Position vector relation between comet and earth

1. Nov 24, 2007

karnten07

1. The problem statement, all variables and given/known data

There is a comet that is at a position between the earth and the sun that has an orbital radius half that of earths, this is the perihelion position for the comet. The comet's and earth's orbits are coplanar. The comet's orbit crosses that of the earth's. When the comet is placed so that it is just crossing the orbit of earth, so at twice the radius it was originally at, how does its position vector change, r?

I assume the origin to be at the center of the sun, so does this just make the position vector twice what it was??

2. Relevant equations

3. The attempt at a solution

2. Nov 24, 2007

Shooting Star

The magnitude of the posn vector when it crosses earth's orbit is twice that of at perhilion. But posn vector has direction as well. Was any other info given?

3. Nov 24, 2007

karnten07

The question says that when the comet is at perihelion it has velocity 2Ve where Ve is the velocity of the earth. I am asked to find the velocity of the comet when it crosses the orbit of earth in terms of Ve. I am trying to show that the angular momentum of the comet at each position is the same, ie. conserved. So i have written each position in terms of moment of inertia x angular velocity and the only component i was unsure about was the position vector and how it differed between the two positions. I see that the magnitude is half that of earths when it is at perihelion, but if we assumed the positions of the earth and comet were in the same position, they would have the same position vector. Obviously this wouldnt happen (at least hopefully, for our sake), but it shouldnt make any difference to the calculation as in this question i don't think i'm given enough information to consider the vector direction.
So in answer to my question, should i just say that the position vector of the comet is twice the value of when it is at perihelion?

4. Nov 24, 2007

karnten07

Oh wait, im then asked to find the angle at which the orbit of the comet crosses that of earth's so i guess i will have to consider the position vector.

The angular momentum i have written as so:

L=Iw
L=mR^2 x (r x v)/[r]^2

I have used the square brackets to indicate that it is the magnitude of r.

So for the comet at perihelion it has L:

=mR^2 x (rc x 2Ve)/[r]^2
where rc is the position vector of the comet at perihelion

For the comet at earth's orbital distance it has L:

=m(2R)^2 x (rce x V)/[2r]^2

where rce is the position vector of the comet when it crosses earths orbit. V is the velocity of the comet as it crosses earths orbit.

These two equations should be equal.

I am going to carry out some calculations according to what i have writtern here, if anyone thinks im going about this the wrong way or sees an error in my equations, corrections would be great. Thanks

5. Nov 24, 2007

karnten07

So i get my equations after cancelling terms as:

rce x Vce = r x 2Ve

Vce is the velocity of the comet as it crosses earths orbit.

If the magnitude of rce is twice that or r, then Vce will equal Ve as it crosses earths's orbit. I have reread the question and it asks for the comets SPEED so i have doen that now. now i have to find the angle at which it crosses earth's orbit. Should i assume that it is the angle from the comets position at perihelion to that at crossing earth's orbit?

Last edited: Nov 24, 2007
6. Nov 24, 2007

Shooting Star

Keep up the effort.

Angular momentum L of comet is m*2Ve*r, where r is radius of earth's orbit, because at perhelion, the velo is perp to the posn vector. Also, L is const in magnitude. You know the energy at the new position in terms of the energy at perhelion. You have to use all this infmn. Try and see up to where can you get.

7. Nov 24, 2007

karnten07

Do you mean L of the comet is m*Ve*r where r is the radius of earth's orbit, because at this point i have calculated the comets velocity to be the same as earths?

Or do you mean m*Ve*r/2 as L of the comet at perihelion?

Last edited: Nov 24, 2007
8. Nov 24, 2007

Shooting Star

A typo -- should be r/2.

9. Nov 24, 2007

karnten07

Okay, i think i'm getting somewhere now. I have L = r xmv

so re/2 * 2Ve *sin 90 = re* Ve *sin theta

So when cancelling it shows that sin theta = 1 so theta = 90

Im not sure if this is right though because the comet isn't travelling in a circle, so perhaps i have calculated the speed of the comet as it crosses earth's orbit incorrectly.

Last edited: Nov 24, 2007
10. Nov 24, 2007

karnten07

I have an equivalence equation of:

2Ve n = Vx sin theta n

where Vx is the speed of the comet at crossing the earths orbit and n is the unit vector. My previous answer for Vx i think was incorrect.
What energy equation do i use to calculate the speed and angle?

11. Nov 24, 2007

Shooting Star

Yes, on RHS, it's not Ve, but the proper v you can calculate by considering that total energy is const. Also, theta is the angle between r and v, not the theta of the posn vector.

Tell me, in which chapter were you given this. I'm asking so that I can help you appropriately.

12. Nov 24, 2007

Shooting Star

Also, whether you are familiar with conic sections, like ellipses, in polar co-ordinates?

13. Nov 24, 2007

karnten07

Yes i see that it is incorrect now. I also see that theta is the angle between r and v. This question relates to the chapter in my notes called gravitation and keplers laws. This chapter came just after a chapter on rotational motion of rigid bodies.

14. Nov 24, 2007

karnten07

There is a mention of conics and ellipses here in my notes. I think i may need to do more reading around about this subject. So i might come back to this question later on. Thanks for your help and if you have any further insights please do tell.

15. Nov 24, 2007

Shooting Star

The path of the comet is an ellipse with e=1/2. We can find the pt of intersection with the circular orbit of earth. But you have taken a very good decision to read up on it before proceeding further. You can come back to it well prepared and you'll get all the help.