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The point of an elliptical orbit closest to the Sun is called the perihelion, and the point most distant from it is called aphelion (Fig). Denoting the distance from the perihelion to the Sun by r_0and the velocity of the planet at the perihelion by v_0 , find the radius of curvature of the orbit at the perihelion and at the aphelion, the distance from the aphelion to the Sun, and the velocity of the planet at the aphelion. Prove that the motion of a planet in an elliptical orbit is only possible, if its total energy is negative. |
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Answer» potential energy `U_0 = - gammamM//r_0` To find the radius of curvature in the perhelion apply Newton.s second LAW `(mv_0^2)/R_0=(gammamM)/r_0^2` form which `R_0=(mv_0^2r_0^2)/(gammaMm)=-(2r_0K_0)/U_0` The radius of curvature at the aphelion is the same as in the perihelion since the ellipse is a symmetrical figure. We have, according to Newton.s second law, `(mv_a^2)/R_0=(gammamM)/r_a^2` The total mechanical energy of the planet, according to the law of conservation of energy is `W=(mv_0^2)/(2)-(gammamM)/r_0=(mv_a^2)/2-(gammamM)/r_a` Eliminating the VELOCITY we obtain `(gammamMR_0)/(2r_0^2)-(gammamM)/r_0=(gammamMR_0)/(2r_a^2)-(gammamM)/r_a` Cancelling out `gammam` we obtain a quadratic equation `r_a^2(2r_(0)-R_0)-2r_ar_0^2+R_0r_0^2=0` The first root of the equation is `r_a = r_0`This means that in this case the ellipse reduces to a circle of radius ro. In this case the radius of curvature is also `R_0 = r_0` , and, consequently, the orbital velocity is `r_(a)=(r_0R_0)/(2r_0-R_0)=-(r_0K_0)/W` Since the DISTANCE from the aphelion to the Sun is a positive quantity we must have `Wlt0` . This means that a planet can move in an elliptical orbit only if the sum of its kinetic and potential energies (i.e. its total mechanical energy) is negative. In particular in the case of a CIRCULAR orbit, `W = - (gammamM)/(2r_0)` |
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