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Solutions to newtonian gravity

by llamascience
Tags: gravity, newtonian, solutions
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llamascience
#1
Dec20-07, 11:13 AM
P: 21
I've only just completed high school, but I have a decent grasp on basic undergrad maths/physics. After hours and hours of paper and mental calculations, I still can't see a way of solving Newton's law of gravitation for the path of a body given inital position/velocity i.e. the second order non-linear differential equation r'' = -GM/(r^2).

Can somebody please put my mind at ease by telling me that it cannot be done or showing me how. T'would be much appreciated as I am getting sick of pages that go from Newton's law to conic sections in one jump.
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Parlyne
#2
Dec20-07, 11:30 AM
P: 546
It can be done; but, it's rather long. The first thing you need to keep in mind is that Newton's 2nd law is a vector equation. So, what you should have there is [itex]\frac{d^2\vec{r}}{dt^2}} = \frac{-GM}{r^2}\hat{r}[/itex], where [itex]\vec{r} = r\hat{r}[/itex]. Then, remember that since the radial direction changes with angular coordinates, [itex]\hat{r}[/itex] has a non-zero time derivative. From these considerations and a good choice of angular coordinates, you should be able to separate this down to two equations in r and [itex]\phi[/itex] and a constraint that all motion will be in a plane. From the two equations, you should be able to get a differential equation for r and one that lets you determine [itex]\phi[/itex] once you know r. The r equation can be solved by changing variables to [itex]u = \frac{1}{r}[/itex]. Hope this helps.
nicksauce
#3
Dec20-07, 11:48 AM
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Try an energy approach (for the 1 dimensional case)

[tex](\frac{dx}{dt})^2 = \frac{C}{x}[/tex]

Squareroot both sides, separate the variables and it should be solvable.

llamascience
#4
Dec20-07, 11:56 AM
P: 21
Solutions to newtonian gravity

Wait up there, is that KE = PE??
nicksauce
#5
Dec20-07, 11:59 AM
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Quote Quote by llamascience View Post
Wait up there, is that KE = PE??
Essentially. More Correct would be KE + PE = Constant, but solving for the C in the expression I gave shouldn't be hard.
llamascience
#6
Dec20-07, 12:50 PM
P: 21
Maybe it's 4 in the morning or maybe im just missing the point entirely, but is C supposed to be a constant? Please explain in more detail if possible.
nicksauce
#7
Dec20-07, 01:21 PM
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Yes it is supposed to be a constant.

Suppose your particle starts from rest, at a distance x_0. Then you have

[tex]
\frac{1}{2}mv^2 - \frac{GM_1M_2}{x} = -\frac{GM_1M_2}{x_0}[/tex]

You can then reduce it into the form

[tex]
v^2 = A + \frac{B}{x}
[/tex]

Where A and B are constants, and then solve by separating x and t.
Shooting Star
#8
Dec21-07, 02:49 PM
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Quote Quote by llamascience View Post
Can somebody please put my mind at ease by telling me that it cannot be done or showing me how. T'would be much appreciated as I am getting sick of pages that go from Newton's law to conic sections in one jump.
I cannot do both, since it can be done, and right now I can't show you how. I may be able to do it, say, tomorrow. Would you like to know the most general approach? Are you comfortable with vectors, just the basic differentiation and stuff like that?
llamascience
#9
Dec22-07, 09:47 AM
P: 21
i'd hope so, or i did not deserve a 97% in my calc course :P

if you dont mind doing so, id like a detailed explanation of how to go about this. its been haunting my mind for the past year and all the replies ive been given have been helpful, but unsatisfactory
Shooting Star
#10
Dec22-07, 12:27 PM
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I'll start with very basic stuff, since I don't know what you know and what approach you want. We'll proceed in small steps.

The general eqn of motion would be mr'' = -GMmr/r^3. Can you prove from this that the motion is in a plane, and also that there is a conserved vector, that is, a vector whose magnitude and direction is constant?

For this, you don't require the inverse square law, but any central force, that is, a force of the form f(r)r. Note that r is the magnitude of r.

[Hint (may be ignored): d/dt(rXr') = rXr''.
llamascience
#11
Dec23-07, 11:38 AM
P: 21
yer, the vector product of the position vector and the velocity is conserved and as this is always perpendicular to the velocity, the motion must be in plane
Shooting Star
#12
Dec23-07, 01:58 PM
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(EDIT: I would like a one line justification for your above statement. That is to say, why is the product of the posn vector and the velo conserved?)

Excellent. I'll just elaborate on this a bit. To review what you've said, we got,

d/dt(rXmv) = 0 => rXp = L, where p = mv is the linear momentum, and L is the constant vector. L is called the angular momentum about the origin.

Since r.L = 0, r must be always perp to L, and so r must lie in a plane.

(I've got to run now, but we'll finish off by next one or two posts.)
llamascience
#13
Dec24-07, 09:42 AM
P: 21
aahhh, gotcha. i knew id seen a similar form before, just couldnt put my finger on it

we never covered angular mechanics to any significant detail in TEE physics and for some reason vectors were practically left out of the course. still, i understand the basics from outside reading

JUSTIFICATION: so as you pointed out[tex]\frac{\partial}{\partial t}(\vec{r}\times\dot{\vec{r}}) = \vec{r}\times\ddot{\vec{r}}[/tex]
so [tex]\frac{\partial}{\partial t}(\vec{r}\times\dot{\vec{r}}) = \vec{r}\times\normalfont{f(r)}\hat{\vec{r}} = f(r)\vec{0} = \vec{0}[/tex]
thus the vector [tex]\vec{r}\times\dot{\vec{r}}[/tex] is time-invariant. then, like you say the position and velocity are perp. to this so the motion is planar
D H
#14
Dec24-07, 11:00 AM
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One minor correction here: Those are total derivatives wrt time, not partial derivatives.

Note that angular momentum L is necessarily conserved whenever the magnitude of the force vector is a function of the magnitude of the distance vector only and the force vector is directed along or against the distance vector. This is called a central force problem.

It is better to work in the center of mass frame rather than a frame with the Sun at the origin. Total energy (kinetic + potential), angular momentum, and linear momentum are all conserved in this frame. While linear momentum is tautologically zero in the center of mass frame, that total energy and angular momentum are constants of motion is the key to solving the problem.
Shooting Star
#15
Dec24-07, 12:31 PM
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Hi D H,

Thanks for correcting llamascience on the use of the partial derivatives. This student has just finished high school and I'm trying to satisfy his/her curiosity in the simplest possible way. Right now, I don't think introduction of reduced mass will be very fruitful. Anyway, he has to solve the same equations for that. So, for the present, we'll take the origin at the Sun and the earth to be very light compared to Sun.

Have a look at post #10 for central forces.

(Hi llamascience, be back with you.)
sadhu
#16
Dec24-07, 10:54 PM
P: 158
I hope you now that angular momentum remain conserved during any motion in gravitational field thus
r*m*v=constant=initial given to it
G*mass of earth*massof body/r+0.5 *m*v*v=constant=initial one....
try if this can help
llamascience
#17
Dec24-07, 11:09 PM
P: 21
actually, for the sake of my picking up something new here, could both of you explain your methods simultaneously?

dont worry about my being just out of high school (australia btw, just to avoid confusion). i dont mean to sound over-confident, but like i say; ive read a lot of university level maths and understand a great deal of it, especially in physics related areas like vector calculus, differential equations etc.

still, continue with your method :)
Parlyne
#18
Dec24-07, 11:27 PM
P: 546
Quote Quote by sadhu View Post
I hope you now that angular momentum remain conserved during any motion in gravitational field thus
r*m*v=constant=initial given to it
G*mass of earth*massof body/r+0.5 *m*v*v=constant=initial one....
try if this can help
This is not quite right. The magnitude of angular momentum is [itex]mrv\sin\theta[/itex], where [itex]\theta[/itex] is the angle between [itex]\vec{r}[/itex] and [itex]\vec{v}[/itex]. In the most general case, there is no reason to think that [itex]\vec{r}[/itex] and [itex]\vec{v}[/itex] are always perpendicular.


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