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Expanding on what Sofia said, the mass above the Earth isn't an isolated system because gravity is acting on it. As she said only approximated isolated systems exit, imagine a lump of iron in a room. There is a magnetic field set up in the room, but because you are inside it you don't know it. The lump begins to move and you measure KE. You would conclude ...


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If you imagine drawing the potential as a function of position and pick a point and zoom in on it, then if you zoom in enough (and the potential is differentiable) then it will look like a line. But the line might not be horizontal, when you assume the potential is spatially constant to first order you are assuming that straight line it looks like happens ...


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Your $E$ is potential energy in the rubber, which transforms to kinetic $K$. So your starting velocity $v$ will be: $E=1/2 mv^2$ From conservation of energy: $0-1/2 mv^2=0-mgh$ $h=v^2/2g$ and $v^2=2E/m$ Confirmed Interestingly enough the rubber does not obey Hook's law, and you need a lot more work if you want to find out what really ...


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I'm not sure whether this is a very advanced question from someone in a college course who knows integral calculus or a high-school student who knows none of it, so let me answer it both ways. The high-school answer should be intelligible to the college student, so let's tackle that first. Basically: You're not taking the average of either. The expression ...


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A particle subjected to only a conservative force field, without other restrictions, will move in a direction which reduces the potential energy of the system. The particle itself does not have potential energy; the particle-field system has potential energy (some may say the field-field system, but let's not nit-pick that yet). So what's important is not ...


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Couple of things to point out here, and I'm not sure whether either of those answer your question, so here are some ideas to help you think about the problem and guide you towards the full answer: The lowest possible value of Earth's gravitational potential is met: a)if the object is at Earth's exact center (r=0, namely there is no mass "below" him (m1=0), ...


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The equation you cited makes use of "big G", the universal gravitational constant. Generally, this equation is used if you want to calculate the attractive force between two bodies, such the moon and the Earth, or a satellite and the Earth, or the Earth and the Sun, or if you want to calculate escape velocity from he Earth's gravitational field. The center ...


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We treat a mass producing gravity as if all the mass is concentrated at the centre. Does that make sense to you? Also potential energy values are relative, it is only the difference in two values of P.E. that counts, not any absolute value. But anything placed on the surface of earth will never fall any deeper into the earth. We can assign a value of ...


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It's s not always 0, but it is safe to assume that it is the relative difference between the plates that is important, not the absolute value.


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This answer has images only (An image is equivalent to 1000 words) If you have glasses red-left / blue-right see image below in black & white 3D. and moreover a colored 3D.


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$dx$ is measured along the $OP$ axis and $ds$ is measured along the surface. The relationship between them follows from simple geometry - $ds$ is at an angle $\pi/2-\theta$ to $dx$. From this diagram, $\frac{dx}{ds}=\sin\theta$. The result follows from rearranging. $ds$ is the width of a strip of the shell of constant surface density: $dx$ is the ...


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Feynman's trajectory The trajectory discussed by Feynman is shown below in red for the blue path, which is a hyperbolic deflection of a small particle around a large star centered at $(0, 0)$. Discussion Feynman's trajectory here trying to answer the question: how much has the speed increased between A and B. He is answering that by saying that there is ...


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Starting from the work around a close curve $$W = \oint \vec F(r)d\vec s = 0$$ now be stokes throem: $$ \oint \vec F(r)d\vec s = \iint_A \vec \nabla \times\vec F(r)d\vec A = 0$$ where $A$ is the area surrounded by the closed curve. Now if you do the math you will see that $ \vec \nabla \times \vec \nabla U( r) = 0 $ for any field $U( r)$ thus we can ...


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The reason for this is that, in your notation, $W=C_1-C_2$. This means that the work performed in moving about a circuit from point $A$ to point $B$ along curve $\mathcal C_1$ and then back along $\mathcal C_2$ is exactly the difference between the work performed in moving the particle along $\mathcal C_1$ versus moving it along $\mathcal C_2$. This is ...


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If a mass M is at a point travelling at a given velocity (= speed and direction vector) then it has a given amount of energy (potential and kinetic energy summed) relative to being stationary* at the same point with respect to a given frame of reference. ie solely by knowing position and velocity vector the PE & KE are defined. (* or at some zero ...


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Here is what I think he means: first we have a planet going around the sun in some orbit, then we change the direction of the velocity to go radially outwards, for example by letting the planet go inside some pipe we put in it's path (notice that a normal planet would never do this, because there are no big pipes in space and also there would be quite a lot ...


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So the principle that systems find themselves at an energy minimum needs to be heavily qualified: it is a consequence of ever-present friction/drag forces having a sign opposite to your velocity, so that the power wasted due to friction is terminally negative. The principle therefore states, "when a system only has two forces, one due to drag and the other ...


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SECTION A : The example in Feynman's Lectures Let a body P (Planet or Particle or whatever) moving in orbit around a center of attraction called $\:\rm{SUN}$, as in above Figure. Suppose that the attractive force $\:\mathbf{f}\left(r\right)\:$ depends continuously only on the distance $\:r\:$ of the body P from the center $\:\rm{SUN}$. Here it's not ...


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The point is that if $\frac{1}{2} mv^2 - GMm/r$ is constant, then $v$ only depends on $r$! This is surprising and very useful; it means that $v$ will be the same no matter what path a planet takes from some $r_1$ to $r_2$. In this case, the two paths he's using are the planet's usual elliptical orbit, and a path that goes straight toward the sun. You don't ...


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It looks like you're confusing vector and scalar $R$. The potential between two charges depends on the magnitude of $R=|\vec{R}|$, not on the vector. So in this case $R_{13}=R_{31}=|\vec{R}_{13}|$.


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People are mistaking mass for force. An object moving does not have more mass, it has inertia, and moving it away from the earth weakens the gravitational attraction between the earth and the object, causing a weight change. However, weight is a measurement of gravitational force on between two bodies containing mass. Mass, however, cannot and does not ...


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It isn't necessary to introduce the effective potential in orbital mechanics but it is really useful. Let's say we have a particle moving in a central gravitational potential. Newton's laws give you a vector equation of motion \begin{equation} m \ddot{\vec{x}} = - \nabla U \end{equation} where $U = - G M m /r$. In a general coordinate system this is a ...


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i only know the effective potential when talking about central forces situations in classical mechanics, there it is defined as: $V'(r) = V(r) + \frac{L^2}{2mr^2}$ with V(r) being the radial potential, the second term can be considered a centrifugal potential which results from considering the azimuthal part of the kinetic energy. $E = V + E_{kinetic} = ...


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Given your question, it seems likely that your misunderstanding comes from a limited sense of vectors, fields, and partial derivatives. So there's a lot of education that we have to cover in a very short time. Multivariate functions When we transition from a function $f(x)$ to a field, which is a function of many variables $f(x, y, z)$, we suddenly have ...


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It's always possible to expand the potential in Taylor series around any local minima (in this example $U(x) $ has local minima at $x_0$ , thus $U'(x_0)=0 $ ) $$ U(x) \approx U(x_0)+\frac{1}{2}U''(x_0)(x-x_0)^2 $$ Setting $ U(x_0)=0 $ and $ x_0=0$ (for simplicity, the result don't depend on this) and equating to familiar simple harmonic oscillator ...


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The work-energy theorem leads us to the following result; \begin{equation} \oint \vec F\cdot d\vec s=0 \end{equation} \begin{equation} \oint \vec F\cdot d\vec s=\underbrace{\int \int }_{\text{surface}}(\nabla \times \vec F)\cdot d\vec n \end{equation} Using the rules of vector calculus there must exist some scalar function such that; \begin{equation} \vec ...


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Let $F$ be a force field. Assuming that the force field is a conservative vector field, then it follows that the line integral of the force field is zero $$\oint_{O} F \cdot dr = 0$$ The del operator $\nabla$ is defined in 3 dimensions as $$\nabla =\left\langle\frac{\partial}{\partial{x}}, \frac{\partial}{\partial{y}}, ...


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On spherical coordinates, the gradient of a general function $V$ is: $$ \nabla V = \frac{\partial V}{\partial r}\mathbf e_r + \frac{1}{r}\frac{\partial V}{\partial\theta}\mathbf e_\theta + \frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}\mathbf e_\phi $$ If $V(r, \theta, \phi)$ only depends on $r$, that is $V = V(r)$, which is exactly the case of the ...



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