Q:Distance of planets/moon

[cookie8 (Vincent)]

When we talk about mean distance within the solar system in astronomical terms, for example distance of moon from earth is 384403 km,are we measuring from surface to surface or centre to centre?
I cannot get an answer even in Wikipedia.

[iceman (Mike)]

I'm guessing it would be mean distance, since they're all on elliptical orbits and the distance would change throughout the year.

[cookie8 (Vincent)]

Hm....I mean surface to surface will be closer than core to core. Which are we measuring?

[bojan]

Measuring surface to surface (Moon.. for obvious reasons), but result is expressed in centre to centre.

[multiweb (Marc)]

Originally Posted by cookie8

Hm....I mean surface to surface will be closer than core to core. Which are we measuring?

I'd assume including or excluding the two radii in the measurement wouldn't make much difference if they take the mean because they're insignificant compared to the distance in between the bodies and also because of the excentricity of the orbits. For moon/earth I'd guess surface to surface? Didn't they install those mirrors for laser measurement during one of the apollo missions?

[michaellxv (Michael)]

I would have thought centre to centre. This article equates the average distance to the semi-major axis.
http://www.universetoday.com/guide-t...earth-to-moon/

[Robh (Rob)]

Distances are expressed centre to centre.
Earth-Moon laser travel time measurement will be surface to surface but the distance will still be expressed centre to centre i.e. radii added back in.

Consider the Sun. The planets orbit the Sun in an elliptical path with the Sun at one focus. The furthest a planet is from the Sun is a+f, where a is the semi-major axis and f is the Sun's offset from the ellipse's centre. The closest a planet is from the Sun is a-f. The average of the two is (a+f+a-f)/2=a, which is the semi-major axis.

Regards, Rob.

[Barrykgerdes]

For astronomical calculations (3D) it is required to work from the centre of the body, Usually the mean centre if there is eccentricity

However for navigational purposes the points aimed at need to be surface objects, actual aiming points. Imagine sending a rocket to the moon for a soft landing knowing the distance was 250,000 miles but not knowing the the height of the terrain!

I have never given any thought to these distances before so it is just what I would assume. I may be wrong.

Barry

[OneOfOne (Trevor)]

I would assume it would be centre of gravity to centre of gravity as this is the point about which the objects "orbit". Generally, this would be the geometric centre, or very close to it, unless the object has some large "dense bits". Gravitational measurements of the Moon show that there are areas of increased density.

[trent_julie]

Hey all,

I had a bit of a hunt around after reading Vincent's original question. It took me a little bit of searching to try and find some text from a reputable source to back up my assumptions or disprove them completely.

Of my understanding, I couldn't understand that it would make sense to measure from surface to surface within the solar system given the rotation of the Earth on its axis. I thought that this mattered. This rotation of Earth would effect that value of distance to other objects, at a rate of up to roughly 15deg/h or 6000km/h assisting or retarding from a fixed point on Earth whilst assuming that the object was staying still in its place of orbit.

It seems when giving the distance from the Earth surface to the Moon surface, it is done by calculating the distance to the mathematical centres at a particular point in time then removing the radius of the Earth and Moon.

Consequently, when stating distances from object to object it is convention to state the distances between the objects mathematical centres rather than an observational frame of reference which could only give a distance from surface to surface at an instant in time e.g. bouncing a LASER of the surface of the moon.

Using the mathematical frame of reference rather than the Observational frame of reference negates my Earth's rotation problem. (in which I am yet to find the equation for how it does effect the observer)

In Dot point 52 of the link below Beth Barber of NASA states the convention for stating distances.
http://helios.gsfc.nasa.gov/qa_earth.html

and for further reading,
http://www.davidcolarusso.com/astro/

Thanks for asking that question Vincent, It made me hunt around for answers rather than be spoon fed answers via podcast. This information hunt has helped me come to the conclusion that the reason that when I Image Jupiter, the reason I have blue or red edges, is not because of my poor image processing, and definitely not because of chromatic aberration, but because of Jupiter is red shifting one side and blue shifting the other.......Jokes.

I am sure the forums local astrophysics buffs will be happy to explain in more depth and if I am utterly incorrect please tell me so!

Cheers,

Trent