Inclination of planetary orbits

In summary, the conversation discusses the calculation of planetary orbit inclinations with respect to the Sun's equator-plane. The JPL Horizons system is recommended as the authority on this topic, with the suggestion to use the inclination and longitude of ascending node to determine the answer. However, it is noted that orbital elements, including inclination, change over time and a complete reference system must be specified for precision. The JPL Horizons system is suggested for its accuracy.
  • #1
Leo Klem
13
0
Does anyone know the inclination of each planetary orbit (in degrees) with respect to the plane of the Sun's equator-plane?All planetary inclinations are known with respest to the ecliptic, and the ecliptic's inclination is known with rescpect to the Sun's equator (7.25 degrees). Such data, however, seem to me not sufficient for a precise calculation of the answer I need.
 
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  • #2
Hi Leo - The authority on these things is the JPL Horizons system.

I'm not sure if you can use the sun's equator, but you'll certainly be able to find the inclination of each planetary orbit with respect to the ecliptic, then do the arithmetic by adding/subtracting the 7.25 deg. (At least it seems that simple at first look . . . )

Note that inclination, like other orbital elements, actually osculates, meaning it changes ever so slightly over time. So look in the osculating elements of your results after you submit your object data. The inclination is given by the quantity "IN", which is in degrees by default.

hth
 
  • #3
What you suggest doesn't work, as I've already pointed out, when stating that those data are not sufficient; because there is no criterion to chose whether each particular orbit inclination has to be added or subtracted with/from 7.25 degrees.
 
  • #4
Use the inclination and the longitude of ascending node.
 
  • #5
D H said:
Use the inclination and the longitude of ascending node.

Yup, that's the answer. You have to specify a complete reference system, and if you throw out the ecliptic as your reference frame, you also throw out Earth's ascending node as the reference zero-degree direction. You can either retain it, or look at each orbit independently in terms of its ascending node, or something more clever.

Without knowing more about what you're trying to do, I'm thinking that's the best we're going to be able to do for you.

I referred you to JPL's horizons because you said you were seeking precision.

You're welcome.
 
  • #6
Thanks to Spacester and DH. I got the answer.
 

Related to Inclination of planetary orbits

What is the inclination of a planetary orbit?

The inclination of a planetary orbit is the angle between the plane of the planet's orbit and the reference plane of the solar system, which is usually the plane of the Earth's orbit around the sun.

How is the inclination of a planetary orbit measured?

The inclination of a planetary orbit is typically measured using a reference plane such as the celestial equator or the ecliptic. This is done by observing the position of the planet in the sky and calculating its angle from the reference plane.

What causes variations in the inclination of planetary orbits?

The inclination of a planetary orbit can vary due to a number of factors, including gravitational interactions with other planets, tidal forces from the sun, and collisions with other objects in the solar system.

Why is the inclination of a planetary orbit important?

The inclination of a planetary orbit plays a crucial role in determining the climate and seasons of a planet, as well as its ability to support life. It also affects the way planets interact with each other and can influence the stability of a planetary system.

How does the inclination of a planetary orbit affect spacecraft missions?

The inclination of a planetary orbit can greatly impact the trajectory and fuel requirements for spacecraft missions. Inclinations that are closer to the Earth's equator are generally easier to reach and require less fuel, while higher inclinations may require more complex and costly maneuvers.

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