The classic problem in gravitation is this: given the initial position and velocity vector of a satellite (or a comet, a planet, etc.), find its maximum and minimum distance from the central body. The key is that two quantities are conserved simultaneously — the total energy and the angular momentum — and together they are enough to close the problem.
Strategy for max/min distance
- Compute the total energy at the initial state A: .
- Compute the angular momentum about the central body: , in magnitude with the angle between and .
- At the points of maximum or minimum distance , so (lever arm ).
- Impose conservation of and conservation of between A and B: this gives a quadratic equation in (or in ).
- The two solutions are the minimum distance (perihelion) and the maximum distance (aphelion).
Note
At the point of closest or furthest approach, the velocity vector is tangent to the orbit and perpendicular to : this is what makes and lets us eliminate the angle.
Why do two solutions come out? In an elliptical orbit there are exactly two points where the velocity is perpendicular to the radius: perihelion (the closest point) and aphelion (the furthest point). At all other points the velocity has a radial component, because the body is approaching or receding from the centre. The quadratic equation captures precisely these two special points.
Two solutions and the sign of
- If : the equation has two positive roots, perihelion and aphelion (ellipse).
- If : a single positive root, the perihelion (parabola).
- If : a single positive root, the point of closest approach (hyperbola).
The worked exercise satellite, aphelion and perihelion applies this strategy step by step.
Links
Topics: Gravitazione Concepts: Orbite · Conservazione dell’energia meccanica · Conservazione del momento angolare Skills: Conservazione dell’energia · Conservazione del momento angolare
Related exercises: True or false on energy and orbits · Speed of a comet at perihelion · Worked exercise: satellite, aphelion and perihelion