Angle of space motion and radial-velocity vector

Sep 2018
7
0
Hello
So I am trying to figure out the angle of a space motion vector with its radial velocity vector. I have to sample equations, but despite plugging in the numbers I am not getting the same answers.

So I have:
tan-1((4.74d(u")/(vr))

With:
d = 5.14 pc
u" = 0.66092"
vr = -26.3km s-1

so:
tan-1((4.74x5.14(0.66092)/-26.3)

= 1.1779

But the answer is supposed to -31.5

I tried convering parsecs and kilometers to meters, but still didn't get the right answer.

What am I missing?

Also, I am not entirely sure what the 4.74 number is, I can't find it in my text book.

Thankyou in advance!
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
It looks like you expressed your answer in radians, whereas the "correct " answer is in degrees.
 
Sep 2018
7
0
Converting the anser to degrees would make it 67.488,

Not -31.5
 

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
4.74 x 5.14 x 0.66092 /( -26.3) = -0.612

tan-1(-0.612)= -31.5 degrees
 
Sep 2018
7
0
So I have caught up with getting to -0.612

but when I do:
tan-1(-0.612)

The answer I am getting is +0.01068
When I convert that to degrees I get:
+0.611

I am still not sure what I am doing wrong?
 
Last edited:

ChipB

PHF Helper
Jun 2010
2,361
289
Morristown, NJ USA
but when I do:
tan-1(-0.612)

The answer I am getting is +0.01068
Remember that the notation

$\tan^{-1} (x)$

means arctangent of x. What you have apparently been calculating is the tangent of 0.612 degrees, but what you need to do instead is calculate the arctangent of 0.612, and express the answer in degrees.