caused by the Earth's gravitational field. If there was no gravity then there wouldn't be any centripetal force and all the air molecules would just float away into space. This is why there is no atmosphere on the moon - because it doesn't have enough gravity to sustain one.

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In some sense yes. Let me explain a little. If we were to take a sealed container of gas and put it into free space far away from other bodies so that the gravitational force on the box is negligible would you agree that there would still be some pressure in the container? If we assume we have an ideal gas then the pressure is simply given by $$P=nk_{B}T$$ ...

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Even though you were wrong, you nailed the critical question at the heart of the necessary understanding. Bold is my emphasis: Modelling this planet at my head then anywhere that effective gravity increases in strength is effectively lower potential energy "down" from anywhere that it doesn't. I would expect this to cause the planet to deform and reshape ...

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If the rope is "radially directed" it means every point has the same angular velocity $\omega$. Assume a length $2\ell$, then you can integrate the force on the rope from $r-\ell$ to $r+\ell$ - gravitational force must equal centripetal force. This gives you an equation for $\ell$ as a function of $\omega$ and $r$. See if that gets you going.

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I would go with the hydrostatic equilibrium condition being the source of the issue here. When there is no polytropic gas when $\xi \gt \xi_0$, there is no need for any hydrostatic equilibrium to be in place - the gravitational potential is free to take on any value, determined (together with boundary conditions) by Laplace's equation: $$\nabla^2\Phi = 0$$ ...

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When we assume hydrostatic equilibrium, the pressure gradient is taken. Maybe the pressure should be 0 when $\xi>\xi_0$, so that constraint should be explicitly applied. This is generally understood as an implicit restriction/constraint of the hydrostatic equilibrium (HSE) condition for starting the Lane-Emden equation: HSE is for the star and not ...

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You should look at it like an asymptote. Yes the proton would accelerate but it would probably accelerate to .999991c or more likely less due to the massive energy required to accelerate something so fast already. Therefore you could always keep accelerating your particle but it would never cross the Light-Barrier.

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It's probably best to not think of single photons as sources of gravitational energy. For one thing, most bulk electromagnetic fields are not eigenstates of the photon number operator. For another thing, the thing that couples to the gravitational field is the energy density of the field. This density is proportional to the intensity of the field, ...

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Air pressure exists because if we place something in a gas, then the molecules/atoms flying around will keep banging into it, and in this way produce a net constant force per unit area. As explained by @Chris2807 in the neat formula $P=n k_{B} T$, this is proportional to how many particles there are (since this is proportional to the amount of "banging" in ...

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You can't shield against gravity. If you could, you could just shield gravity from one side of a wheel and make a perpetual motion machine. Space elevators have been considered, but you still have to fight against gravity and friction. It just means that you can just climb a rope instead of using a rocket.

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A partially reflective Dyson sphere is equivalent to asking what happens if we artificially increase the opacity of the photosphere - akin to covering the star with large starspots - because by reflecting energy back, you are limiting how much (net) flux can actually escape from the photosphere The global effects, depend on the structure of a star and ...

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Your intuition is correct. At that scale, the planet behaves essentially like a liquid, and it's shape will equalize into an ellipsoid where: The local gravity vector is everywhere perpendicular to the surface, and The downward force felt by a person standing on the surface is the same everywhere. But I don't think you could get an eccentricity as ...

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