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Astronomy news on Phys.org

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What is your question?Mikael17 said:

In the main text the question is "Is it possible?". The answer of that question is a three-letter word. "Yes"

In the headline, the question is "How". This takes a bit longer equation.

Now, note that for each apside, the speed is tangential - the radial component of speed is zero at apsides.

The total energy of a body is E=m(v

Therefore, given r

E/m=v

L/m=v

This is two equations for two unknowns. Could be solved. Given one pair of v and r, which are a solution, find the pair of v and r which are the other solution.

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I don't see how you can do it without the mass of the primary (the ##M## in post #2).

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Very much this. However, given the mass of the primary and (indirectly) the angular momentum the full effective potential is known as well as thd classical turning point. It is then just a matter of finding the other classical turning point, which amounts to finding the roots of a second degree polynomial.Ibix said:I don't see how you can do it without the mass of the primary (the ##M## in post #2).

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If the OP was asking a well-defined question one could assume that "perihelion" and "aphelion" is a reference to the Sun as primary mass.Ibix said:I don't see how you can do it without the mass of the primary (the ##M## in post #2).

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If you know the primary mass (e.g. Sun) then yes. It follows more or less directly from the vis-viva equation that holds for all two-body keplerian orbits.Mikael17 said:

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Indeed. Using vis-viva and solving for one apsis distance gives a direct "symmetric" relationship of this distance as a function of the other apsis distance and speed. This distance is then either positive, infinite or negative for elliptic, parabolic or hyperbolic orbits, respectively, so one equation covers all cases.snorkack said:

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The case of positive distance further divides into the cases where the found apsis distance was bigger than given apsis distance (found was apoapsis), found apsis distance was equal to given apsis distance (neither was apsis after all because the orbit was not elliptic) or the found apsis distance was smaller than given apsis distance (found apsis was periapsis).Filip Larsen said:Using vis-viva and solving for one apsis distance gives a direct "symmetric" relationship of this distance as a function of the other apsis distance and speed. This distance is then either positive, infinite or negative for elliptic, parabolic or hyperbolic orbits, respectively, so one equation covers all cases.

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Yes, but my point is that since you end up with an equation that is valid for all cases you don't really need to "worry" about the orbit classification unless you specifically want to know or verify that too. Also, the full orbit classification is more of a mathematical thing since in practice there are only either elliptical or hyperbolic orbits with the rest being "degenerate" (i.e. only approximately true) cases.snorkack said:The case of positive distance further divides into the cases where the found apsis distance was bigger than given apsis distance (found was apoapsis), found apsis distance was equal to given apsis distance (neither was apsis after all because the orbit was not elliptic) or the found apsis distance was smaller than given apsis distance (found apsis was periapsis).

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The maths is actually no worse if the satellite mass is non-negligible, but you do need to know the satellite mass and you do need to take care about how you defined perihelion distance. You want to use the satellite-to-barycenter distance, and this will be significantly different from the satellite-to-primary distance in this case.

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To elaborate on this with the teeniest amount of math. The two-body problem with a Kepler potential separates into the motion of the barycenter (free motion so rectilinear) and a Kepler potential problem for the separation vector of the masses with gravitational potential ##- GM/r## with ##M = m_1 + m_2## being the total mass of the system (the potential energy is obviously ##- Gm_1m_2/r##, but the mass entering in the kinetic energy for the problem for the separation vector is the reduced mass ##m_1 m_2/M##).Ibix said:

The maths is actually no worse if the satellite mass is non-negligible, but you do need to know the satellite mass and you do need to take care about how you defined perihelion distance. You want to use the satellite-to-barycenter distance, and this will be significantly different from the satellite-to-primary distance in this case.

The separation from the barycenter of mass ##m_2## is a factor ##m_1/M## of the separation of the masses (just by definition of the barycenter) if you prefer looking at the apsides of that rather than the apsides of the two-body separation.

If you are interested in the separation apsides rather than the mass-barycenter apsides, everything is just as above with the primary mass replaced by the total mass.

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Filip Larsen said:If the OP was asking a well-defined question one could assume that "perihelion" and "aphelion" is a reference to the Sun as primary mass.

Helios (Greek for sun) is an easy one and 'Gee' (a greek term for Earth) is also familiar but there are some others like

Star | periastron, apoastron | -astrons |

I'd like a general term for apo and peri points of an orbit.

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What is wrong with the one we have?sophiecentaur said:I'd like a general term for apo and peri points of an orbit.

https://en.m.wikipedia.org/wiki/Apsis

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Oh for a content addressable memory. Thanks! I promise to use it whenever possible .Orodruin said:What is wrong with the one we have?

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