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Well known Kepler laws state that earth revolve around sun in an elliptical path with sun at one of the focii. My question is rather simple. Why so?

I mean for an equilibrium of earth that is it is not accelerated towards or away from sun its gravitational force must be balanced by centrifugal force (pseudo force as frame corresponding to earth)

But following an elliptical path it distance from sun won't be same and hence the gravitational force and hence how is this?

I suspect may be it is due to presence of other celestial bodies but that seems too vague and escaping from the truth

Also maximum of derivations in gravitation are done by assuming circular path of earth around sun. (atleast at my level)

Please give me the real physics behind it?

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marked as duplicate by Emilio Pisanty, stafusa, M. Enns, John Rennie newtonian-mechanics Sep 30 '17 at 4:46

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ What do you mean by "maximum of derivations in gravitation are done by assuming circular path of earth"? $\endgroup$ – Floris Sep 29 '17 at 14:43
  • $\begingroup$ Many of them like proof of time and orbit relationship of kepler $\endgroup$ – Pranjal Rana Sep 29 '17 at 14:45
  • $\begingroup$ There's a story that when asked why planets move in ellipses, to answer that question, Newton invented Calculus. Kepler didn't know why, he just worked out what the orbits did. Kepler wasn't even a fan of ellipses cause they seemed as counter-intuitive to him as everyone else in his day. He just worked out and published the math. Newton worked out why. futurism.com/… $\endgroup$ – userLTK Sep 29 '17 at 15:16
  • $\begingroup$ Given that there has to be an orbit even the smallest perturbation to a circular orbit will result in an elliptical orbit so the chance of an orbit being exactly circular is negligible. $\endgroup$ – Farcher Sep 29 '17 at 15:37
  • $\begingroup$ Why do you think the earth isn't accelerating towards the sun? It is constantly accelerating towards the sun. The direction of the acceleration and the direction of the movement need not be the same; I suspect you are confusing them. $\endgroup$ – Eric Lippert Sep 29 '17 at 17:15
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What you are missing is that Earth speed is varying along its orbit, contrary to a circular one.

More precisely, at any point of the orbit, the acceleration of the Earth has a component tangent to the orbit, and a component perpendicular to the orbit. For a circular orbit, the former is always zero, and the latter is therefore exactly equal to the gravitational force, i.e. a centripetal acceleration. That was in an inertial frame, so now if we take a frame rotating along with the Earth, that's the picture you had in mind: the centrifugal force, which is opposite to the centripetal acceleration in the inertial frame, compensates the gravitational force.

But for an elliptic orbit, the component of the acceleration tangent to the orbit is only zero at the apogee and perigee. Since the gravitational force is equal to the sum of the tangent and normal component of the acceleration, we can say that part of the gravitational force bends the trajectory toward the Sun, and part accelerate the Earth along the trajectory (or decelerate it, depending on the position on the orbit).

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There are many mathematical derivations of this online; let me give you the intuitive explanation that I think you are looking for.

I will look at two points on the ellipse: when the Earth is closest to the sun (perihelion) and when it is furthest (aphelion). Throughout, we assume conservation of angular momentum - that is, the product of the orbital velocity and the distance are constant.

When the Earth is at aphelion, we will assume velocity $v_a$, and distance to the Sun $r_a$. We know the force of gravity will be proportional to $\frac{1}{r_a^2}$. The centripetal force needed for a circular orbit is $$F_c = \frac{mv_a^2}{r_a}$$

If we express this in terms of the angular momentum $L=mv_ar_a$ (which is constant) we get

$$F_c = \frac{L^2}{mr_a^3}$$

But we know the force of gravity scales with the inverse square, not the inverse cube: so when we are far away, the force will be "stronger than is needed" and the Earth will "fall towards the Sun". When we are close, the force is not strong enough and we "don't fall fast enough for a circular orbit".

The result is an elliptical orbit. This was already proven by Newton (incidentally, the elliptical result only follows if the power law is inverse square - that makes for an interesting large scale proof that gravity is indeed following that law at large scales).

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  • $\begingroup$ Fantastic job Floris, I am agreed but one small doubt flickers..... If the orbit is circular then what is the use of aphelion and perhelion. $\endgroup$ – Pranjal Rana Sep 29 '17 at 15:15
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    $\begingroup$ When the orbit is circular, the velocity is just right and there is no "closest" or "furthest" point... $\endgroup$ – Floris Sep 29 '17 at 15:19
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There are essentially five solutions to the two body problem, where the bodies are attracted to each other by gravity.

  • A hyperbolic path - the two bodies are moving away from each other too quickly for gravity to bring them into a closed orbit.
  • A parabolic path - the two bodies are moving away from each other just quickly enough for gravity not to be able to bring them into a closed orbit. This is a kind of "boundary case" between the hyperbolic path and the elliptical orbit.
  • An elliptical orbit - the general case when gravity between the two bodies limits the distance between them.
  • A circular orbit - a very special variant of the elliptical orbit, where the direction of the bodies' velocity away from each other is exactly perpendicular to the displacement between them, and the magnitude of that velocity is exactly the right amount to balance out the gravitational attraction.
  • A collision course where the bodies move directly towards each other, although they may start off moving away from each other.

For the parabolic solution or the circle solution, conditions need to be exactly right. That is, the initial relative velocity of the bodies needs to be a precisely calculated magnitude, in a precisely calculated direction. This would require an enormous coincidence. That's why the other three solutions (hyperbola, ellipse or collision course) are the only ones that every happen in nature.

In particular, if there were ever a satellite or a planet orbiting a planet or a star in an exactly circular orbit, it would only take a very slight tug from a third body to perturb it into an elliptical orbit.

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