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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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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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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 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 $$m \ddot{\vec{x}} = - \nabla U$$ where $U = - G M m /r$. In a general coordinate system this is a ...

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

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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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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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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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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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$dx$ 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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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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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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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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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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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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