Imagine you have two homogeneous spheres with the same diameter of $d=0.1 m$. They have the same mass $m = 1 kg$. The distance between the centers of mass is $r= 1 m$. Their electrical charge shall be disregarded. At $t=0$ the spheres do not have any relative motion to each other. Due to gravitation they will accelerate and start moving towards each other. After some time they will touch each other.   How to calculate analytically the time it takes the two spheres to meet each other. I'm not interested in a numerical solution.   I have already tried several ways but I don't get to a solution.    Imagine that the 2 spheres have different masses and diameters.  $m_{1}=2 kg$, $m_{2}=5 kg$,  $d_{1}=0.03 m$, and $d_{2}=0.3m$.  How to calculate analytically when and where the 2 spheres are going to meet?    How do you calculate the second problem taking the theory of relativity into account? I know that it will not change the result that much but I am interested in the mathematical solution.
612
The common understanding is that, setting air resistance aside, all objects dropped to Earth fall at the same rate. This is often demonstrated through the thought experiment of cutting a large object in half. The halves clearly can't then fall more slowly just by being sliced in two.  However, I believe the answer is that when two objects fall together, attached or not, they do "fall" faster than an object of less mass alone does. This is because not only does the Earth accelerate the objects toward itself but the objects also accelerate the Earth toward themselves. Considering the formula: $$ F_{\text{g}} = \frac{G m_1 m_2}{d^2} $$  Given $F = ma$ thus $a = F/m$, we note that the mass of the small object doesn't seem to matter as when calculating acceleration the force is divided by the $m$ term, its mass. However, this overlooks that the force is actually applied to both objects, not just to the smaller one. The acceleration on the second, larger object is found by dividing $F$, in turn, by the larger object's mass. The two objects' acceleration vectors are exactly opposite, so closing acceleration is the sum of the two:  $$ a_{\text{closing}} = \frac{F}{m_1} + \frac{F}{m_2} $$  Since the Earth is extremely massive compared to everyday objects, the acceleration imparted on the object by the Earth will radically dominate the equation. As the Earth is $\sim 5.972 \times {10}^{24} \, \mathrm{kg} ,$ a falling object of $5.972 \, \mathrm{kg}$ (just over 13 pounds) would accelerate the Earth about $\frac{1}{{10}^{24}}$ as much, which is one part in a trillion trillion.  Of course in everyday situations, we can for all practical purposes treat objects as falling at the same rate because of this negligible difference—which our instruments probably couldn't even detect. But I'm hoping not for a discussion of practicality or what's measurable or observable, but what we think is actually happening.  Am I right or wrong?  What really clinched this for me was considering dropping a small Moon-massed object close to the Earth and a small Earth-massed object close to the Moon. This made me realize that falling isn't one object moving toward some fixed frame of reference, but that the Earth is just another object, and thus "falling" consists of multiple objects mutually attracting in space.
65185
From universal law of gravitation, gravitational force exerted on a body of mass m by another body of mass M is                 $$ \mathbf F = \frac{GMm}{x^2}  $$ where x is the distance between the centres of both the objects.  So, work done by gravitational force in bringing the object of mass m from infinity to a distance r from the centre of body of mass M is           $$ W = \int_\infty^r \vec{F(x)}.\vec{dx}$$                   $$=\int_\infty^r \frac{GMm}{x^2}\hat x.\vec{dx}$$ (where $\hat x$ is the unit vector in the direction in which the body of mass M is attracting the body of mass m, i.e. the direction of $\vec{dx}$ which results the angle between both vectors $0$) $$ =\int_\infty^r \frac{GMm}{x^2} {dx}\ cos0$$  $$  = - GMm\left(\frac{1}{r}-\frac{1}{\infty}\right) $$ $$= -\frac{GMm}{r}$$  Now, we know that $$W=-(∆U)$$ $$-\frac{GMm}{r} = -(U_r - U_\infty)$$ $$-\frac{GMm}{r} = (U_\infty - U_r)$$ Since, Zero of potential energy is at infinity by convention, so $U_\infty$ = 0 $$-\frac{GMm}{r} = -U_r$$ $$\frac{GMm}{r} = U_r$$  I get potential energy at a distance r as positive, then why is it that gravitational potential energy is $$-\frac{GMm}{r}$$  What is wrong in my derivation?
104030
Usually when on Earth we can use as an approximation:  g = a = 9.18 $\frac{m}{s^2}$ s = vi + 1/2 a$t^{2}$  However, how would I determine the distance traveled in situations where the the acceleration due to gravity is changing non-negligibly because the distances are so large? I know that the following equation has to be used somehow:  a = $Gmr^{-2}$  I am not sure exactly how though. For example, if given the information that an object is at a certain distance and is being attracted by a larger object of a given mass, the method should calculate where the object will be at a given time.
118540
It is well-known that measurements of the velocity profile in galaxies are not compatible with Newtonian laws. A way to circumvent the problem is to assume that galaxies are surrounded by a spherical halo of Dark Matter. The density of mass of this Dark Matter can then be computed from the velocity profile with Newton's laws.   My question is the following: the formation of a galaxy involves essentially only the gravitational interaction. According to Newton (and General Relativity too), the trajectory of a particle does not depend on its mass. As a consequence, I do not understand how usual matter and Dark Matter could have followed different trajectories leading to such different mass distributions nowadays. Can somebody explain this (apparent) paradox ?
108258
So I was thinking about something for the past while  Consider a large spherical foam-ball with homogeneous density. Where a foam ball is defined as an object that can absorb matter with 0 friction (example: a gravitational well without an object inside). This is a purely theoretical construct.  If the foam-ball has a radius R and a charge Q. What charge must the foam-ball have, such that there is well defined sphere or horizon such that any object with a negative charge (even if it is the slightest bit) must travel faster than C in order to escape the field of the ball.  Aka what charge would turn this into a black-hole for all objects with opposite sign on their charge?  Once again I assume the foam-ball is a single particle, that doesn't repel itself... its just a very large homogenous 'thing' with charge.
68485
In the Wikipedia article , it says     After Newtonian gravitation was found to be inconsistent with special relativity, . . .   I don't see how Newtonian gravitation itself is inconsistent with special relativity.  After all, Newton's Universal Law of Gravitation is completley analogous to Coulomb's Law, so it would seem that, if there were an analogous "gravitational magnetic field", one could formulate a theory of gravitation in exact analogy with Maxwell's Theory of Electromagnetism, and of course, this would automatically be consistent with Special Relativity.  What about this approach to gravitation does not work?  The only problem I could see with this is the lack of evidence for a "gravitational magnetic field".  That being said, gravity is "weak" as it is, and my guess is that it would be extremely difficult to set up an experiment in which a massive body were moving fast enough for this "magnetic field effect" to be obsevable.  EDIT:  As has been pointed out to me in the answers, Newton itself is inconsistent with SR.  Similarly, however, so is Coulomb's Law, yet, electromagnetism is still consistent with SR.  Thus, how do we know that it is not the case that Newton's Law is the special "static" case of a more general gravitomagnetic theory exactly analogous to Maxwell's Theory:  Let $\mathbf{G}$ be the gravitation field, let $\rho$ be the mass density, and $\mathbf{J}$ be the mass current density, let $\gamma _0$ be defined so that $\frac{1}{4\pi \gamma _0}=G$, the Universal Gravitation Constant, let $\nu _0$ be defined so that $\frac{1}{\sqrt{\gamma _0\nu _0}}=c$, and suppose there exists a field $\mathbf{M}$, the gravitomagnetic field, so that the following equations hold: $$ \vec{\nabla}\cdot \mathbf{G}=-\frac{\rho}{\gamma _0} $$ $$ \vec{\nabla}\cdot \mathbf{M}=0 $$ $$ \vec{\nabla}\times \mathbf{G}=-\frac{\partial \mathbf{M}}{\partial t} $$ $$ \vec{\nabla}\times \mathbf{M}=\nu _0\mathbf{J}+\gamma _0\nu _0\frac{\partial \mathbf{G}}{\partial t} $$ where theses fields would produce on a mass $m$ moving with velocity $\mathbf{v}$ in our inertial frame the Lorentz force $\mathbf{F}=m\left( \mathbf{G}+\mathbf{v}\times \mathbf{M}\right)$.  This theory would automatically be consistent with SR and would reduce to Newton's Law of Gravitation for the case of gravitostatics:  $\frac{\partial \mathbf{G}}{\partial t}=\frac{\partial \mathbf{M}}{\partial t}=\mathbf{J}=0$.  (To be clear, you can't just set the time derivatives equal to $\mathbf{0}$ as it seems I have done.  Upon doing so, you obtain the corresponding static theory, which is technically incorrect (as you can easily see because it won't be relativistically invariant), but is nevertheless often a useful approximation.)  My question can thus be phrased as:  without appealing directly to GR, what is wrong with this theory?
29836
Let us assume the universe filled with positive charge. About a particular point, all the positive charged particles will be symmetrical. Now consider a sphere of radius $r &lt; \infty$ and apply Gauss law. According to Gauss law the electric flux through the sphere will be some finite value as it contains some positive charge. But by symmetry the electric field through the sphere and hence electric flux will be zero. This is a contradiction.
17632
There is something I don't understand about the  $H$, as it seems to clump two concepts together that I can't quite unify in my head. On the one side, we have  $$V = D H$$  which means that for a given distance $D$, there is a certain amount of new space created over time - and $H$ is simply the factor that makes this relationship work. So, for example, say we have two points 1 Mpc apart, this would mean they recede at about 70 km/s away from each other  (given our current approximation of $H$).  Now the thing I can't wrap my head around is that   $$T =\frac{1}{ H}$$  is also the age of the universe. Contrary to claims made on, say, Wikipedia this means that $H$ cannot possibly have been a constant throughout the past 13 billion years, because mathematically $1/H$ means that $H$ must be continually shrinking as the universe ages.  So if $H$ did start out as some huge value and is now shrinking over time, doesn't this mean the expansion of the universe is slowing down? Because if $H$ is shrinking, I'll get a lower value of $V$ today than I'll get tomorrow. Shouldn't the notation then be more like  $$V = DH(t)$$  So which one is it? If $1/H$ is simply the solution for $D=0$, how can we use it as the expansion-velocity-per-unit-of-distance at the same time? What's worse, how can literature say $H$ has probably been more or less constant forever and simultaneously assert that $1/H$ is the current age of the universe? What am I missing?
69540
Using a "successive impact model" (as if each ball were seperated from the other ones), I produced the following animations:    You can see any combination of balls with masses of 1 or 2 (left) or 1 and 4 (right).  Unfortunately, I do not have any Newton cradle to make some experiments, and I'd like to compare my results with observations. I contacted the author of , which is the most instructive one I found about asymmetric Newton cradle. In particular, he writes that the "OoO" configuration behaves as the "ooo" one, which is not what I simulate. But the author is not sure whether who is wrong because he did the experiments a long time ago.  I am especially interested in asymmetric Newton cradle with simultaneous impacts: left and right balls collinding at the same time.   Please let me know if you have any idea on where to find such experimental data.  Edit Just of few general remarks to avoid extending the comments.    Energy and momentum are conserved. However, they are sufficient to ensure uniqueness of the post-impact velocities only when there are 2 balls. The successive impact model precisely consists in propagating the impact ball after ball, and therefore leads to a unique solution (which conserves momentum and energy globally). From the waves point of view, changing the mass of a ball can be done by increasing the diameter, or the density (or a combination of both). This simple model cannot account for such subtleties of course, but maybe it does not matter when the balls are "small enough". Even if I think it's quite basic, following WetSavannaAnimalakaRodVance's comment, I am ready to hand out my Mathematica code to anybody who wants it. I'll check for good ways to share it.
33184
Why is the moment of inertia (wrt. the center) for a hollow sphere higher than a solid sphere (with same radius and mass)? I have completely no idea and I am inquiring about this as it is an interesting question that popped in my head while doing physics homework.
30620
If we consider a case where a closed circular wall is present, and inside the boundary, just adjacent to the inner wall, two particles are placed touching the surface of inner boundary such that the distance between both particles tends to 0. Particle A and B have identical masses, shapes and physical properties. Particle B applies a constant force on particle A and particle B is in uniform circular motion. Thus, particle A also moves in a circle. And here, centripetal force is not required to make particle A move in a circle.   What's going on?
83781
I'm building a physics simulator for a graphics course, and so far I have it implementing gravitational and Coulomb forces. I want to add collisions next, but I'm not exactly sure how to go about doing it.   A quick summary of how this is working so far is: All objects are spheres of a set radius, mass and charge. The mass's and charges of the spheres are treated as point charge/mass for the calculation. Every step in time (about 1/50th of a second) the forces acting on each object are calculated in a nice big nested for loop that figures out the coloumb and gravitational force between 2 objects, for every set of 2 objects, and then they are summed together. I use this net force to determine the acceleration, and the rest is fairly obvious from there.   What I want to add in is collisions. I can deteremine pretty easily if a collision is happening (if the distance between them &lt;= radius of one + radius of other), what I am not so certain of is how I should add in the collision force (and would need to do it component wise). I want elastic collisions, though for now I'd just be happy with getting conservation of momentum  The information I have easily available are: the velocity of each object (and I mean velocity, speed and direction), the position of each object, the mass/charge (charge obviously not needed here) and the Net force calculated so far for each object for the next step in time. I dont need exact formulas (would prefer if they arent exact), just need a nudge in the right direction.
19773
What's the cleanest/quickest way to go between Einstein's postulates [] of   Relativity: Physical laws are the same in all inertial reference frames. Constant speed of light:  "... light is always propagated in empty space with a definite speed $c$ which is independent of the state of motion of the emitting body."   to Minkowski's idea [] that space and time are united into a 4D spacetime with the indefinite metric  $ds^2 = \vec{dx}^2 - c^2 dt^2$.  Related to the question of what is the best derivation of the correspondence are: Is the correspondence 1:1? (Does the correspondence go both ways?) and are there any hidden/extra assumptions?    Edit  Marek's answer is really good (I suggest you read it and its references now!), but not quite what I was thinking of.   I'm looking for an answer (or a reference) that shows the correspondence using only/mainly simple algebra and geometry. An argument that a smart high school graduate would be able to understand.
39442
If two bodies are close, both will get attracted to each other and collide. Under what conditions will the two bodies start revolving around their common center of mass? I understand that such bodies represent the gravitional two-body problem, but I want to know what are the initial conditions that allow it so settle into such a configuration (without going into the maths)
120255
Calculate the potential and the intensity of the gravitational field at a distance $x&gt; 0$ in the axis of thin homogeneous circular plate of radius $a$ and mass $M$.    Could anybody describe how to calculate this? Slowly and in detail. I'm helpless.  Answer is: potencial $\phi = - \frac {2 \kappa M}{a^2}(\sqrt{a^2+x^2}-x)$ and intesity $K = \frac {2 \kappa M}{a^2} \left( \frac{x}{\sqrt{a^2+x^2}}-1 \right)$
66864
The picture shows an accelerating spaceship with two clocks inside it. It is so far away from all other bodys that gravity is of no importance.   Will the bottommost clock be slower than the topmost one? Or will both clocks have the same speed?
23124
Consider a system of the two identical positive point charges situated in free space (isolated from the influence of any other external fields) as shown in the attached diagram. Particle $1$ is at $(a,a,0)$ and particle $2$ is at $(0,a,0)$. Their velocities, at the considered instant of time, are as shown in the diagram ($\mathbf{v}_1$ along the $+x$ axis, $\mathbf{v}_2$ along the $+z$ axis).    Now, by applying the Biot-Savart law, we find that the magnetic field due to particle $2$ at the position of particle $1$ is along $+y$ axis, which means that the force acting on particle $1$ is along $+z$ axis according to Fleming’s left-hand rule. A similar analysis shows that there is no magnetic force on particle $2$ as the magnetic field of particle $1$ should vanish at positions (relative to particle $1$) located along its velocity vector.  Now, if we observe the net torque on the system of two particles about the $y$ axis then it is non-zero and is directed along the $-y$ axis. Here, there are no external forces or torques acting on the isolated two-particle system, and yet the net torque, as well as the net force on the particles, are nonzero. Why? Also, Newton's third law of motion seems to be broken in this scenario. Why?    Edit $1$  I have come to know from the responses that the electromagnetic field itself takes away some momentum and angular momentum about the considered axis. However, I think that if I consider only two charged particles as my system then the Abraham-Lorentz force can be assumed to be acting upon the system, and that is sufficient to make sure that we have considered the momentum being carried away by the electromagnetic field itself.   Even after considering the action of the Abraham-Lorentz force for the two-particle system, the scenario breaks both Newton's third law and the conservation of linear and angular momentum. This is because the Abraham Lorentz forces do not exactly counterbalance the force and the torques, on the two-particle system under consideration, due to the magnetic field.    Edit $2$  The previous edit was a result of confusion and misunderstanding on my part. The force associated with the momentum carried away by the electromagnetic field as a result of the electromagnetic interaction described in the question is the Lorentz force itself which simply doesn't obey the third law of motion. The Abraham-Lorentz force is a different story. It is associated with the momentum carried away by the radiation emitted by the accelerated charged particles. This is an additional force apart from the Lorentz force and corresponds to an additional carriage of momentum by the electromagnetic field. The momentum carried away by the electromagnetic field in correspondence with the Lorentz force doesn't correspond to radiation.