Understanding Hyperbolic Orbits: The Relationship Between Foci and Trajectory

Click For Summary
The discussion centers on understanding hyperbolic orbits, specifically the placement of the orbited body and the trajectory equation. Participants clarify that if the Earth is at one focus, a comet's path can be described by the equation r - P = -2a. The validity of using r - P = 2a depends on the focus from which distances are measured. It's emphasized that the definitions of r and P are crucial in determining the correct equation for hyperbolic trajectories. Overall, the relationship between the foci and trajectory equations is key to understanding hyperbolic orbits.
WWCY
Messages
476
Reaction score
15

Homework Statement


Screen Shot 2017-11-26 at 3.46.21 AM.png


I'm rather confused as to where the orbited body is placed, and where the orbiting trajectory lies in this figure.

Is it right for me to say that if I placed say, the Earth on the left hand focus, a comet with a hyperbolic trajectory will travel a path defined by the equation ##r - P = -2a##?

Thanks in advance.

Homework Equations

The Attempt at a Solution

 

Attachments

  • Screen Shot 2017-11-26 at 3.46.21 AM.png
    Screen Shot 2017-11-26 at 3.46.21 AM.png
    8.4 KB · Views: 1,227
Physics news on Phys.org
WWCY said:

Homework Statement


View attachment 215615

I'm rather confused as to where the orbited body is placed, and where the orbiting trajectory lies in this figure.

Is it right for me to say that if I placed say, the Earth on the left hand focus, a comet with a hyperbolic trajectory will travel a path defined by the equation ##r - P = -2a##?

Thanks in advance.

Homework Equations

The Attempt at a Solution

Looks right.
 
haruspex said:
Looks right.

Hi, thanks for clearing that up!

Edit: Is r-P = 2a also a valid hyperbolic orbit?
 
WWCY said:
Hi, thanks for clearing that up!

Edit: Is r-P = 2a also a valid hyperbolic orbit?
Not if the sun is at the negative focus.
 
haruspex said:
Not if the sun is at the negative focus.

But if I do want to use r - P = 2a, and place my planet on the right hand focus, wouldn't I need to redefine r to originate from the right focus and P from the left? This gives r - P = -2a again.

Thank you
 
WWCY said:
But if I do want to use r - P = 2a, and place my planet on the right hand focus, wouldn't I need to redefine r to originate from the right focus and P from the left? This gives r - P = -2a again.

Thank you
It depends how you are defining r and P.
If you define r as distance from the left focus and P that from the right focus then for the gravitational source at the left focus it is r-P=-2a; for the gravitational source at the right focus it's r-P=2a.
But if you define r as distance from gravitational source and P as distance from the other focus then it is r-P=-2a whichever side the source is.
 
haruspex said:
It depends how you are defining r and P.
If you define r as distance from the left focus and P that from the right focus then for the gravitational source at the left focus it is r-P=-2a; for the gravitational source at the right focus it's r-P=2a.
But if you define r as distance from gravitational source and P as distance from the other focus then it is r-P=-2a whichever side the source is.

Thanks!
 

Similar threads

Replies
7
Views
2K
Replies
4
Views
4K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 86 ·
3
Replies
86
Views
8K
Replies
22
Views
4K