I'm just trying to prove that gravitational force on an object within a hollow sphere is zero when these are the only two objects under consideration within the system. I know some multivariable calculus, I just want someone to walk me through the integration deriving the mass element dM and the integration of thin rings composing the hollow sphere. It would also be nice if you could show me doing it one way using the solid angle and one way without using the solid angle. Thanks.
1475
It is quite easy to derive the gravitational field intensity at a point within a hollow sphere. However, the result is quite surprising. The field intensity at any point within a hollow sphere is zero.   What exactly is the reason behind this? Except for, of course, the mathematics behind it. Is there any logic why the field intensity should be zero within a sphere? For example, it is logical to say that the field intensity would be zero at the center, as all the intensities cancel out. However, this cannot be the case for any point within the sphere.
12580
The height of a string in a gravitational field in 2-dimensions is bounded by $h(x_0)=h(x_l)=0$ (nails in the wall)  and also $\int_0^l ds= l$. ($h(0)=h(l)=0$, if you take $h$ as a function of arc length) .  What shape does it take?  My try so far: minimise potential energy of the whole string,  $$J(x,h, \dot{h})=\int_0^l gh(x) \rho \frac{ds}{l}=\frac{g \rho }{l}\int_0^l h(x) \sqrt{1+\dot{h}^2} dx$$  With the constraint  $$\int_0^l \sqrt{1+\dot{h}^2} dx- l=0$$ If it helps, it's evident that $\dot{h}(\frac{l}{2})=0$.  Generally, this kind of equation is a case of a constrained variational problem, meaning that the integrand in  $$\int_0^l \frac{g \rho }{l}h(x) \sqrt{1+\dot{h}^2} +\lambda(\int_0^l \sqrt{1+\dot{h}^2} dx- l)dx$$  Must satisfy the Euler Lagrange equation. The constraint must also be satisfied.  But, in truth, by this point I am clueless. $\lambda$ is worked through $\nabla J  = \lambda \nabla(\int_0^l \sqrt{1+\dot{h}^2} dx- l)$. I have tried this , but get nonsensical answers.  Is this method the best? If so, in what ways am I going about it wrongly thusfar?
110123
This is my first year as a physics student, and I've never learned about vectors past a basic level, so this is confusing me. When we have gravity on an inclined plane, we separate it into two components, which I understand.  However, consider the image below, and there's a box at point A. When separating the gravity components, you draw a triangle AGC (sorry, the G and D are on top of each other and difficult to distinguish). AG becomes the force of gravity in the y-direction, and GC becomes the force of gravity in the x-direction. Then you do the trig functions from there. However, when I tried this myself, I drew triangle ADF instead and tried the trig functions from there. It didn't work. I'm having trouble understanding why you can't compute the trig functions from AGF. The only partial solution I came up with was that force of gravity in the y-direction can't be the hypotenuse, as the force of gravity in the y-direction is always less than the force of gravity. But I think I'm missing something more.
81697
If you have a conducting hollow sphere with a uniform charge on its surface, then will the electric field at every point inside the shell be 0. The reason the electric field is 0 at the center is clear from the symmetry of the sphere, but for a point at a certain distance from the center, shouldn't a net electric field exist?  I saw a similar question on the site but it was vaguely explained.
80169
What if a person is put between two identical dense spherical bodies (well, not as dense as black holes, but definitely 100 times more than earth) with mass $M_1=M_2=10^{30}kg$, exactly in the middle between them ( vertical axis parallel with "human height vector" and the line connecting two spheres' centres passes exactly trough person's  centre of mass). Let's say $M_1$ and $M_2$ are mounted onto some very strong poles and they are not approaching each other, but are stationary at $1$ meter.   I am aware that the forces would cancel out at exactly the middle, but human body is not symmetrical and it's mass is not equally distributed across it's volume. Would the human in question be torn apart or would it just deform? What would happen if the human body was completely symmetrical and with equal mass distribution, and could withstand such forces without being deformed? Would the person maybe spin?  I am not sure if this will affect the overall outcome, but let's  say the mass of the person in question is $m=80~kg$.
118043
Let's say that we want to measure the gravity force in 1D, 2D, 3D  and higher spatial dimensions.   Will we get the same force strength in the first 3 dimensions and then it will go up? How about if we do this with Electromagnetic force?   I've include the Electromagnetic force just to see if I can find an analogy to the gravity force behavior.
64718
I suspect an error in my “Classical Dynamics” lecture notes but my lecturer doesn’t agree with me so I need your help!  We assume that in the $n$-body problem, the force between particle $i$ and $j$ can be derived from a potential i.e  $$\vec{F_{ij}}=-\vec{\nabla_i}V_{ij}$$  where $\vec{\nabla_i}$ is the gradient with respect to position vector $\vec{r_i}$   $V$ would usually only depend on $\Vert\vec{r_i}-\vec{r_j}\Vert$  My lecturer then defines the total energy of the $n$-body system as  $$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{ij}V_{ij}$$  However, I am convinced that this is wrong since we’d be double counting potentials. My lecturer says we should indeed be double counting potential. In my opinion, the expression should read   $$E=\sum_i \frac{1}{2}m_i\vert\vec{\dot{r_i}}\vert^2 + \sum_{1\le{i}\lt{j}\le{n}}V_{ij}$$  For the second equation we also get:  $$\dot{E}=\sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{1\le{i}\lt{j}\le{n}}\vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}}  + \vec{\nabla_j}V_{ij} \bullet \vec{\dot{r_j}} = \sum_im_i\vec{\dot{r_i}}\bullet\vec{\ddot{r_i}} + \sum_{i}\sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \bullet \vec{\dot{r_i}} = \sum_i \vec{\dot{r_i}} \bullet \Bigl(m_i \vec{\ddot{r_i}} + \sum_{j\neq{i}} \vec{\nabla_i}V_{ij} \Bigr) = \sum_i \vec{\dot{r_i}} \bullet \Bigl( m_i \vec{\ddot{r_i}} - \vec{F_{ij}^{tot}} \Bigr) = 0 $$  So this seems reasonable to me. Is this actually correct?  We also have an example of the two-body problem in the notes where we only use $V_{12}$ as the potential, not $V_{12}+V_{21}$
18167
Let thing of a bird standing still in a box on top of a weighing machine that shows a mass $m_0$. Now, imagine that the bird is flying, still in the same box and the same weighing machine shows a mass $m_1$. As the bird when flying, is applying a force towards the weighing machine, could we deduce that $m_0 = m_1$? I'm asking this question because saying that these masses are equal makes me as uncomfortable as saying that they are not equal and can't figure the right answer. So does $m_0 = m_1$ and why?
63366
If we call the velocity of the mass $u$, then we can say: $u \cos θ = v$ and hence $u= \frac v{\cos θ}$. But why can't we write  $v \cos θ= u$ and add the contribution from both of the strings: $u= \frac{2v}{\cos θ}$ ? Why is it only $u= \frac v{\cos θ}$? Please note that it is not a homework based  question. I just want someone to solve the doubt arising in my mind.
66253
Question: Let $c:I^2\rightarrow\mathbb{R}^3$ be the singular $2$-cube given by $$c(s,t)=\left(\frac{1}{2}s^2,st,\frac{1}{2}t^2\right)$$Let $x=(x,y,z)$ denote the cartesian coordinates on $\mathbb{R}^3$ and let $\omega$ be the $1$-form on $\mathbb{R}^3$ given by $$\omega=xy^2dz$$  i. Compute $c^*\omega$  ii. Compute $c^*d\omega$  iii. Compute $\int_cd\omega$  iv. Without using stokes theorem compute $\int_{\partial c}\omega$  Answer:   i. $(c^*\omega)(s,t)=(\omega\circ c)(s,t)=\omega\left(\frac{1}{2}s^2,st,\frac{1}{2}t^2\right)=(\frac{1}{2}s^2)(st)(d(\frac{1}{2}t^2))=(\frac{1}{2}s^3t)(tdt)=\frac{1}{2}s^3t^2dt$  ii. $d\omega=d(xy^2dz)=y^2dx\wedge dz+2xydy\wedge dz+xy^2d(dz)=y^2dx\wedge dz+2xydy\wedge dz$  $c^*d\omega=(st)^2d(\frac{1}{2}s^2)\wedge d(\frac{1}{2}t^2)+s^3td(st)\wedge d(\frac{1}{2}t^2)$  $=(st)^3ds\wedge dt+d^3t^3ds\wedge dt=2(st)^3ds\wedge dt$  I am stuck on parts iii. and iv.  Is there a formula for these types of problems?  I have not seen an example of one of these computed before  What is the name of this type of problem so that I can look up methods for it?  Any help would be greatly appreciated
22795
Let's say we have a solid with uniform mass distribution, but we don't know its shape. However, we know the moment of inertia of the solid with respect to as many axes of your choosing as you want (e.g. you know $I_{Ox}$, $I_{Oy}$ and $I_{Oz}$). Is it possible to deduce the shape of the solid based on the moment of inertia $I_{Ox}$, $I_{Oy}....$? And if so, what would be the minimum number of axes required to do it ?  Do note that the mass is supposed to be evenly distributed, otherwise we could have a cube with the bulk of the mass concentrated in a sphere within, for example, and thus the different moments of inertia would not take into  account the "lightweight corners" of the cube.  Then, there is the parallel axis theorem, which reduces the scope of the problem, as we could therefore only choose axes going through the center of mass, for  example. Other than that, I'm completely stumped.
123104
The area of ​​the portion of the sphere $ x^{2} + y^{2} +z^{2} = 1$ located inside of the cylinder $x = x^{2} + y^{2}$, and above the plane $z = 0$.  I'm stuck, so any tip will be helpful  Thanks in advance!
117846
My book introduces the force of gravitation as a non-contact force between two bodies of mass $M_1$ and $M_2$ separated by a distance $r$ . Then it says it is directly proportional to the product of masses $M_1 M_2$ and inversely proportional to $r^2$. Then writes the the force of gravitation as  $$ F = G\dfrac{M_1M_2}{r^2}. $$  But why does it take square of $r$ and not another power? What is the cause of taking $r^2$? Why not another power of $r$?
26851
Perhaps this is trivial: how does one prove that the work done by a homogeneous gravitational field on a system of particles is equal to the work done on a point mass with the total mass of the system, located at the center of mass of the system.
27685
Let there be a hollow sphere (Throughout the question we will ignore the thickness of the sphere) which is positively uniformly charged,q of radius,  R. Suppose there to be a test positive charge, q' at point P which is r distance away from the hollow sphere(r>R, P is outside of hollow sphere).  Now I have an doubt on evaluating Electric Potential,V at point P.  Both source charge,q and test charge,q' are +ve, so source charge will repel the test charge which will make the charge, q' to go to infinity.I have to apply external force, F' to displace the charge, q' from infinity to point P against the equal but repulsive force, F.In short, F' is  displacing the q' from infinity to P.  So External Force, F' will do +ve work and Repulsive Force, F will do -ve work. This way the Electric Field,E produced by hollow sphere is Opposite to Displacement of test charge(i.e.,θ=180).  Now we know that E=k(q/(r^2))  [Please Notice that infinity is the lower limit of integration]  Also V=-∫r∞(E dr cosθ)=-∫r∞(E dr cos 180)=∫r∞(E dr)  At last I ended up to V=- k(q/r). I know that V=k(q/r).  I just want to know why I am having this shitty negative sign in my equation.
107676
I am trying to derive  in GR as an exercise: a spherically symmetric gravitational field is static in the vacuum area. I managed to prove that $g_{00}$ is independent of $t$ in the vacuum, and that $g_{00}*g_{11}=f(t)$.  But the next question is: Show that you can get back to a Schwarzschild metric by a certain mathematical operation. I am thinking at a coordinate change (or variable change on $r$) to absorb the $t$ dependence of $g_{11}$, but I can't see the right one. Does someone has a tip to share?
45177
I'm having trouble understanding how a  (not the  that is generated as a result but the material itself) work. The particles are aligned in a specific direction to give rise to force but I don't see how this alignment gives rise to "attraction" or force.