Not enough information given?(Momentum Question)

[PolyFX]

Homework Statement

A n object with mass 5e23 kg travels around a star in a nearly circular orbit in the xy plane. Its speed is nearly constant at 4.1e4 m/s.

Diagram accompanying the question
http://img197.imageshack.us/img197/7682/68232428.jpg

It asks to find the momentum at point D and also it asks to find the change in momentum going from D to A.

Homework Equations

change in momentum = pf - pi where pf = final momentum and pi = initial momentum
momentum = (gamma) x (mass) x (velocity)

The Attempt at a Solution

I was stuck right from the beginning because the example in the textbook had the exact same diagram but gave atleast one position vector of any of the points. In this question, a position for the object is not given. I then tried to do trigonometry (long shot?), by making each of the points contained within a right angled triangle. With a 45-90-45 triangle, I assumed that the vector D had a postion vector of (1,0,0) and therefore A had a position vector of (0,-1,0). This is obviously not the answer because I had not even utilized the given information in the question.

Maybe there is enough information given and I am just missing something?

-Thanks in advance.

 

[saunderson]

Hi PolyFX,

my definition of the momentum of a single particle is

$$\vec P = m \, \vec v = m \, \dot{ \vec r}$$​

You wrote

momentum = (gamma) x (mass) x (velocity)

What do you mean with "(gamma)"?

D had a postion vector of (1,0,0) and therefore A had a position vector of (0,-1,0)

I don't think that you mean (1,0,0) is the position vector of D! (1,0,0) is the unit tangent vector of the circle in point D. Maybe you have learned that velocity is also tangential!?

with best regards

 

[PolyFX]

Saunderson, I just wanted to thank you for pointing me in the right direction. I figured out how to do the question today and got the correct answer. Turns out, I had to multiply the mass by the speed to get the momentum at point D (positive x co-ord). The negative momentum of point d was the y co-ordinate for point A. Oh and as for the gamma sign, I use it to stand for the proportionality factor but it turns out I didn't need it for this question because its value was very close to 1.


Either way, I just wanted to thank you for taking your time to help.