8/1/2023 0 Comments Elliptical orbit![]() ![]() (This is an alternativeĭerivation.) Hyperbolic (and Parabolic?) Orbits What is its potential energy at thatĭepends only on the length of the major axis. Through the point B at the end of the minor axis. L 2 2 m 2 = G M b 2 2 a , find the speed of the planet at it goes The substitution v 1 = L / m r 1, v 2 = L / m r 2 in this equation gives The major axis, we simply add the total energies at the two extreme points:ġ 2 m v 1 2 − G M m / r 1 + 1 2 m v 2 2 − G M m / r 2 = 2 E. To prove that the total energy only depends on the length of Ī simple generalization of the result for circular orbits. Rate area is swept out, and that rate is L / 2 m, so: Remember T is the total area of the orbit divided by the ![]() We’re now ready to find the time for one orbit T. L 2 2 m 2 = G M ( 1 r 1 + 1 r 2 ) = G M b 2 2 a. We can immediately use the above result to express the The arithmetic mean of r 1, r 2 and the semi-minor axis b is Pythagoras’ theorem applied to the triangle F 1 B C gives Point B at the end of the minor axis is a. (from the “string” definition of the ellipse) the distance from the Sun to Recall that the Sun is at a focus F 1 of the elliptical path (see figure below), and To make further progress in proving the orbital time T depends on a but not on b , we need to express r 1, r 2 in terms of a and b. The area of the ellipse is π a b (recall it’s a circle squashed by a factor b / a in one direction, so π a 2 becomes π a b ), and the rate of sweeping out of area is L / 2 m , so the time T for a complete orbit is given by: Using the angular momentum equation to write v 1 = L / m r 1, v 2 = L / m r 2 , and substituting these values in this equation Rearranging, and dropping the common factor m ,ġ 2 ( v 1 2 − v 2 2 ) = G M ( 1 r 1 − 1 r 2 ). Labeling the distance of closest approach r 1 , and the speed at that point v 1 , the furthest point r 2 , the speed there v 2 , we haveġ 2 m v 1 2 − G M m / r 1 = 1 2 m v 2 2 − G M m / r 2 = E. We’ll derive the results for a planet, beginning with theĮnergy and angular momentum are the same at the two extreme points of the orbit: That the orbits are elliptical, they are fairly easy to prove. Orbits of planets, asteroids, spaceships and so on -and, given These results will get you a long way in understanding the Semimajor axis, not on the length of the minor axis:Įlliptical orbit depends only on the length a of the semimajor axis, not on the Time to go around an elliptical orbit once depends only on the length a of the ![]() Smallest ellipse, so we can figure out, from the results given below, how muchīasic relevant facts about elliptical orbits: This then immediately gives us the major axis of this ( Important Exercise: Sketch the orbits of earth and Mars, and thisĮlliptical trajectory-then check your sketch with the applet!) Remembering it is an ellipse with the Sun at one focus, the smallest ellipse weĬan manage has the point furthest from the Sun at Mars, and the point nearest Visualizing the orbit of the spaceship going to Mars, and ![]() Spaceship (or a planet) in an elliptical orbit, both the total energy and the orbital time depend only on the length of Will take the ship exactly half way round the ellipse. Path because, as will become apparent, following the most fuel-efficient path Ship in this elliptical path, and we can calculate the time needed if we know We can calculate the amount of fuel required Spaceship’s trajectory to Mars will be along an elliptical path. Ignoring minor refinements like midcourse corrections, the Lifting fuel into orbit is extremely expensive. It is crucial to minimize the fuel requirement, because The two important questions (apart from can I get back?) are: Think about an astronaut planning a voyage from earth to Lecture), once that is established a lot can be deduced without further fancy AndĪlthough proving the planetary orbits areĮlliptical is quite a tricky exercise (the details can be found in the last Spaceship leaving earth and going in a circular orbit won’t get very far. Deriving Essential Properties of Elliptic Orbitsįrom a practical point of view, elliptical orbits are a lot Try aiming for Mars yourself with this applet. Previous index next Elliptic Orbits: Paths to the Planets ![]()
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