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Keplerian motion

  • Thread starter peter456
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  • #1
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Homework Statement



1. Determine the motion of single test particle (zero mass) falling from infinity as a function
of time passing a unit distance from a central mass. (Zero total energy case of
Keplerian motion.)


Homework Equations


I'm completely stuck


The Attempt at a Solution



I'm completely stuck
 

Answers and Replies

  • #2
HallsofIvy
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First, what do you mean by a zero mass test particle "falling"? If it has 0 mass then it has no kinetic energy, no potential energy, and there is no gravitational force on it. Why would it move at all?
 
  • #3
tiny-tim
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Welcome to PF!

Homework Statement



1. Determine the motion of single test particle (zero mass) falling from infinity as a function
of time passing a unit distance from a central mass. (Zero total energy case of
Keplerian motion.)
Hi peter456! Welcome to PF! :smile:

Do you mean that it has negligible mass (so that it doesn't affect the position of the central mass)?

This will be a hyperbola whose closest point is distance 1 from the central mass.

Kepler's laws apply to hyperbolas as well as ellipses … what are they? :smile:
 
  • #4
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yes it has negligible mass, i just don't understand the question. Apparently i'm supposed to find the equation of the particles position as a function of time.
 
  • #5
tiny-tim
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yes it has negligible mass, i just don't understand the question. Apparently i'm supposed to find the equation of the particles position as a function of time.
Use polar coordinates, for a curve going through (1,0) at t = 0, and with speed 0 at t = ±∞.

And, as I said, what are Kepler's laws? :smile:
 
  • #6
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yeh i know what keplers laws are, i just don't really understnad how to apply them to this case, where do i start? what do you mean use polar coordinates?
 
  • #7
tiny-tim
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  • #8
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r and θ. :smile:
could you please start me off becuase i am completely baffled with this question.
 
  • #9
dynamicsolo
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Does the problem actually require you to use Kepler's Laws? I ask because they won't actually be very useful in this problem. The relation of total mechanical energy of the central mass-test mass system to the semi-major axis of the trajectory requires the vis-viva equation, which is not one of the three laws. The total mechanical energy of the system is zero (as the problem states), so what conic section will the test mass follow?

You will also need to determine the angular momentum of the system for this case. As a central force, gravity will exert no net torque "on the orbit", so that angular momentum is constant. You will need the general definition of angular motion in order to work out the relationship between distance and velocity (angular or linear). What is that definition?
 
  • #10
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Does the problem actually require you to use Kepler's Laws? I ask because they won't actually be very useful in this problem. The relation of total mechanical energy of the central mass-test mass system to the semi-major axis of the trajectory requires the vis-viva equation, which is not one of the three laws. The total mechanical energy of the system is zero (as the problem states), so what conic section will the test mass follow?

You will also need to determine the angular momentum of the system for this case. As a central force, gravity will exert no net torque "on the orbit", so that angular momentum is constant. You will need the general definition of angular motion in order to work out the relationship between distance and velocity (angular or linear). What is that definition?
I'm thinking the orbit will be a parabola because the energy is zero, not too sure. As for the angular momentum i don't know?
 
  • #11
dynamicsolo
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Yes, the zero-energy case has parabolic motion. You will find the polar-coordinate equation for the conic sections helpful. The general definition of angular momentum I am thinking of is

L = r · mv · sin(theta) ,

where r is the distance from the center of the central mass, mv is the linear momentum of the test mass (the central mass is assumed to be stationary), and theta is the angle between the "position" and momentum vectors.

The closest approach that the test mass makes is "one unit", which means that the mass of the central mass and the gravitational constant G are buried in the distance definition. A useful point to consider is that, at closest approach, the velocity or linear momentum vector will be perpendicular to the position vector.
 
  • #12
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Yes, the zero-energy case has parabolic motion. You will find the polar-coordinate equation for the conic sections helpful. The general definition of angular momentum I am thinking of is

L = r · mv · sin(theta) ,

where r is the distance from the center of the central mass, mv is the linear momentum of the test mass (the central mass is assumed to be stationary), and theta is the angle between the "position" and momentum vectors.

The closest approach that the test mass makes is "one unit", which means that the mass of the central mass and the gravitational constant G are buried in the distance definition. A useful point to consider is that, at closest approach, the velocity or linear momentum vector will be perpendicular to the position vector.
i've calculated an equation which describes the motion of the test particle, not sure if it's correct though. 4(1 - x) = y^2, the problem is it's a sideways parabola which doesn't make sense since the particle is released from above with the central mass below it. If this equation was correct how would i go about calculating x and y as a function of time?
 
  • #13
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help anyone?
 
  • #14
tiny-tim
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i've calculated an equation which describes the motion of the test particle, not sure if it's correct though. 4(1 - x) = y^2, the problem is it's a sideways parabola which doesn't make sense since the particle is released from above with the central mass below it. If this equation was correct how would i go about calculating x and y as a function of time?
Hi peter456!

Show us how you got that equation. :smile:
 
  • #15
dynamicsolo
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I think I need to ask at this point what level your course is? "Keplerian motion" is used in this context simply to mean that we are following the motion of a small object following a trajectory in the presence of a (effectively infinitely massive) central body. The trajectory will be a conic section, which is why the reference is made to Kepler.

However, the means needed to solve your problem came about in Newton's time and afterwards, since you need the ideas of mechanical energy and angular momentum, which arose well after Kepler.

You will also find it a lot more convenient to use polar coordinates because we are interested in motion in a gravitational field, which depends on radial distance from the central mass. While the problem can be solved in rectangular coordinates, it is awkward to do so.

You can save some trouble by calling the mass of the test object 1 mass unit, the closest approach to the central mass 1 time unit, and the speed it has at that point 1 velocity unit. (This will also make the constant angular momentum equal to 1 unit.)

You will want the polar equation for a parabola

r = 2a / (1 + cos phi),

where a is the distance of closest approach (here, we can set a = 1), and phi is the angle that the vector from the central mass to the test mass makes to the direction from the central mass to the direction of closest approach.

If you put this together with the equation for mechanical energy conservation, you can find the speed of the test mass as a function of distance from the central body, and thus as a function of the angle phi. What remains is to find the distance and velocity as a function of time.

What I'm not clear on from the problem statement is how much they want: are they asking for position as a function of time, speed as a function of time, the velocity (vector) as a function of time, or all of this?
 
  • #16
dynamicsolo
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OK, the last piece you'll need is the expression for the conservation of angular momentum (which Kepler's Second Law is an expression of, even though the concept didn't exist in 1609...):

L = m·v·(d phi/dt) ,

using the definition for phi in the previous post. For our set of units, this will give us

v · (d phi/dt) = 1 .

The pieces we have now can be used to find a separable differential equation for t as a function of phi (which I think is invertible, though not pretty...). Putting that together with the conservation of energy equation and the equation for the parabolic trajectory will lead to r(t) and v(t).
 
  • #17
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I think I need to ask at this point what level your course is? "Keplerian motion" is used in this context simply to mean that we are following the motion of a small object following a trajectory in the presence of a (effectively infinitely massive) central body. The trajectory will be a conic section, which is why the reference is made to Kepler.
Yeh correct, i tihink i'm suppose to use ideas from Newton, rather than Keplers.

What I'm not clear on from the problem statement is how much they want: are they asking for position as a function of time, speed as a function of time, the velocity (vector) as a function of time, or all of this?
I think they want the position as a function of time, i.e x(t) and y(t).
 
  • #18
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Hi peter456!

Show us how you got that equation. :smile:
Well since we know the total energy is zero, we know the orbit of the test particle will be a parabola. So using we polar coordinates we have:

r = L/[1 + cos(theta)]

plugging, r=1 at theta = 0 (at closest approach, i think, this part i aint sure about) we get L equals 2. Hence:

r = 2/[1 + cos(theta)]

now converting back to cartesians, we get:

4(1 - x) = y^2
 
  • #19
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i really need nelp!!!!
 
  • #20
dynamicsolo
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I've been on the road for three days (with intermittent 'Net access) since this thread started and I thought someone else would have jumped in on this by now. Anyway...

You have the polar equation for position as a function of angle from the direction of closest approach. You know that the total mechanical energy is zero, so, in the units you're using, this tells you that

(1/2)·1·(v^2) - (1/r) = 0 , or v^2 = 2/r .

Conservation of angular momentum tells us that, again in the units we're using,

L = 1·(r^2)·(d theta/dt) = 1

(which is related to the "areal velocity" being constant, which is essentially Kepler's Second Law). [I have to take back one of the equations I wrote earlier.]

So you have

(d theta/dt) = 1/(r^2) = [1 + cos(theta)]^2 / 4 .

This gives us a separable differential equation relating the angle theta to time:

(d theta)/{[1 + cos(theta)]^2} = (1/4) dt .

You would integrate this to get time as a function of theta. This would have to be inverted and put together with the polar equation for r as a function of theta to get the function r(t).
[I didn't say this is pretty; the cases for elliptical or hyperbolic motion look a bit worse...]

This is usually the form in which people write the result. Speed is related to radial distance through the mechanical energy equation. If you want to write the position in Cartesian coordinates, you would have

x(t) = r(t) cos[theta(t)], y(t) = r(t) sin[theta(t)] ,

since we set this up so that the direction of closest approach points along the positive x-axis.
 
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  • #21
14
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This is extremely messy.

o = theta

how the hell do you rearrange for o?

(1/6)tan^2(o/2) + (1/2)tan^2(o/2) = t

this is crazy!!!
 
  • #22
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i made a mistake, it's really:

(1/6)tan^2(o/2) + (1/2)tan(o/2) = t
 
  • #23
tiny-tim
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i made a mistake, it's really:

(1/6)tan^2(o/2) + (1/2)tan(o/2) = t
It's a quadratic equation …

put tan (o/2) = x. Then (1/6)x^2 + (1/2)x - t = 0 …

solve it in the usual way. :smile:
 
  • #24
dynamicsolo
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i made a mistake, it's really:

(1/6)tan^2(o/2) + (1/2)tan(o/2) = t
Well, I finally went and looked it up -- and found that I'd pretty much told you right! And you had the proper insight into what to do next by going over to half-angles to get rid of the '1' term in the denominator of the left-hand side of the differential equation. There is an error, however, in that you should now have

(1/6)tan^3(o/2) + (1/2)tan(o/2) = t . [Check your integration.]

Congratulations! You have now arrived at what is called Barker's Equation!

Maybe a couple of comments would be worthwhile at this point. You are correct in that this is rather messy: analysis of gravitational trajectories generally is! The case of parabolic motion, however, is at least soluble analytically, which is to say that the equations for position and velocity as a function of time can be written down (which is probably why you were given this problem). The general equation for elliptical or hyperbolic motion is not soluble analytically: the results must be found numerically. (There is a whole literature on efficient ways to solve what is called the 'Kepler equation'. I got pointed to the next reference I'm giving you by Peter Colwell's Solving Kepler's Equation Over Three Centuries.)

You can find the rest of the solution of the parabolic motion problem in Archie (A.E.) Roy's Orbital Motion, pages 89-92 in the third edition (1988). Some other sources will also discuss this. You have to solve a cubic equation in (theta/2). Shall I describe the rest or do you want to try looking it up? (It sure doesn't look obvious as far as how to finish this off...)
 
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