# Celestial sphere

Status
Not open for further replies.
V

...

O

#### origin

##### Guest
Re: Calculating angles on the Celestial sphere

verizen":31vmkdr2 said:
Hello, I am having trouble with a question in my textbook and was wondering if anyone could help me out with it. Here's the following question, as followed...

I want you to Calculate the angle subtended on the Celestial sphere by the separation of Venus and the Sun when the Earth and Venus are separated by about 90 degrees from an observation point on the Sun. Assume circular orbits for Venus and the Earth show your work.

I don't understand how to calculate or the which method to use.

I came up with this parallax, it's not exactly accurate but hopefully you have a better idea then what I've generated..

o<Venus
|
O<-Sun ------------------------- |90 O<-earth

I interpret the question the same way you do. So you have basically figured it out. You have a right triangle. You know the length of the 2 legs (or can look them up), and you know your trig functions (I assume) so you have all you need to calculate the angle.

A

#### aphh

##### Guest
Re: Calculating angles on the Celestial sphere

Since you have Sun in your equation you can not use spherical trigonometry based on the equatorial coordinates relative to the Sun, the ecliptic or ecliptica. In your calculation Sun would reside at the origin, so you need to use trigonometry of the triangles.

However, just in case you ever need to calculate the angular distance of two locations on the celestial sphere knowing the coordinates of the two locations, you would use spherical trigonometry like this

cos x = cos z * cos a * cos b + sin a * sin b

where x is the angular distance between the two locations.
z is the angular distance of the two locations at the equator of your coordinate system.
a is the declination (or elevation) of location a from the equator of your coordinate system.
b is the declination (or elevation) of location b from the equator of your coordinate system.

So let's say you wanted to calculate the angular distance between stars Alkaid and Alioth in the sky (stars in the handle of Big Dipper). You find the coordinates of the stars to be;

Alkaid has declination of 49.3 degrees and Right Ascension (location of known starting point, the vernal equinox, on the celestial equator) of 195.8 degrees.

Alioth has dec 56.0 degrees and r.a. 180.1 degrees respectively.

What you first calculate is their distance z at the equator of the celestial sphere, that would be 195.8 - 180.1 = 15.7 degrees.

Now you have everything to plug into the spherical trig equation cos x = cos 15.7 * cos 49.3 * cos 56.0 + sin 49.3 * sin 56.0 and you'll get the angular distance between these two stars or locations on the celestial sphere to be 11.6 degrees in the sky.

You can use this most basic spherical trigonometric formula to calculate distances also on earth, when you know the coordinates of the two locations. Hope this helps.

V

#### verizen

##### Guest
Re: Calculating angles on the Celestial sphere

Thanks for the help I appreciate it ALOT, I'm trying to understand it piece by piece because I haven't done cos sin laws since grade 12 and I'm in University now :| So in this case we wouldn't have to use the method of parallax=(360/2pie)*baseline/distance which gives a distance = baseline * 57.3/parallax?

V

#### verizen

##### Guest
Re: Calculating angles on the Celestial sphere

...

A

#### aphh

##### Guest
Re: Calculating angles on the Celestial sphere

verizen":4i1jp2j3 said:
o<Venus
| \
O<-Sun ------------------------- |90 O<-earth

If that's your situation, why not simply use the distances of Earth and Venus from the Sun, and calculate the angle using tangent like this,

tangent Sun - Venus = (distance Sun - Venus) / (distance Sun - Earth) ?

So if the distances were

Earth = 1 AU
Venus = 0.72 AU

0.72 / 1.00 = 0.72, take inverted tangent (tan^-1), and you'd get 38.8 degrees separation between Venus and the Sun from Earth's perspective (as per your scenario).

Tangent is the trig function that fits your bill, as it is TOA, meaning opposite side of the angle over adjacent side. You need to refresh your basic use of trig functions on right triangle and Pythagora's law, before proceeding.

V

#### verizen

##### Guest
Re: Calculating angles on the Celestial sphere

aphh":feah1vkn said:
verizen":feah1vkn said:
o<Venus
| \
O<-Sun ------------------------- |90 O<-earth

If that's your situation, why not simply use the distances of Earth and Venus from the Sun, and calculate the angle using tangent like this,

tangent Sun - Venus = (distance Sun - Venus) / (distance Sun - Earth) ?

So if the distances were

Earth = 1 AU
Venus = 0.72 AU

0.72 / 1.00 = 0.72, take inverted tangent (tan^-1), and you'd get 38.8 degrees separation between Venus and the Sun from Earth's perspective (as per your scenario).

Tangent is the trig function that fits your bill, as it is TOA, meaning opposite side of the angle over adjacent side. You need to refresh your basic use of trig functions on right triangle and Pythagora's law, before proceeding.

Yeah about that diagram it was written incorrectly the planet venus is suppose to be aligned with earth and the sun being the observation point...

O

#### origin

##### Guest
Yeah about that diagram it was written incorrectly the planet venus is suppose to be aligned with earth and the sun being the observation point...

Really? That seems hard to believe, because the angle would be zero, unless I am missing something....

V

#### verizen

##### Guest
origin":wgjkcst9 said:
Yeah about that diagram it was written incorrectly the planet venus is suppose to be aligned with earth and the sun being the observation point...

Really? That seems hard to believe, because the angle would be zero, unless I am missing something....
No your right with your statement, but the way the diagram was displayed is incorrect the earth is supposed to be 90degrees from Venus.

------------------------------ O>Venus
-------------------------------[
-------------------------------[
O>Sun----------------------[90__________O>Earth

Venus is 90degrees in separation from the Earth...Not above the sun

O

#### origin

##### Guest
I am completely confused at this point. I haven't the faintest idea what the question actually is. Good luck...

S

#### SpaceTas

##### Guest
Reading the question, your 1st diagram seems right. Then you can use flat trig to get the angle as per a latter post by aphh.
There is a deliberate red herring about the celestial sphere, but you better say something about why angle you calculate is the same as that on the celestial sphere. Or maybe the person setting the question is just being very precise.

Status
Not open for further replies.

Replies
0
Views
142
Replies
0
Views
173
Replies
0
Views
252
Replies
0
Views
134
Replies
0
Views
257