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The problem is in your assumption that the force is $F = 2GMm/r^2$. This is true for the force on a point mass from a sphere or another point mass, but not otherwise. What you need to do is sum each the force on each particle from every other particle. For a continuum object, $$\vec F = \int \rho \vec g\, dV$$ where $\rho$ is the density and $\vec g$ the ...


7

This is not altogether correct. In particular, it does not matter what the pressure of the room is (unless you reach extreme values), because actually what is holding the water column is not some "high" value of the pressure at the bottom of it, it is the fact that the pressure at the top of the water column (at the water-finger interface, or in the air ...


5

In your picture 1, the gravitational force is calculated incorrectly. The formula you have used only applies between pointlike masses. You have to divide the object into elements, calculate the contributions of each element and sum up. The picture 2 is only the first step in the whole process, so actually not even your picture 2 is generally correct.


4

Earth has few and relatively tiny sattellites other than the moon specifically because of this large moon. Note the mass ratio of our moon to our planet. It is the highest in the solar system by a large amount. This one large sattellite will over time sweep up and aggregate other smaller sattellites. Put another way, we probably did have other smaller ...


4

During a supernova, a star blasts away its outer layers; this actually reduces the mass of the star significantly. Any star or planet has an escape velocity - the slowest an object must be traveling for it to escape the gravitational field of the star/planet. For Earth, this is 11.2 km/s. (Note that this value doesn't account for any atmospheric effects.) ...


3

The gravitational field $\mathbf g$ equals, by definition, negative of the gradient of a correspondonding potential $\Phi$; \begin{align} \mathbf g = -\nabla\Phi. \end{align} Therefore, it suffices to produce a gravitational potential $\Phi$ whose value is zero at a point but whose gradient is non-zero at that point. This is straightforward to do. Let a ...


3

Your thought experiment of dropping an iron ball and a feather need not be in water; in fact, it is more commonly considered in air, but the pertinent facts are the same. All objects, regardless of their mass or composition, are accelerated identically by gravity. But within a particular medium, the acceleration of particular objects might be impeded by ...


3

It seems like they were able to rigorously prove the existence of N-body choreographies by using interval Krawczyk method to show that a minimum exist to the variational problem solved in the subspace of the full phase space satisfying some symmetry conditions. Following the links given I found this paper where they explain the method. It's not exactly a ...


3

Here is an example for the Sun. The figure below plots a (reliable) estimate for the interior density profile of the Sun, $\rho(r)$. So for a given radius $a$, the mass interior to that radius is given by $$ M(a) = \int^{a}_{0} 4\pi r^2 \rho(r)\ dr $$ And of course the gravitational field strength assuming spherical symmetry will be $$g(a) = - G ...


2

The planets are attracted towards the sun, as you would expect from the gravitational force. The planets don't fall into the sun, though, because their velocities are at right angles to that force. The planets end up being pulled by the sun into a circle. A planet's speed is constant, but its direction changes. I think it's easiest with a picture: ...


2

If you are really concerned about an axiomatic treatment, here is a method that takes the integral definition as a postulate, and then recovers the point mass situation as a special case: define the "interaction energy" of a pair of nonoverlapping mass distributions $\rho_1,\rho_2$ as ...


2

As you said, if feather has a stronger decelaration, is due to air friction (not because iron is more dense than feather or something). The same should hold in absence of gravity: if you give a boost to these balls, then the feather one will stop first. From this last consideration, I'd say that this medium does not affect gravitational force, but just it ...


2

A Lagrange point is a position relative to a system of two-body gravitational objects at which a third negligible small object, if its velocity is correct too, would not move relative to the two other objects. The two gravitational objects will orbit around the center of mass of both objects together (also called the barycenter). I am not sure, but I also ...


2

Your expression for the velocity looks right; but we have to get a few other things taken care of. First - the center of the marble doesn't move from 0 to 2R, it moves from r to 2R-r - so the potential energy due to this is smaller than mg(2R) which is what you had in your expression. On the other hand, you need to take account of the energy of the sphere ...


2

The formula $F=G \frac{m_1 \cdot m_2}{r^2}$ is valid only for point masses. However, it can be applied to non-point masses if its spherically symmetric. Enter Shell Theorem: 1.A spherically symmetric body affects external objects gravitationally as though all of its mass were concentrated at a point at its centre. So, when a spherically symmetric ...


2

As the formula you wrote suggests, it depends on $r$, the distance between the satellite and the centre of mass. $3.1\:\mathrm{km/s}$ is for geostationary orbit, and $8000\:\mathrm{m/s} = 8\:\mathrm{km/s}$ is for low Earth orbit. Please see Wikipedia for more details.


1

I assume this question is about classical mechanics. Therefore gravitational force doesn't have a limit. According to conservation law of energy (P.S. Gravitational energy $E_p = -\frac{GMm}{r}$): $\frac{1}{2}mv^2-\frac{GMm}{r}=0,\ when\ r\rightarrow\infty$ Therefore, the escape velocity is $v=\sqrt{\frac{2GM}{r}}$.


1

Let's use a spring scale in water. Putting both the feather and the iron ball of same masses we notice that the spring shows the same reading(say $a_0$) for both feather and iron ball. Hence the gravitational force on both the feather and iron are same. Now let's repeat this experiment in air(or oil or some other material). We notice that the spring scale ...


1

Even though the people who replied are correct, fun fact: the speed that gravity is propagating is very slightly affected by mediums. Here are 2 pages from Kip Thorne's lectures at Les Houches. Gravitational waves are affected by a medium through a dispersion relation, but the effect is ridiculously small and is neglected.


1

There are multiple forces involved in your thought experiment. Namely, the gravitational force and the 'force' of air resistance. This does not mean that the presence of the air affects the gravitational force. As another thought experiment, consider two planets orbiting in space. Somewhere between these two planets there is a point where the ...


1

Your answer (1) is the correct one. It is actually quite simple if you think in terms of conservation of energy. What you have described is a simplified version of a two body problem. Note that strictly speaking, both the doughnut $(D)$ and the ball $(B)$ will move towards each other. But without outside influence, their combined center of mass should be ...


1

1) is correct. The wrong reasoning about 2) is that what you have in mind is probably Newtons Law for point masses. When the sphere is close to the doghnut the gravitational force will be more complicated, but still point towards the centre of the doughnut due to the symmetries in the situation. It will however stay finite, because all points of the sphere ...


1

The harsh answer is, "by solving Einstein's field equations". That is an extremely more difficult problem than using Newton's law -- the equations take up at least a page when you write them out. However for the case of light near a black hole, you can treat the black hole as unaffected by the light, and instead solve the null geodesic equation. This is ...


1

There follows my try to decompose the solution into a minimal amount of calculation and apart from that only geometrical considerations. The centripetal acceleration is $a_c=\frac{v^2}R$. It is directed towards the center. We define the $z$ coordinate as starting at the top and pointing vertically downwards (see the following Figure). The conservation ...


1

A mass moving in a circle has centripetal acceleration $v^2/r$ directed toward the center of the circle. You can get $v$ from potential energy. The mass here has two forces on it. Gravity is constant and down. The reaction force of the surface (assuming no friction) is normal to the surface. When the sum of these two forces becomes less than centripetal ...


1

For your first question: Yes gravity is mediated in all directions equally. As for your second question: I assume you are asking about elliptic orbits. If this is the case then, from the law of conservation of angular momentum, when the planet is closer to the sun it has greater velocity, and when its further out it travels slower because $\vec{L}$ = ...


1

UPDATE I finally got it right. My thanks goes to @NeuroFuzzy, who pointed me in the right direction. According to wiki's Legendre polynomial solution for the elliptic integral, "an exact solution to the period of a pendulum is:" $$T=2\pi\sqrt\frac{\ell}{g}\sum\limits_{n=0}^\infty ...


1

This is really a comment to Mathbreaker's answer, but it's hard to do formulae in comments. If you simply solve: $$ T_m = 2\pi\sqrt{\frac{\ell}{g}}(1+\frac{1}{4}\sin^2(\tfrac{\alpha}{2})) $$ for $g$ you get: $$ g = \frac{4 \pi^2 l (4 + \sin^2(\tfrac{\alpha}{2}))^2}{16 \tau^2} \tag{1} $$ We use the identity: $$ \sin^2(\tfrac{\alpha}{2}) = \tfrac{1}{2} - ...


1

This concept has gotten consideration from NASA. In the NASA MagLifter concept, a 300-600 miles per hour speed on a superconducting magnetic levitation track appoximately 2.5 miles long and going up a mountain to about 10,000 feet is proposed. The option of using a helium filled tunnel to reduce drag was also proposed.


1

Yes, what you have formulated is fine. The pressure acting on the water from the bottom of the straw will be equal to the weight of the water times the cross section area. So $101325\pi r^2$ is the force acting from below also. That is precisely why there is an equilibrium and the water is not falling.



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