Planets are spherical (roughly) in shape because a spherical configuration gives the minimum gravitational potential energy.   How does one prove that the sphere is the shape that gives minimum gravitational potential energy for a volume of uniformly dense material?
6028
Is it possible to use the Lagrangian formalism involving physical terms to answer the question of why all planets are approximately spherical?  Let's assume that a planet is 'born' when lots of particles of uniform density are piled together, and constrained only by the Newtonian gravity of the $N$ particles with no external gravity.    My question is: Is it possible to use the variational calculus starting from this situation to prove that planets should be spherical?
98546
What level of error do I encounter when approximating the earth as a sphere? Specifically, when dealing with the location of points and, for example, the great circle distances between them.  Are there any studies on the average and worst case error compared to an ellipsoid? I'm wondering how much accuracy I'd be sacrificing if I go with a sphere for the sake of easier calculations.  My particular scenario involves directly mapping WGS84 coordinates as if they were coordinates on a perfect sphere (with the  defined by the IUGG) without any transformation.
8745
Say you have a hollow sphere with a uniformly distributed charge on the surface. Why is the electric field everywhere inside the sphere zero?  For the centre, its easy to add the vectors from the surface charges and show they sum to zero because of symmetry.   But can you show me how the field cancels out to zero for points other than the centre using vector addition?
21106
I was doing the question :     A uniform 8.40-kg spherical shell, 50.0 cm in diameter has four small   2.00-kg masses attached to its outer surface and equally spaced around it. This combination is spinning about an axis running through the   center of the sphere and two of the small masses (see ). What friction torque is needed to reduce its angular speed   from 75.0 rpm to 50.0 rpm in 30.0 s?   I understand how to get $\alpha$, and how to get $\tau$ from $I$ and $\alpha$. What I do not understand is their computation of the moment of inertia. The question did not provide any information regarding the shapes of the masses, but the solution (which writes $I=\frac{2}{3}MR^2+2mR^2$) seems to assume that the masses are point particles. Thus, the two masses along the rotational axis have a moment of inertia of 0 ($r=0$), and each of the other two masses has a moment of inertia of $I=mr^2 =mR^2$, where $R$ is the radius of the sphere and $r$ is the distance from the axis.  But why are the masses considered to be point particles? I wanted to use the parallel-axis theorem to determine the moment of inertia about the rotational axis; not knowing the shape of the masses made this impossible, since I couldn’t find the moment of inertia about the CM. If I knew $I_{CM}$, then couldn’t I use $I_{axis}=I_{CM}+mR^2$ for the two masses not on the rotational axis and $I_{CM}$ for the two that are? That would make $I_{total}=\frac{2}{3}MR^2+4I_{CM,\:mass}+2mR^2$ Would that not be a better way of solving this problem?
76612
A question occurred to me when browsing SE this evening, just curious.  What specifies the inside of a 3D construct? If I have a hollow sphere, what's to say that the world isn't the inside, and the center of the sphere isn't the outside. Picture the Asylum in the Hitchhiker's Guide to the Galaxy series, if you have read that.  Additionally, if the inside of a 3D object is defined as what is smaller than the outside, what happens when there is a a sphere with a volume equivalent to 51% of the universe?
90607
When you are inside some non uniformly and non radially polarized sphere, why do the dipoles from outer shells don't affect you (the potential there is only due to inner dipoles from what I have seen)?
60061
Say we have $N$ spheres indexed as $1,2,3,\dotsc, N$ such that all of them have identical weight apart from one, and we don't know if that one is heavier or lighter. We have to determine which sphere has the odd weight using just a balance scale.  We could solve this problem by weighing repeatedly, but I am interested in a solution involving weighing as few times as possible, so my question is what is the optimal algorithm for this task?
25860
This might be a completely wrong question, but this is bothering me since many days ago. Given the mass (Sun) curves the space around it, gravitation is the result of such curved space (Correct me if I am wrong, source: A Documentary film). Given any point on a circle with center same as center of the mass, curvature in the space should be equal (Intuition).   Planets rotate around the Sun because of the curve in the space they should follow a circular path and the distance between planet and Sun should be at a distance.  Given the fact that earth has a elliptical orbit around the sun, and the distance between Earth and Sun varies according to position of the earth. Why do we have a elliptical orbit not a circular orbit.
115608
In most popular explanations of General Relativity, both in print and film/television, gravity is demonstrated using an example of a 2 dimensional plane being flat, then when putting a heavy object in it it curves the plane into a funnel shape and you can visualize how other objects roll down the funnel into the gravity well simulating something like the orbits of the planets around the sun as an example.  However this demonstration relies on the notion of "down", or even the actual gravitational attraction on Earth, so that the objects in orbit roll or fall down the well.  This kind of explanation seems to rely on a force in order to explain that the force of gravity is just the curving of spacetime. At the very least I find this to not really explain how the theory of General Relativity actually works.
35601
Almost all of the orbits of planets and other celestial bodies are elliptical, not circular.   Is this due to  by other nearby massive bodies?  If this was the case a two body system should always have a circular orbit.  Is that true?
84522
I had to do a problem, and part of it was to find the mechanical energy of satellite orbiting around mars, and I had all of the information I needed. I thought the total mechanical energy would be the kinetic energy+ the potential energy, or $KE+PE$. However, I had the answer sheet and it said that I had to do $KE-PE$, because when you integrate $Gm_1m_2/r^2$ you get a negative sign. I can see why it works out mathematically, but I don't understand why you are actually losing energy when orbiting in a gravitational field.
3100
So I have been trying to understand why there is a negative sign in the following$\Delta U= -\int\vec{F}\cdot d\vec{l}$  I wanted to try to understand it by trying to derive the gravitational potential energy $E_p=-\dfrac{GMm}{r}$ I start by defining potential energy to be $0$ at $\infty$ and see what I get if I let gravity do work on the mass until a distance r away from a planet  $W_G=\int_{\infty}^{r} \vec{G}\cdot{d\vec{l}} = \int_{\infty}^{r} G\cos{\theta}dl = \int_{\infty}^{r} G dx$ G is the gravitational force so:  $W_G= \int_{\infty}^{r} -\dfrac{GMm}{x^2}dx = \Bigg[\dfrac{GMm}{x}\Bigg]$ applying limits of integration gives $W_G = \dfrac{GMm}{r}$  Thus I am confused why the minus sign, from the derivation I wrote it seems to have to be $\Delta U= \int\vec{F}\cdot d\vec{l}$ at least for gravity. I also did the same derivation for a mass on earth ($W_G = -mgh$ Where $\vec{G}$ is considered to be constant) and the result indeed is the correct one $\Delta U= -\int\vec{F}\cdot d\vec{l}$ so my confusion arises, why don't i get the same result for both cases.   I am also wondering if I carry out the same derivation but for elastic potential energy, what would I get... Thank you for any help!
29133
In mechanics problems, especially one-dimensional ones, we talk about how a particle goes in a direction to minimize potential energy. This is easy to see when we use cartesian coordinates: For example, $-\frac{dU}{dx}=F$ (or in the multidimensional case, the gradient), so the force will go in the direction of minimizing the potential energy.   However, it becomes less clear in other cases. For example, I read a problem that involved a ball attached to a pivot, so it could only rotate. It was then claimed that the ball would rotate towards minimal potential energy, however $-\frac{dU}{d\theta} \neq F$! I think in this case it might be equal to torque, which would make their reasoning correct, but it seems like regardless of the degrees of freedom of the problem, it is always assumed that the forces act in a way such that the potential energy is minimized. Could someone give a good explanation for why this is?  Edit: I should note that I typed this in google and found this  where it states that minimizing potential energy and increasing heat increases entropy. For one, this isn't really an explanation because it doesn't state why it increases entropy. Also, if possible, I would like an explanation that doesn't involve entropy. But if it is impossible to make a rigorous argument that doesn't involve entropy then using entropy is fine.  As a side note, how does this relate to Hamilton's Principle?
83634
Foreword  This question was inspired by initial mistakes in . I wanted to explore the strange circle with $A&gt;\pi r^2$ and got lost into geometrical jungle.  A  is usually described by it's height $h$, radius of the base $a$ and radius of the sphere $r$. Besides we have a relation $a^2+(r-h)^2=r^2$ so the cap actually only has two degrees of freedom.  When considering a case of circle on a sphere (like the Earth), it's the surface of a spherical cap, but we can't see the height or other classic parameters of this spherical cap. Instead we encounter parameters like area $A$, perimeter $C$ and diameter $d$. I will try to write formulae for these: $$A=2\pi rh$$ $$C=2\pi a$$ $$d=2r\theta$$  Where $\theta$ is introduced with $\sin\theta=\frac{a}{r}$ or $\cos\theta=(1-\frac{h}{r})$.  As the cap only has two degrees of freedom, it should be a good problem with unique solution if only two values are known.  A-C Problem  The simple example: $A$ and $C$ are given. I show this only as example before the other tasks.  We can easily find $a=\frac{C}{2\pi}$ and then solve:  $$A=2\pi rh = 2\pi r (r-\sqrt{r^2-a^2})$$  to obtain  $$r=\frac{A}{2\sqrt{\pi(A-\pi a^2)}}$$  Parameters $a$ and $r$ are now known and it is straightforward to find $h$, $d$, $\theta$ and anything else.  C-d Problem  $C$ and $d$ are given.  First step once again is easy: $a=\frac{C}{2\pi}$. The knowing of $a$ suggests that we might prefer the $\theta$ definition with the sine. We obtain: $$d=2r\theta=2r\arcsin\frac{a}{r}$$  That's where I got stuck. Is there an easy way to solve this? Is there a hard way to solve this?  I tried solving numerically with Wolfram Mathematica and it went into complex numbers. When solving the geometrically equivalent $\sin\frac{d}{2r}=\frac{a}{r}$ I got some roots, but the search seemed very unstable and not finding the root closest to initial value. Besides it was finding a lot of different roots while I am pretty sure that there should only be one $r$ for each pair of $d$ and $a$. What would be the right formulation to solve this problem numerically?  A-d problem  $A$ and $d$ are given. This seems the hardest one. We obtain a system of equations:  $$A=2\pi r h$$ $$d=2r\arccos\left(1-\frac{h}{r}\right)$$  (The first equation suggested solving the problem in $h$ and $r$).  Is this system solvable analytically? Or numerically? My experiments with Wolfram Mathematica never converged.   Of course, I could express $h$ or $r$ from the first equation and leave only the second in one of these forms: $$d = 2r\arccos\left(1-\frac{A}{2\pi r^2}\right)$$ $$d = \frac{A}{\pi h}\arccos\left(1-\frac{2\pi h^2}{A}\right)$$  But I have no idea how to solve these either.
47375
Why is it that, at equilibrium, certain potentials are minimised? That is, for a system at constant temperature and pressure, the Gibbs free energy is minimised, and for fixed volume and temperature, the Helmholtz free energy is minimised. I haven't been able to find a reason for this.
4729
In popular science books and articles, I keep running into the claim that the total energy of the Universe is zero, "because the positive energy of matter is cancelled out by the negative energy of the gravitational field".  But I can't find anything concrete to substantiate this claim. As a first check, I did a calculation to compute the gravitational potential energy of a sphere of uniform density of radius $R$ using Newton's Laws and threw in $E=mc^2$ for energy of the sphere, and it was by no means obvious that the answer is zero!  So, my questions:   What is the basis for the claim &ndash; does one require General Relativity, or can one get it from Newtonian gravity? What conditions do you require in the model, in order for this to work? Could someone please refer me to a good paper about this?
50722
Often in physics, Objects are approximated as spherical. However do any perfectly spherical objects actually exist in nature?