Newtonian Gravitation and stuff

In summary: 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.
  • #1
mooncrater
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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)##?
 
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  • #2
What does 'semi' mean in semimajor?
 
  • #3
Semi means ##half##.
So semimajor axis=half the length of major axis.
 
  • #4
mooncrater said:
Semi means ##half##.
So semimajor axis=half the length of major axis.
Right. How does that affect your equations?
 
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  • #5
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.
 
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  • #6
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.
 
  • #7
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.
 
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  • #8
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).
 
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  • #9
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.
 
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  • #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.
 
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What is Newtonian gravitation?

Newtonian gravitation is a theory proposed by Sir Isaac Newton in the 17th century to explain the force of gravity between objects. It states that every object in the universe attracts every other object with a force that is directly proportional to their masses and inversely proportional to the square of the distance between them.

How does Newtonian gravitation differ from Einstein's theory of general relativity?

Newtonian gravitation is a classical theory that only works for objects moving at low speeds and in weak gravitational fields. Einstein's theory of general relativity is a more comprehensive and accurate theory that takes into account the effects of high speeds and strong gravitational fields.

What is the equation for calculating the force of gravity between two objects?

The equation is F = G * (m1 * m2) / r^2, where F is the force of gravity, G is the gravitational constant, m1 and m2 are the masses of the two objects, and r is the distance between them.

Can Newtonian gravitation be applied to celestial bodies?

Yes, Newtonian gravitation is commonly used to calculate the orbits of planets, moons, and other celestial bodies in our solar system. However, for more accurate predictions, the theory of general relativity is needed.

What are some real-world applications of Newtonian gravitation?

Newtonian gravitation is used in many fields, such as aerospace engineering, astronomy, and navigation. It is also the basis for the laws of motion and has played a crucial role in the development of modern physics and technology.

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