Newtonian Gravitation and stuff

AI Thread Summary
The discussion revolves around calculating the angular momentum of a planet in an elliptical orbit around the sun, with specific maximum and minimum distances from the sun denoted as r1 and r2. Participants clarify the use of the semi-major axis, defined as half the sum of r1 and r2, and its significance in determining velocity using the formula v = √(GM/a). There is uncertainty regarding which radius to use for angular momentum calculations, leading to a focus on the relationship between velocity and the angle between the radius and velocity vectors. The conversation also touches on the conservation of energy and angular momentum as valid methods for solving the problem, with references to geometric properties of the ellipse, such as the latus rectum. Ultimately, the discussion emphasizes the need for a deeper understanding of the relationships involved in elliptical orbits to accurately derive angular momentum.
mooncrater
Messages
215
Reaction score
18

Homework Statement


The question says that :
A planet of mass ##m## moves along an ellipse around the sun so that its maximum and minimum distances from the sun are equal to ##r_1## and ##r_2## respectively. Find the angular momentum ##M## of this planet relative to the centre of the sun.

Homework Equations


##mv^2/r=GMm/ r^2##(I don't know how to use fractions in latex)

The Attempt at a Solution


What I did:
##r_1+r_2=a##=semi major axis of the ellipse
And using the given equation:
##v=√(Gm_s/a)## since the semimajor axis length is used in this formula.
Now to find the angular momentum ##M=mvr##
But I don't know which ##r## iinto be used here. The answer is ##m√(2Gm_sr_1r_2/a)##. So can we say that ##r_{eff}=√2(r_1r_2)##?
 
Physics news on Phys.org
What does 'semi' mean in semimajor?
 
Semi means ##half##.
So semimajor axis=half the length of major axis.
 
mooncrater said:
Semi means ##half##.
So semimajor axis=half the length of major axis.
Right. How does that affect your equations?
 
  • Like
Likes mooncrater
mooncrater said:

Homework Statement


The question says that :
A planet of mass ##m## moves along an ellipse around the sun so that its maximum and minimum distances from the sun are equal to ##r_1## and ##r_2## respectively. Find the angular momentum ##M## of this planet relative to the centre of the sun.

Homework Equations


##mv^2/r=GMm/ r^2##(I don't know how to use fractions in latex)
For fractions you can use: \frac{numerator}{denominator}

The Attempt at a Solution


What I did:
##r_1+r_2=a##=semi major axis of the ellipse
And using the given equation:
##v=√(Gm_s/a)## since the semimajor axis length is used in this formula.
Now to find the angular momentum ##M=mvr##
But I don't know which ##r## iinto be used here. The answer is ##m√(2Gm_sr_1r_2/a)##. So can we say that ##r_{eff}=√2(r_1r_2)##?
Hmm. What conditions must be satisfied in order for the expression mvr to yield the angular momentum? hint: r and v are the magnitudes of two vectors. I'm dubious about the validity of using the velocity when the radius is equal to the semimajor axis.
 
  • Like
Likes mooncrater
haruspex said:
Right. How does that affect your equations?
Actually I saw in resnick walker, that for a circular orbit the ##r## put in the equation is the radius of the path. Whereas for an elliptical path it's semimajor length is put. I don't know the exact reason why it is done.
gneill said:
For fractions you can use: \frac{numerator}{denominator}

Hmm. What conditions must be satisfied in order for the expression mvr to yield the angular momentum? hint: r and v are the magnitudes of two vectors. I'm dubious about the validity of using the velocity when the radius is equal to the semimajor axis.
Thanks now I know how to write those fractions. I think you're trying to give me a hint to solve the question by energy conservation and angular momentum conservation. If so, then I should add that I know ##that## method. The problem occurring is that if I use the equation given in the question then is there something like ##r_{eff}## for an elliptical path? If so then does it have a value ##√2r_1r_2## where ##r_1## and ##r_2## are the shortest and longest distances of the body from the sun.
 
mooncrater said:
Actually I saw in resnick walker, that for a circular orbit the ##r## put in the equation is the radius of the path. Whereas for an elliptical path it's semimajor length is put. I don't know the exact reason why it is done.
I'm trying to draw your attention to your equation r1+r2=a= semimajor axis.
 
  • Like
Likes mooncrater
mooncrater said:
Actually I saw in resnick walker, that for a circular orbit the ##r## put in the equation is the radius of the path. Whereas for an elliptical path it's semimajor length is put. I don't know the exact reason why it is done.
Me neither. It doesn't make sense to me unless it was a particular case of a circular orbit where the semimajor axis is identical to the radius.
Thanks now I know how to write those fractions. I think you're trying to give me a hint to solve the question by energy conservation and angular momentum conservation. If so, then I should add that I know ##that## method. The problem occurring is that if I use the equation given in the question then is there something like ##r_{eff}## for an elliptical path? If so then does it have a value ##√2r_1r_2## where ##r_1## and ##r_2## are the shortest and longest distances of the body from the sun.
The only thing that springs to mind as a possibility would be the geometric mean of the radii, and that's just a thought without investigation.

The formula for the velocity with respect to distance is derived from energy conservation, making use of the total mechanical energy of the orbit. Granted, the velocity at that instant is given by ##\sqrt{\frac{GM}{a}}## as you state, which is a nice neat expression, but I can't see where this is helpful without knowing the flight path angle at that instant (so you can account for the non right angle between the velocity and radius vectors when calculating the angular momentum).
 
  • Like
Likes mooncrater
haruspex said:
I'm trying to draw your attention to your equation r1+r2=a= semimajor axis.
Oops... that was a typo... I didn't notice it... it is actually
##r_1+r_2=2a##
 
  • #10
gneill said:
The only thing that springs to mind as a possibility would be the geometric mean of the radii, and that's just a thought without investigation.

The formula for the velocity with respect to distance is derived from energy conservation, making use of the total mechanical energy of the orbit. Granted, the velocity at that instant is given by ##\sqrt{\frac{GM}{a}}## as you state, which is a nice neat expression, but I can't see where this is helpful without knowing the flight path angle at that instant (so you can account for the non right angle between the velocity and radius vectors when calculating the angular momentum).
I think what you mean to say is that though we have calculated the velocity of the body but we don't know the difference in angle between the radial vector and the velocity vector which is needed to find the angular momentum of the body with respect to sun. Therefore using my method to solve this problem is absurd due to which the only remaining way to solve it is the conservation of energy and angular momentum. Am I correct here?
 
  • #11
mooncrater said:
I think what you mean to say is that though we have calculated the velocity of the body but we don't know the difference in angle between the radial vector and the velocity vector which is needed to find the angular momentum of the body with respect to sun. Therefore using my method to solve this problem is absurd due to which the only remaining way to solve it is the conservation of energy and angular momentum. Am I correct here?
Well, I wouldn't go so far as to say absurd. It may be possible that there's some simple relationship for the flight angle when the radius vector magnitude is equal to the semimajor axis. But I'm not aware of such. There are other approaches to the problem involving the geometry of the ellipse (one I know of involving the latus rectum). But the conservation of of energy approach is perfectly serviceable and relatively straightforward.
 
  • Like
Likes mooncrater
  • #12
gneill said:
The only thing that springs to mind as a possibility would be the geometric mean of the radii, and that's just a thought without investigation.

gneill said:
There are other approaches to the problem involving the geometry of the ellipse (one I know of involving the latus rectum).

It just occurred to me that the Latus Rectum of the ellipse is in fact the harmonic mean of the peri and apo apses. That is, in this case:

##p = 2 \frac{r_1 r_2}{r_1 + r_2} = \frac{r_1 r_2}{a}##

So, must have been a long-unused little gray cell that got tickled when I thought of the geometric mean possibility; Right train of thought but wrong mean! I won't give away the relationship between the angular momentum and p (since that would be giving away too much), but you should be able to work it out easily once you've obtained the answer by another method.
 
  • Like
Likes mooncrater
Back
Top