Problem w/ Gravitation, Orbits

  • #1
Physics_Boi
20
2

Homework Statement


Given in the picture attached below[/B]


Homework Equations


?

The Attempt at a Solution


I had no idea how to tackle this problem.[/B]
 

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Answers and Replies

  • #2
Dewgale
100
9
Have you learned about angular momentum, and the conservation of angular momentum? If so, what can this tell you about the problem?
 
  • #3
Physics_Boi
20
2
Wouldn’t the angular momentum be changing because there is a gravitational force acting up on the object?
 
  • #4
Dewgale
100
9
Remember, it's a torque that changes angular momentum, not a force. Is there any torque being applied in this scenario?
 
  • #5
Physics_Boi
20
2
Now that I think about it, the gravitational force acts between the center of masses, so there wouldn’t be a torque. Does that mean I would calculate the moment of inertia and conserve angular momentum to find the velocity?
 
  • #6
gneill
Mentor
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You really need to show relevant equations. In particular I'd concentrate on the relationship between the semi latus rectum (which is where point P sits), the eccentricity, and the perihelion distance. After that you need to look through your collection of equations for one that gives you the orbit velocity at a given distance.

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
Yep! Though you can probably just approximate the planet as a point.

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
I tried using Keplers law, but I got a different answer. First, using the characteristics of the ellipse, I determined that the length of the semi minor axis is a * sqrt(0.75). Using the Latus Rectum relationship, I found the distance from the focus and the point P to be 0.75a. By using Keplers Law (T^2 is proportional to R^3), I found that V^2 is inversely proportional to R^3. After substituting values: (v/v1)^2 = (0.75/0.5)^3, I got that v = (3/2)^(3/2) * v1. Can you explain to me what I am doing wrong?
 
  • #10
Dewgale
100
9
How did you find v^2 to be inversely proportional to R^3? Kepler's law probably isn't going to be of much use in this case.
 
  • #11
Physics_Boi
20
2
I thought that V was inversely proportional to time. This led me to substitute (1/V) as T into Kepler’s Law.
 
  • #12
Dewgale
100
9
I thought that V was inversely proportional to time. This led me to substitute (1/V) as T into Kepler’s Law.

##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 average speed over the course of the orbit, which doesn't help you when analyzing a given portion of the orbit.
 
  • #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
Like I said earlier, you need Conservation of Angular Momentum

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
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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.
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).

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
As for finding the angle theta, wouldn’t that always be 90 degrees because the velocity is tangent to the ellipse? Also, I am not familiar with eccentricity or perihelion radius, so would there be any way to solve this?
 
  • #17
Dewgale
100
9
Theta is the angle between the velocity and the line from the sun to the planet, not from the center of the ellipse to the planet. So it's not always 90 degrees - one can clearly see at the first point that it is a different value.
 
  • #18
Physics_Boi
20
2
I looked at the picture and the distances, but I don’t understand how I would be able to calculate the value of the angle.
 
  • #19
gneill
Mentor
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I think, at this point, we need to inquire about what level of course this question is coming from so that we can tailor our support accordingly.

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?
 
  • #21
Physics_Boi
20
2
I am preparing for the U.S. Physics Olympiad exam currently. This is one of the problems from the F=ma exam, which is the preliminary test for USPHAO. https://www.aapt.org/physicsteam/2018/
 
  • #22
Physics_Boi
20
2
I found the angle of the velocity to be 63.5 degrees, but when I solved using angular momentum I got a different answer. What I did to find the angle was set up the right triangle (between the two foci and the point), and use inverse tangent to find one of the angles, and use basic addition and manipulation of angles to figure out the rest.
 
  • #23
Dewgale
100
9
I just realized that you aren't given the distance from the sun to P (I thought you had been, my apologies) - so I'm not sure how you found the angle without finding that distance. Did you derive the distance?
 
  • #24
Physics_Boi
20
2
Yes, I used the relationship a^2 - b^2 = c^2 for an ellipse, where a is semi major axis length, b is semi minor axis length, and c is focal length. I knew a and c, so I substituted and just solved for b^2. Using the Latus Rectum properties, b^2/a is the y-coordinate.
 
  • #25
Dewgale
100
9
Yes, I used the relationship a^2 - b^2 = c^2 for an ellipse, where a is semi major axis length, b is semi minor axis length, and c is focal length. I knew a and c, so I substituted and just solved for b^2. Using the Latus Rectum properties, b^2/a is the y-coordinate.

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
That was the conjugate angle. The angle in the triangle I got was a little more than 50 degrees. I then used the fact that the other two angles were equal, and that all the angles added up to 180 degrees (since they form a line) to get the angle of 63.5 degrees.
 
  • #27
Physics_Boi
20
2
What I did was that I considered the top angle of the triangle. Theta = arctan (a/0.75a) = 53.1 degrees. Then, the remaining two angles must add up to 180 - 53.1 = 126.9 degrees. Since they are the same, 126.9/2 = 63.4 degrees. Am I doing something wrong?
 
  • #28
Dewgale
100
9
Apologies, I wasn't thinking of the triangle correctly. As far as I can tell you're doing things correctly, but remember, you want half of the latus rectum, not the whole latus rectum - which is what that formula is for, so never mind about that. It still doesn't give the right answer, so I'll think about this a bit more. Please let us know if you figure it out.

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
Okay, so K-12 level. That helps.

I found the angle of the velocity to be 63.5 degrees, but when I solved using angular momentum I got a different answer. What I did to find the angle was set up the right triangle (between the two foci and the point), and use inverse tangent to find one of the angles, and use basic addition and manipulation of angles to figure out the rest.
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.
 
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  • #31
Dewgale
100
9
Gneill, he was working from the geometric fact that if you draw a line from the two focii to a point on an ellipse, the angle between the tangent at that point and each line is the same, following from the link I'd posted in post #20. One can then easily derive the angle between the position vector and the velocity if one knows the length of the semilatus rectum.
 
  • #32
Physics_Boi
20
2
Ok thank you for the help! I plugged in the values into the equation, and solved. I got the correct answer.
 

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