- #1

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter Physics_Boi
- Start date

- #1

- #2

Dewgale

- 100

- 9

- #3

Physics_Boi

- 20

- 2

- #4

Dewgale

- 100

- 9

- #5

Physics_Boi

- 20

- 2

- #6

gneill

Mentor

- 20,947

- 2,892

Angular momentum is not going to help much here because at the outset you don't know the angle that the velocity vector makes with the radius vector at point P. You might get there if you can work out the eccentricity and find the slope of the ellipse at the latus rectum.

- #7

Dewgale

- 100

- 9

Gneill, is it not the case that, even after finding the tangent line to the ellipse at that point, one would need to use conservation of angular momentum to find the velocity at a later time? As such, I'm not sure saying it won't help is really accurate. There's just an intermediary step before finding the initial angular momentum.

- #8

Physics_Boi

- 20

- 2

Ok thank you for the help!

- #9

Physics_Boi

- 20

- 2

- #10

Dewgale

- 100

- 9

- #11

Physics_Boi

- 20

- 2

- #12

Dewgale

- 100

- 9

##T## is not time in Kepler's Laws -- it's period, i.e. the amount of time to complete a full orbit. It's inversely proportional to the

- #13

Physics_Boi

- 20

- 2

Then what concepts would I have to use? Is there any specific idea that would be useful?

- #14

Dewgale

- 100

- 9

I assume you haven't learned about cross product yet, but you should know that for a point particle, the angular momentum is given by ##L = m v r \sin(\theta)##, where m is the mass, v is the speed, r is the distance from the origin (in this case the sun), and ##\theta## is the angle between the velocity and the line pointing from the sun to the planet. As was previously said, to use angular momentum, you need to find that angle for your initial position - it's obviously 90 degrees at the end.

- #15

gneill

Mentor

- 20,947

- 2,892

If you can get the eccentricity and then find the slope at the latus rectum, I suppose you could go that route. It seems cumbersome to me. (although I'm absolutely willing to be shown otherwise).Gneill, is it not the case that, even after finding the tangent line to the ellipse at that point, one would need to use conservation of angular momentum to find the velocity at a later time? As such, I'm not sure saying it won't help is really accurate. There's just an intermediary step before finding the initial angular momentum.

There are relationships between the perihelion radius, the semi latus rectum, the eccentricity, and the semi major axis that would be a more direct route. Just add the expression for the speed at a given radius (knowing the semi major axis) and you're there.

- #16

Physics_Boi

- 20

- 2

- #17

Dewgale

- 100

- 9

- #18

Physics_Boi

- 20

- 2

- #19

gneill

Mentor

- 20,947

- 2,892

I wouldn't expect high school physics students to have compiled a suitable list of relevant equations, yet I would expect that from first year college students majoring in an astrophysics related course of study.

So if I may be so bold, may I enquire as to what course this problem has arisen?

- #20

Dewgale

- 100

- 9

https://www.mathopenref.com/ellipsetangent.html

- #21

Physics_Boi

- 20

- 2

- #22

Physics_Boi

- 20

- 2

- #23

Dewgale

- 100

- 9

- #24

Physics_Boi

- 20

- 2

- #25

Dewgale

- 100

- 9

Your angle you derived is off - you want the conjugate angle to what you got. Remember, you want the angle betwen the Latus Rectum and the velocity, which you can see is not the angle in the top left of your right-angle triangle, but rather the bottom right (because it's 90-##\alpha##, where ##\alpha## is what you got.)

- #26

Physics_Boi

- 20

- 2

- #27

Physics_Boi

- 20

- 2

- #28

Dewgale

- 100

- 9

Edit: You're making some calculation error. Taking 0.75a for the semilatus rectum, you can show using conservation of momentum in a fairly straightforward way that ##v_2 = (3/\sqrt{5})v_1##.

- #29

Physics_Boi

- 20

- 2

Ok I will try it and let you know. Thank you for the help!

- #30

gneill

Mentor

- 20,947

- 2,892

No idea what you actually did there. How did you find the slope of the ellipse? And with respect to what? You need to post details of your work if we're to help.

Consider the following:

##r_p = \frac{p}{1 + e} = a(1 - e)##

These are basic relationships that would be handy to know if you're doing astrophysics.

The above relates the perihelion distance (##r_p##) to the semi latus rectum (##p##), the eccentricity (##e##), and the semi major axis (##a##) .

In the problem statement you are given ##r_p## in terms of ##a## (##r_p = 0.5 a##). So with the above relationships you have immediate access to ##e## since we know ##r_p## in terms of ##a## from the problem statement.

Plug in the givens and you find that

##e = \frac{1}{2}##

From there you find p, the semi latus rectum. ##p = \frac{3}{4}a## . I'm happy to report that it's the same value that you found.

Consider also that the speed of a body on orbit is given by

##v = \sqrt{\mu\left(\frac{2}{r} - \frac{1}{a}\right)}##

Where ##\mu## is the gravitational constant of the primary (the star). It'll drop out in further calculations.

So now you have a speed and radius for one point on the orbit (##v_1## at the latus rectum). You also have the radius of the perihelion, A little algebra should bring you to the desired conclusion.

- #31

Dewgale

- 100

- 9

- #32

Physics_Boi

- 20

- 2

Share:

- Last Post

- Replies
- 7

- Views
- 683

- Replies
- 5

- Views
- 294

- Replies
- 6

- Views
- 246

- Replies
- 5

- Views
- 539

- Last Post

- Replies
- 3

- Views
- 515

- Last Post

- Replies
- 12

- Views
- 633

- Replies
- 18

- Views
- 627

- Replies
- 9

- Views
- 740

- Replies
- 3

- Views
- 657

- Replies
- 25

- Views
- 432