# Revolution of Earth around the Sun

We study the orbit of many planets around the sun modelling it as a uniform circular motion. Is this due to the fact that empirically we measured that the magnitude of the earth's velocity (for example) is constant and the path around the sun is very well approximated by a circle? Or there are more deep and theoretical reasons (without moving out from Classical Mechanics)?

• You can also study as an elliptical mition May 13 '19 at 13:39
• The planets' orbits are ellipses, but the eccentricity is mostly small (apart from Mercury), so a circle is an ok crude approximation, and a lot easier to calculate than an ellipse. But if you want to predict planet positions in the sky, circular approximations are not very useful. May 13 '19 at 13:39
• Wikipedia has a graph of how the eccentricities of the orbits of the rocky planets will change over the next 50,000 years: en.wikipedia.org/wiki/Orbital_eccentricity May 13 '19 at 13:44
• The answers here may be useful : physics.stackexchange.com/q/479055/207455
– user207455
May 13 '19 at 13:54

There is a lot more to the story of planetary motion than that. The circular model is a simplified version presented in elementary level physics texts because it is solvable without a lot of calculus or differential equations.

Kepler's Laws state that they are ellipses with the sun at one focus. So the data demonstrates this.

Newton used this and his laws of mechanics to figure out the nature of the force that would generate Kepler's laws as a solution. This led to Gravity, in its Newtonian form as an inverse square law. It turns out that all conic sections are possible solutions for astronomical bodies traveling near the sun.

However, even in Newton's lifetime there was a problem with Mercury. Mercury's orbit deviates from predictions and could not be predicted or modeled. This eventually led to Einstein's general relativity and a new law of gravity that looks Newtonian for very weak fields. Einstein accurately predicted Mercury's orbit using his laws.

As for "modeling" paths due to gravity there are several things to consider. Every planet will pull on the other planets so all sources need to be included to be exact. For the most part is seems that this is not necessary. However modeling satellite and missile trajectories in low altitude orbit requires detailed models of the Earth since is it not a perfect sphere, or uniform mass density. This is noticeable.

In short, how you "model" an orbit depends on they type of system you are looking at. You may be able to use circles or you may need relativistic many body approximations.

In this case if we consider there's only one force between the sun and earth is gravitational force between them then the net torque on Earth is 0.So total angular momentum of Earth is conserved(w.r.t to a fixed axis).So it's not necessary that planetary orbit will always be circular.

• Hi and welcome to Phys.SE! Conservation of angular momentum in case of central force confines the planetary motion to a plane and does not specifically tell anything about the shape of the orbit. May 13 '19 at 15:04
• Actually, the mathematics behind this can explain shape of orbit . May 13 '19 at 15:13
• Yes, the shape of the orbit depends on the angular momentum. What I meant to say was, the conservation of it does not tell anything about its value except that it is some constant, right? In other words, eccentricity $\epsilon = \sqrt{1+\dfrac{2EL^2}{GMm^2}}$ depends on $L$. But, the conservation of $L$ says nothing about if $\epsilon$ is $0$ or $1$ or any other positive value, right? May 13 '19 at 15:30