The gravitational force between two masses is given by: $$\vec{F_g}=G\frac{m_1m_2}{r^2}$$ and the electrostatic force between two charged particles by: $$\vec{F_e}=k\frac{q_1q_2}{r^2}$$  Both the forces have inverse square dependence on the separation or distance although the two types of forces are totally different. One can definitely ask if both charge and mass are totally different properties of matter, why does the force generated by them have such similar formulas?     My question might seem similar to . But that question is more specific if moving masses would generate some kind of field.   My question is, is the similarity in their formula just a coincidence or can we conclude something from it?
43886
It seems paradoxical that the strength of so many phenomena (, ) are calculable by the inverse square of distance.   However, since volume is determined by three dimensions and presumably these phenomena have to travel through all three, how is it possible that their strengths are governed by the inverse of the distance squared?  The gravitational force and intensity of light is merely 4 times weaker at 2 times the distance, but the volume of a sphere between the two is 8 times larger.   Since presumably these phenomena would affect all objects in a spherical shell surrounding the source with equal intensity, they travel in all three dimensions. How come these laws do not obey an inverse-cube relationship while traveling through space?
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 $r$?
1740
The simplest equation expressing mass–energy equivalence is the famous $E=mc^2$ where $c$ represents the speed of light. Compare this with $E_K = \frac{1}{2}mv^2$.   Since $E=mc^2$ can be applied to rest mass ($m_0$) and rest energy ($E_0$) to show their proportionality as  $E_0=m_0c^2$, I ask whether this resemblance is just a coincidence created by the need of any equation to be homogeneous for units, or are the two equations fundamentally related?  I know about  but I could not find the answer I'm looking for there.
103221
According to my text book:  The Lorentz Force F on a charged particle moving in an electromagnetic field is given by:  $$F= qE + q(v\times B)$$  then it states that "it is to be pointed out that only the electric force does work, while the no work is done by the magnetic force which is simply a deflecting force."  How can this by true won't the the magnetic force have a considerable magnitude, will it now impart kinetic energy to the charge? Any explanations will help a lot. Please keep the terminology simple, thanks.
465
What's the difference between magnetic field and electric field lines?  Does $d\vec B$ point in the direction you would experience a force if you were a moving charged particle at that point? I know for a fact the electric field and electrostatic force are parallel, but since $F = q\ \vec v \times \vec B$ for magnetic forces. Does that actually mean though that if, say, in the following graphic:    Where the dot is a wire going in and out of the page, that the $d\vec B$ vectors are not showing how a moving charged particle will move in the field?   Basically, I feel justified in thinking electric field lines communicate how a test charge will move at any given point in space.  How can I juxtapose this with magnetic field lines>
97469
Say we have 2 charges set up like above. Why is $V_3=0$ ?   I can understand mathematically that the scalars fields cancel out, but I don't understand it physically. If I place a positive charge in between the 2 charges, it has electric potential energy because it is feels a force towards the negative charge and repelled by the positive and gains kinetic energy. However because the Electric Potential is 0, thus the electrical potential energy is 0.  How can $V_3 =0V$ if a charge placed in between the two charges will gain kinetic energy?
19429
We can think that the electric field and the gravitational field operate similarly in the sense that the forms of their governing laws (namely, Coulomb's law and Newton's law respectively) are strikingly similar. The only difference one can point out is that while the electric charges come in two flavors, the gravitational masses come in just one.  Now, I have read that when a charged particle moves, the electric field lines associated with are distorted in a fashion because of the finite time required for the information about the change in the position of the charge to get propagated. And, I have led to the understanding that this is the cause of the existence of the magnetic field (and that if calculus is used it can be proven mathematically).   So (if this is true then) why doesn't the same thing happen to the gravitational field? Why is there nothing like a gravitational magnetic field? Or, is there?  Note  I have changed the language and the tone of the question massively. Although the question was fairly well received, I believe it was really ill-posed. As pointed out by ACuriousMind in the comment, the ''reason''described here behind the existence of the magnetic field is something that cannot be found a good support for. But still, due to the similarity between the equations describing their static behavior of electric and gravitational fields, one can still ask as to whether a boost would create some sort of a gravitational magnetic field if the original frame only had a static gravitational field. As the accepted answer points out, the answer is, roughly, a yes but the Maxwell-type equations for gravity aren't as well-behaved as are the original Maxwell equations of electromagnetism--one should note. In particular, the equations for gravity take the Maxwell-like form in appropriate weak limits only in some appropriately chosen gauges--and aren't Lorentz covariant.
14136
I'm far from being a physics expert and figured this would be a good place to ask a beginner question that has been confusing me for some time.  According to Galileo, two bodies of different masses, dropped from the same height, will touch the floor at the same time in the absence of air resistance.  BUT Newton's second law states that $a = F/m$, with $a$ the acceleration of a particle, $m$ its mass and $F$ the sum of forces applied to it.  I understand that acceleration represents a variation of velocity and velocity represents a variation of position. I don't comprehend why the mass, which is seemingly affecting the acceleration, does not affect the "time of impact".  Can someone explain this to me? I feel pretty dumb right now :)
19257
I have learned that $$\mathbf{F} = m \mathbf{a}$$ where both the force $\mathbf{F}$ and acceleration $\mathbf{a}$ are vectors. This makes sense, since both force and acceleration have a direction.  On the other hand, the kinetic energy $$K = \frac12 mv^2$$ looks completely different. It doesn't seem to depend on the direction. How are these two concepts related?
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!
113343
Almost every book on physics that I read have some weird and non-clear explanations regarding the potential energy. Ok, I do understand that if we integrate a force over some path, we'll get a difference in some origin-function values ($\int_{A}^{B} Fdx = U(B) - U(A)$). This function is the potential energy. Of course, whether we can define this term or not depends on the force.  Now, here's an example of explanation (to be more precise - lack of explanation) regarding the GPE from one of the books:     "...When a body moves from some point A to point B, gravity is doing work: $U_A-U_B=W_{A \to B}$. The magnitude can be calculated using an integral:   $W_{A \to B}=\int_{r_A}^{r_B} F(r) dr = \int_{r_A}^{r_B} \left(-\frac{GMm}{r^2} \right)dr=(-\frac{GMm}{r_A})-(-\frac{GMm}{r_B})=U_A-U_B$      ...      Thus,  when $r_A&gt;r_B$, the magnitude is positive and therefore $U_A&gt;U_B$. In other words: when the distance between the bodies is being increased - the gravitational potential energy of the system is also being increased.   Giving absolutely no explanation on why all of a sudden they put a minus sign into the integral.  From another book:     Work done by Coulomb's force: $W_{el}=\int_{r_1}^{r_2}\frac{q_1q_2dr}{4 \pi \epsilon_0 r^2}=\frac{q_1q_2}{4 \pi \epsilon_0 r_1}-\frac{q_1q_2}{4 \pi \epsilon_0 r_2}$      ... Calculating the work gravity is doing is no different from the calculation of the work done by an electric field, with two exceptions - instead of $q_1q_2/4 \pi \epsilon_0$ we should plug $G M m$, and we also should change the sign, because the gravitational force is always a force of attraction.   Now, this is not satisfying at all. So what if it is an attraction force? How this should influence our calculations, if the work is defined as $|F| |\Delta x| \cos \theta$, so the sign only depends on the angle between the path vector and the force vector? Why they put a minus sign? Is is just a convention or a must thing to do?  Some say the sign is important, others say the opposite. Some explain this as a consequence of that we bring the body from infinity to some point, while others say it is a consequence of an attractive nature of the gravitational force. All of that is really confusing me.  Also, in some of the questions like "what work is required to bring something from point $A$ to point $B$ in the field of gravitational/electric force", the books sometimes confuse $U_A-U_B$ and $U_B-U_A$ - as I understand it - the work that I must do is always $U_B-U_A$. However the work that the force that is being created by the field do is always $U_A-U_B$, am I correct?
33841
If the strength of gravitational influence exerted by a body is derived from its mass and energy then is it true that a moving object which has some kinetic energy should also produce stronger gravitational pull than a static/slower one?  If that's correct then would it be also possible for an object moving fast enough to produce gravity so strong that it would collapse into a (fast moving) black hole?
24767
In electromagnetism, we can re-write the electric field in terms of the electric scalar potential, and the magnetic vector potential.  That is:  $E = -\nabla\phi - \frac{\partial A}{\partial t}$, where $A$ is such that $B = \nabla \times A$.  I have an intuitive understanding of $\phi$ as the electric potential, as I am familiar with the formula $F = -\nabla V$, where $V$ is the potential energy.  Therefore since $E = F/q$, it is easy to see how $\phi$ can be interpreted as the electric potential, in the electrostatic case.  I also know that $F = \frac{dp}{dt}$, where $p$ is momentum, and thus this leads me to believe that $A$ should be somehow connected to momentum, maybe like a "potential momentum".  Is there such an intuitive way to understand what $A$ is physically?
122606
Ok,this is a silly question. But for quite long I have been thinking about this. What is gravitation all about?? My book directly writes since force of gravitation is directly proportional to the masses and inversly proportional to the square of the distance ie. $$F = G\frac{M_1 . M_2}{r^2}$$ . So is this gravitation all about? What is the cause of it? The book then says     We are not in the scope in classical approach so as to discuss the cause of this. However,as a matter of fact that if there are more than one mass,then there is gravitation as it is their fundamental property.( Cause of gravitation can be attributed to exchange of non-classical particles between the masses) .    Then it ended the talkings. What are those non-classical particles? Why and how are they exchanged? So far I have studied physics , I never found such weird phenomenon like this. I want to know what is the cause of gravitation and what is the cause of exchange of those so-called non-classical particles? And how do they do it?
23500
Related to this question:   $E=mc^2$ - energy equals mass. So, if an object has gravitational potential energy relative to another object, it should have additional mass to represent that extra energy.  So, if I have an apple in my hand, it should have quite a large amount of gravitational potential energy relative to a black hole at the edge of the observable universe. I did a quick calculation of how much this gravitational potential energy would be (using Newton's Law of Gravitation) and it came out to be the equivalent of around 500 tons (and that's just for one black hole!).  So, why does my apple not weigh a lot more than it appears to?  EDIT: Here's my calculation of the GPE:  Assumptions:   The apple has mass $m_a=0.3 kg$ The black hole has a mass equal to 1000 solar masses, $m_b=2\times10^{33} kg$ Distance between the apple and black hole is 60 billion light years   Obviously $GPE=mgh$ can't be used, because the gravitational acceleration is not constant and varies with distance, so it has to be integrated.  I used Newton's Law of Gravitation to provide the acceleration:  $$ F=\frac{G{m_1}{m_2}}{h^2} $$  A small change in height produces a small change in energy, which can be used to integrate:  $$ \begin{eqnarray*} \delta E&amp;=&amp;F \delta h\\ \int_0^EdE&amp;=&amp;G{m_a}{m_b}\int_1^R\frac{1}{h^2}dh \end{eqnarray*} $$  I'm integrating the force starting at 1m from the center of the black hole, as the force goes to infinity at $h=0$. It's a little crude, but it should provide a conservative estimate.  $$ \begin{eqnarray*} E&amp;=&amp;G{m_a}{m_b}(1-\frac{1}{R})\\ E&amp;\approx&amp; G{m_a}{m_b}\;\;\;\;\;\;(R&gt;&gt;1)\\ G&amp;=&amp;6.673\times 10^{-11}\;\;N(\frac{m}{kg})^2\\ G{m_a}{m_b}&amp;=&amp;4\times 10^{22}\;\;Nm\\ E&amp;=&amp;mc^2\\ \Delta m&amp;=&amp;\frac{4\times 10^{22}}{(3\times 10^8)^2}\\ &amp;\approx&amp;444,000\;\;kg \end{eqnarray*} $$  So, a little short of 500 tons, but still pretty big!
31362
You can have two electrons that experience each other's force by the exchange of photons (i.e. the electromagnetic force). Yet if you compress them really strongly, the electromagnetic interaction will no longer be the main force pushing them apart to balance the force that pushes them towards each other. Instead, you get a a repulsive force as a consequence of the Pauli exclusion principle. As I have read so far, this seems like a "force" that is completely separate from the other well known forces like the strong, electroweak and gravitational interaction (even though the graviton hasn't been observed so far).   So my question is: is Pauli-repulsion a phenomenon that has also not yet been explained in terms of any of the three other forces that we know of?  Note: does this apply to degenerate pressure too (which was explained to me as $\Delta p$ increasing because $\Delta x$ became smaller because the particles are confined to a smaller space (Heisenberg u.p.), as is what happens when stars collapse)?
50825
Newton's law of universal gravitation: "Newton's law of universal gravitation states that every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them."  Coulomb's law: "The magnitude of the Electrostatics force of interaction between two point charges is directly proportional to the scalar multiplication of the magnitudes of charges and inversely proportional to the square of the distances between them."  How did Sir Isaac Newton and Sir Charles Augustine De Coulomb come to know that the force, gravitational or coulomb's,  is inversely proportional to the square of the distance between two point masses or charges, why didn't they just say that the force is inversely proportional to the distance of two bodies from each other or two charges? There must have been something that made them formulate these inverse-square laws.