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Yes it is theoretically possible, as discussed in the other answers and indeed we already do a variation of harvesting a planet's orbital kinetic energy in the space navigation manoeuvre called "gravity assist" or "slingshot" to boost a spacecraft's speed without expending propellant. Here one makes one's spacecraft "collide" with a planet (i.e. make a very ...

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If there is no resistance then there will be no net torque applied about the center of mass. So any initial rotational speed will remain. The rotation center is going to be the center of mass. The effect of gravity will be to accelerate the center of mass, and it will have no effect on the rotational motion of the body. See this accepted answer for a ...

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One can consider to harvest power at Gibraltar strait (Moon/Sun energy from tides). Any turbine to put in space is not feasible and, if not, then soon it will be locked to the motion of the Moon, and then useless.

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... you aren't envisioning a giant tunnel in space that the moon would pass through, are you? Like the crude drawing below? We don't have the materials strong enough to attach it to earth with. We don't have enough fuel to launch that into space The turbines would... collide with the surface of the moon to generate electricity? Even if all this worked, ...

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Theoretically it is possible. After all, what law of physics would it violate? However, any energy that we harvested from the moon's orbit would be removed from the moon's potential energy, causing it to move closer to earth. If we harvested enough energy this way, the moon would collide with Earth!

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Hint: conservation of energy gives $\frac{1}{2}m (\dot{x}^2+\dot{y}^2)+mgy(x)=mgy(x_0)$. But $\frac{dy}{dt}=\frac{dx}{dt}\frac{dy}{dx}$, and $dx=\frac{dx}{dt}dt$. You can use these relations to solve for $\dot{x}$ in terms of $x$, and then your integral of $\frac{dx}{\dot{x}}$ must be your integral of $dt$.

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In a simple universe where the only thing that had gravity on you was the Earth, there is no such limit. As you seem to have seen from Newton's law of gravitation, there is no finite $R$ such that $R^{-2}$ is zero, and that's the only thing that changes. In real life, you can find places where other objects in space have equal but opposite gravitational ...

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This logic when applied to the electrostatic force of 1/r led to the discovery of quantum mechanics. When nuclei and electrons around them were visualized, because that is what the data said, the big question was "why the electrons do not drop on the nucleus and the whole thing becomes neutral"? An electron dropping into the nucleus would classically emit ...

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Gauss law is useful only in the cases of high symmetry systems like sphere, infinitely long (or very long and thin) cylinder, or infinite plane. You can't even apply it to curved sides of "real", short cylinder. You need to show somehow that gravitational field is the same everywhere on you Gauss surface, so that integral in the Gauss law turns out to be a ...

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Ignoring the Coriolis effect, which is velocity dependent. I get for an x-y Cartesian coordinate system, with the y component toward the North Pole and x component pointing out from the equator at the position you are measuring the effective g=G , the following (in ordered pair vector representation); G= ...

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Does our universe have an even distribution of matter in every direction In physics we have to define our terms. In this case, universe, matter and the dimensions in which we define direction, and define what we mean by uniformity. If we take universe to be the observable universe, dimension of the order of billions of light years and units of matter ...

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I am refering to large scales I hope and unfortunately I am not schooled in the terms and there defintion's used by physicists and the community I am merely approaching this from a educated layman point of view and with the application of some reason. Within contained heated fluids there can be a movement of flow in a particular direction but the contents 0f ...

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The distribution of matter is highly uneven in the "local" universe. Dark matter appears to be concentrated in and around galaxies and in clusters on scales of tens to thousands of kpc, and probably forms even larger filamentary structures which lead to the large scale structure we see on very big scales (tens of Mpc). All around us we see galaxies in ...

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The universe is as far as we know, generally uniform. That is, if we "zoom out" it all looks more or less the same.

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You actually don't need to take any time derivatives here. Since the energy, $E$, is a constant, you only need to know it at one moment in time (say at t=0 as an initial condition), and you know it at all other times. Thus, just solve for $\dot{x}(t)$ in terms of the other quantities ($x,E,...)$ in the second equation that you have to get the equation of ...

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No, the Earth's gravity varies over its surface because (a) the Earth is not spherical, (b) it does not have a uniform (or spherically symmetric) distribution of density and (c) of course the "surface" is at different heights. You can find a local gravitational strength (and direction - because the local gravitational vector does not necessarily point to ...

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No, acceleration due to gravity varies, depending solely on the altitude on which Gravitational force is working, the approximate* value can be determined using equations: g(eff, h) = g*[(r/r h)^2] H being the altitude, r being earth's mean radius, g effective is the acceleration due to gravity at an altitude h above sea level. Thanks :) Approximate because ...

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If the Earth were a perfect sphere, it would be the same everywhere on Earth's surface. This is known as the shell theorem. It's not too hard to show mathematically, but you can think of it as the fact that all the mass that is close to you balances roughly with the mass that is far away. If you were to tunnel into the Earth, however, you would only ...

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Like you say, it would be a very difficult problem if we had to resort to using Newton's laws to calculate the forces and accelerations at every point along the trajectories. Thankfully, the universe has conservation laws which make this problem immensely easier. Try using the conservation of mechanical energy and the conservation of total momentum.

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You can't prescribe the motion in the general case. You need to build a simulation. That is calculate the acceleration of each body at every instant and integrate over a small time period to find the change in position and velocity. Then re-calculate the accelerations given the new positions (and velocities) and repeat for many small time steps.

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Gravity only acts on the vertical component, not the horizontal component. The force of gravity speeds an object down on decent as gravity makes the ball fall to earth.

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as we know the energy of an elctron increases as it moves away from the nucleus..that means potential energy is directly depending on distance between electron and nucleus..but when we derive an mathematical expression we get energy inversly proportrional to the radius...to compensate diz we add MINUS sign to show that lesser negative enrgy means more energy ...

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For Newton's shell theorem, it is advisable that you consider just that: shells. Consider that your solid sphere is composed of multiple thin spherical hollow layers or shells having a thickness $dr$ nested in each other. If you want to find force experienced by a point at P, any shell enclosing P can be neglected. This is because gravitational field inside ...

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Acceleration is any change in velocity, whether it's the direction of the velocity or its magnitude. In the case of a perfect orbit it is the direction, the tangential velocity of the satellite should keep a constant magnitude and just change direction as the satellite orbits. When the velocity is slightly less than what is needed for a perfect orbit, it ...

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You are specifically asking about a first order correction to the formula. Starting from $$F = \frac{GMm}{(R+h)^2}$$ for the projectile at height $h$, we can rearrange this as $$F = \frac{GMm}{R^2}\frac{1}{(1+\frac{h}{R})^2}$$ When $h\ll R$ we can use a first order Taylor expansion to write $$F = \frac{GMm}{R^2}\left(1-\frac{2h}{R}\right)$$ Finally, ...

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The equation $$\ddot{r} = -\mathbf{g},$$ is valid iff $\dfrac{h}{R_e} <<1$. The gravitational force is : $$\mathbf{F} = -m\dfrac{GM_e}{R^2}\mathbf{\hat{R}}.$$ Now one defines $r = R - R_e$ with $\dfrac{r}{R_e} <<1$. Then one has : \begin{align} \mathbf{F} &= -m\dfrac{GM_e}{R^2_e(r+R_e)^2}\mathbf{\hat{R}}\\ & = ...

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The key to flow of macroscopic particles is to make sure they don't "clump" or jam. I don't know what is propelling them down your pipe (gravity? Air flow?) but that will affect the answer. In general, adding some vibration keeps particles flowing freely; a larger pipe diameter with minimal obstructions / bends is the other thing. Making the pipe diameter ...

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I don't think creating a vortex effect will increase the velocity of flow, because it would direct movement in a direction other than forward. Smooth bore gunbarrels, for example, have greater muzzle velocity than rifled gunbarrels (assuming all other variables are equal). You might want to maximize laminar flow and to minimize turbulence inside the tube. ...

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You can get this more "intuitively" (idiosyncratically): the flux of this force in closed surface is equal to the quantity of source inside (is a Gauss's Law). This source could be a mass or a charge. The physical picture is: the pressure applied in a closed surface by the field-force is proportional to the quantity of source inside. You can get the ...

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Let me expand on (and correct a minor error) in what I said in the alluded-to thread. Unfortunately, I do not know the mentioned article (although "The Moon's Twin", published in 1989 in "The Magazine of Fantasy and Science Fiction", discussing the Jupiter/Io system, might be it). Therefore, I can't directly address it, but I think I can say three major ...

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When you have a lightly damped oscillator, there is a small correction to the resonant frequency. This is derived in detail on the wiki page for the harmonic oscillator. The form they give is $$\omega = \omega_0\sqrt{1 - \zeta^2}$$ Where the $Q$ (quality factor) of the oscillator is given by $Q=\frac{1}{2\zeta}$.

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You see, when you have a pendulum with friction you account it by including a force $\vec{F}_r=-b\vec{v}$. Then your differential equation for the pendulum is $$ml\ddot{\theta}=-mg\theta-bl\dot{\theta}\iff\ddot{\theta}+\frac{b}{m}\dot{\theta}+\frac{g}{l}\theta=0$$The solution of this differential equation depends on the values of $b$, $m$ and $l$ but since ...

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For simplicity, consider a perfectly circular orbit; the gravitational acceleration is always at a right angle to the velocity vector. This means that the speed cannot change despite the fact that there is constant acceleration. Note that for the speed to change, there must be a non-zero component of acceleration parallel (or anti-parallel) to the velocity ...

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There are private space companies that will allow you to take a trip to space(it costs around $250,000). Take a trip and go in orbit around the earth or go to another space station. Another solution would be to go to the Zero-G place as Lasse said or fall farther as Rob Jeffries said which are both much less expensive. 1 Establish a longer free fall time, for example in an aircraft or space station. 2 A neutral object can be induced a non-zero charge when placed in an electric field. The charges or dipoles within that material will simply rearrange or rotate to aline slightly. An electric field will be generated, which will counteract the current field. Have a look at dielectrics. The gravitational constant$G$is... A constant. Just like the ... 8 Firstly, the gravitational field inside the Earth, decreases with depth. To a first approximation, you can use the shell theorem for spherically symmetric mass distributions to argue that the gravitational field at some depth is due only to the mass enclosed within a sphere interior to that depth. If we further make the crude assumption that the Earth's ... 2 Q1. In the case of a uniform spherical distribution you cannot sense anything further away from the center than you are. This is directly derived from Gauss' law for gravity. At the center you do not feel any net gravitational force from earth at all. Think about it this way: earth is pulling you up from all directions exactly the same way, so all the ... 0 Yes, it would work. The trouble is becoming stationary. The earth is moving, and the trees, sky, and even you are moving with it. 1 Assuming you can stay stationary, you will move 180 degrees in longitude. Unless you started at the equator, you won't be at the opposite side of the world. For example, if you start at San Francisco, CA (37.788 N, 122.466 W) You would be at 37.788N, 57.534E, in Northern Iran near Turkmenistan. Why can't you stay stationary? You need somehow to ... 0 You can not become stationary. what ever you do, you are moving with the earth as it rotates. (I think no more explanation is needed.) 0 When two object orbit each other, the shape of their orbit is independent of their relative mass. In fact they will each have the same shape of orbit, but scaled by the mass of the other. So if you have two objects of mass$m$and$2m$respectively, then the former will have an orbit (circle or ellipse) that is exactly twice as big as that of the other. ... 3 To say that the orbit becomes more circular the greater the Sun's mass is not true. Instead, the eccentricity (i.e. how much the shape of an orbit varies from being circular) is governed by a couple of factors. If you have a planet orbiting about the Sun with a mass much less than that of the Sun, and you know the following for an instantaneous point in the ... 4 No. The shape of the orbit, i.e. how elliptical it is, does not depend on the relative masses of the two bodies. All objects in the solar system orbit around the centre of mass of the solar system. For obvious reasons, namely that the Sun contain far and away most of the mass of the solar system, the centre of mass of the solar system is quite close to the ... 2 Yes, objects with mass all attract to each other and move each other of course, except that the star doesnt change it's theoretical orbital shape depending on it's mass, it probably just experiences small tidal forces that aren't much bigger than it's own centrifugal forces. The oscillation of the star position is a complex dynamic based on it's surrounding ... 3 One has to realize that Kepler's laws are a mere approximation. The motion of planets around the sun is a two-body problem. In case of such two-body problems, both the bodies revolve around the center of mass. But it turns out that Sun is much much heavier than the planets. So the center of mass of the system is very close to the Sun and Hence it is a good ... 2 For two objects to remain in a stable circular orbit, the force acting on them must be equal to the centripetal force corresponding to their rotation. $$F=\frac{mv^2}{r}$$ or in terms of angular velocity $$F=m\omega^2r$$ where$r$is the radius of orbit in this case. As the gravitational force acting on the two stars is the same.$F\$ is equal in both ...

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I will be assuming that the system has some angular momentum about the center of mass initially. If the system has no angular momentum then both the stars would accelerate towards each other and end up colliding. The problem is a two body problem. In such cases, both the stars would revolve around the center of mass. If the distance between two stars is ...

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Does Sphere C have any effect of on the gravitational force between A and B? Nope. C does not make any difference to the force between A and B, but its introduction has the effect that the net force on A has contributions from both B and C. (Likewise for the other spheres too.) This is because gravitation obeys the principle of superposition. Is ...

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The sphere will stay in-between because net force is zero. However, the spheres will all lump together on sphere C because while C will not move, A and B will.

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