Since bodies like planets moving around sun perform uniform circular motion where gravitation provides the necessary centripetal force,  so the orbits should be circular   for the gravitational force to be central but as we know the paths are not perfectly circular but elliptical. Why this is so?
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.
64857
I have heard from  that stable orbits (ones that require a large amount of force to push it significantly out of it's elliptical path) can only exist in a three spatial dimensions because gravity would operate differently in a two or four dimensional space. Why is this?
18428
For the earth sun system, the force of gravitation is an internal force ( forces* a newtons third law pair)  and this INTERNAL force  is causing the earth to accelerate. . it is providing the necessary centripetal force.   Reasoning behind the question:  centripetal forces dont do any work and hence dont cause a change in mechanical energy, dont cause a change in kinetic energy therefore and dont cause a change in velocity.   QUESTION question 1) but the earth is undergoing a change in velocity?  so, is the the gravitational force on the earth an external force? but ive considered our system containing the sun too. shouldn't that be an internal force then?  2) in the earth sun system if for simplicity we call the earth's movement uniform circular motion, is momentum conserved in this system? how?  i suppose ive mixed up too many things
105979
Suppose There's a particle of mass m moving with speed $v_{0}$, at certain moment a force starts acting on it (centripetal force) and it start uniform circular motion  Now every other reference states that centripetal force only brings about change only in direction  But plugging-in things in the equations paint a different picture  suppose  normal/centripetal acceleration = $A$  then $|v| = \sqrt{v_{0}^2+\frac{(A t)^2}{1}}$  we can certainly see that factor of $t$ will bring change in magnitude of velocity   Secondly, why the velocity is always tangential to the circle, but how can we mathematically show such? Intuitively for me if there's constant acceleration toward a certain then at some point of time, the body should 'slack' and fall toward the direction of acceleration   Here' a picture for a reference:
32707
I'm a class 12th student and this thought just struck me from nowhere.  Assume a situation where every human on this planet turns towards west. All of them start walking simultaneously in west direction (exactly west i.e. antiparallel to Earth's rotation and none of their paths cross each other).  As Earth spins from west to east would this activity cause the earth's rotation speed to increase?   (There is frictional force acting on the earth due to motion by nearly a billion people which should produce enough torque to increase angular velocity of the earth.)   p.s. If the magnitude of friction force not enough to rotate the earth then assume each person to weigh more than 100 kg.  Also if it does happen would earth remain in the same orbit around the sun with just increased angular velocity about axis? (or would path around the sun also change due to this? )  I would really appreciate an easy solution cause I am still in class 12th.
23451
If a planet was suddenly stopped in its orbit supposed to be circular, show that it would fall onto the sun in a time $\left( \frac { 1 }{ 4\sqrt { 2 }  }  \right) $ times its time period.   I am getting stuck with this problem. I have applied conservation of energy and I think it may solve it. Please help. Here's my research on the problem:   Firstly, ${ T }^{ 2 }=\frac { 4{ \pi  }^{ 2 }{ r }^{ 3 } }{ GM } $. Assuming $v$ to be the velocity of the planet at a distance $s$ from the sun, we get: $\frac { m{ v }^{ 2 } }{ 2 } +\left( \frac { -GMm }{ s }  \right) =-\frac { GMm }{ r } $, where $r$ is the radius of the planet's orbit.
12664
Why do planets revolve around the Sun if there is a force called frictional force? Why do planets not stop rotating while the objects in motion stop after a while? Why are they continuously moving? Why are planets not affected by frictional force?
103640
For instance, for a two-body system, there will always be a point in-between the two in which the forces of gravity completely cancel, and if I were there then I would experience no net force. I believe there is also a point, or a region, for three body systems as well. Assuming we treat the stars or planets as simply point-masses, will there always be, for any number of objects, a point, or a region, of equilibrium? Or no?
20246
Will an object be pulled towards the centre linearly if the net force provided for a circular motion is larger than the required centripetal force? And why?  For example, if the object in a circular motion which is connected by a string is pulled towards the centre by hand.
20601
Well, this question has been puzzling me for kinda long time, many people believe that orbiting astronauts feel weightless because they are "beyond the pull of Earth's gravity"...How far from the Earth would a spacecraft have to travel to be truly beyond the Earth's gravitational influence? If a spacecraft were really unaffected by Earth's gravity would it remain n orbit? If so, what is the real reason for weightlessness in orbit?
88030
When discussing gravitation between two bodies of finite size, for instance Earth around the Sun, we suppose the mass of Earth and the Sun to be perfectly localized at the center of each body. Is this a "useful" "good" "approximation" or the "absolute truth"?
23004
I'm sorry if the answer is obvious for you guys, but why don't we all (including buildings, road, people, the ground) collapse to the center of the earth because of gravity? Is it because we have velocity, just like the earth not falling to the sun (or electrons not falling to the protons)? But that analogy doesn't sound right because those involve wide empty space.
43945
In curved spacetime particles follow timelike geodesics, which should have maximal proper time (at least locally). I thought this path usually corresponds to a global maximum, and there are only strange exceptions. Propably I was wrong, or I really don't understand something, because it doesn't seem to be true for the case of orbitals in Schwarzschild geometry.  The Schwarzschild metric is (with +--- signature)  $$ g_{\mu\nu} = \textrm{diag}\left(1-\frac{2GM}{r},-\left(1-\frac{2GM}{r}\right)^{-1},-r^2,-r^2\sin^2\theta\right) $$  An observer with fixed $r,\theta,\phi$ coordinates will age according to $d\tau_1 = \sqrt{g_{tt}}dt$. An observer, which orbits the center at the same constant radius with $\theta=\pi/2$ will age according to $d\tau_2 = \sqrt{g_{tt}-\omega^2r^2\sin^2\theta}dt$, so he ages slower during the same amount of coordinate time $dt$. If the two observers start from the same point, they will meet again after a full rotation of the orbiting observer, and the orbiting observer will age less. What I don't understand is that the orbit is the geodesic (if the angular velocity is set properly), so why didn't he age more? He should have been on the path with maximal proper time.
40207
In a circular motion, like the horse rides in a merry-go-round, there is only one force that acting on it, the centripetal force that pulls the ride inward. I'm struggling to answer why the horse rides swing outward instead?
76934
Except for Mercury, the planets in the Solar System .  Is this property special to the Solar System?  :     Most exoplanets with orbital periods of 20 days or less have near-circular orbits of very low eccentricity. That is believed to be due to tidal circularization, an effect in which the gravitational interaction between two bodies gradually reduces their orbital eccentricity. By contrast, most known exoplanets with longer orbital periods have quite eccentric orbits. (As of July 2010, 55% of such exoplanets have eccentricities greater than 0.2 while 17% have eccentricities greater than 0.5.) This is not an observational selection effect, since a planet can be detected about equally well regardless of the eccentricity of its orbit. The prevalence of elliptical orbits is a major puzzle, since current theories of planetary formation strongly suggest planets should form with circular (that is, non-eccentric) orbits.   What is special about the Solar System that orbits of planets here are nearly circular, but elsewhere they are moderately or highly eccentric?
116893
In science documentaries that touch on general relativity, it is often said that gravitational pull isn't an actual a pull (as described by classical physics), but rather one body travelling in a straight line on the curved space around another massive body.  I don't know if I misunderstood that or if it's just an oversimplification meant to help the uninitiated understand the concept better, but something doesn't seem to follow. If the Earth travelling around the Sun is just moving along a straight line in the curved space, shouldn't light also be trapped in orbit around the Sun?  I expect the actual equations in the theory also take into account a body's velocity, not just the curvature.
19360
A lot of my problems have objects moving in circular paths with tangential and normal components of acceleration.  If the tangential component is non-zero though, the speed is changing so the radius of its path would also be changing.  So how can its path be considered circular? The size of of its radius of curvature of constantly changing.