Path Calculation of point mass in gravitational field

• VVS
In summary, vanhees71 uses classical mechanics to derive a formula for the path taken of a body in a gravitational field. He uses conservation of energy and conservation of angular momentum to derive a diferential equation describing the motion. He gets stuck when he tries to solve the equation of motion, but he is able to get Kepler's laws for the motion of planet around the sun using just the right coordinates for the symmetry of the problem at hand.
VVS
Hi,

I am trying to derive a formula for the path taken of a body in a Gravitational Field using Classical Mechanics.

I have attached a pdf file with my derivation so far. Basically I used conservation of energy and conservation of angular momentum to derive a diferential equation describing the motion.

Right now I am stuck. I don't know how to proceed.

Attachments

• gravitation.pdf
198.6 KB · Views: 254
Your derivations are all correct, as far as I can see, and it's in fact a very clever trick to use the conservation laws at hand to get the equations of motion.

You'll find your way by just another trick! Use just the right coordinates for the symmetry of the problem at hand and you'll succeed. As you've already used, due to the spherical symmetry of the force, angular momentum is conserved and thus the motion goes entirely in a plane, you've choosen to be the $xy$ plane.

Now it's more convenient to introduce polar coordinates
$$\vec{x}=\begin{pmatrix} x \\ y \end{pmatrix} = r \begin{pmatrix} \cos \varphi \\ \sin \varphi \end{pmatrix}.$$
Plug this into the expressions for the energy and modulus of the angular momentum. With the latter expression you can eliminate $\varphi$ and $\dot{\varphi}$ completely from the energy expression. Then you get an equation of motion for $r(t)$ alone by taking the time derivative of the energy expression again.

One final trick is to consider this equation in terms of $u(\varphi)=\frac{1}{r(\varphi)}$. This leads to an equation for $u(\varphi)$, you can easily solve. The result are Kepler's Laws for the motion of planet around the sun (in the approximation to neglect the center-mass motion of the sun and treat it as fixed central potential).

If you encounter further problems, let us know!

Thank you for the quick reply.

I am going to redo the derivation in polar coordinates again and then I will get back to you.

Hey vanhees71 I redid the derivation as you suggested with polar coordinates. It really does make life easier. I attached it in the pdf file.

Do I have to differentiate the energy equation again wrt to t? Wouldn't that give me terms in r'''?

Moreover I couldn't eliminate theta' and therefore I didn't get an equation only in r, r' an r'' as you suggeste I would.

How can one solve a differential equation like that?

Could you please give me a hint?

Attachments

• gravitation2.pdf
203.6 KB · Views: 280
Wait! The trick is to calculate $E$ and $L$ (which you did right!). So you have
$$L=m r^2 \dot{\theta}=\text{const}$$
Now you use
$$\dot{\theta}=\frac{L}{m r}$$
to elimininate $\dot{\theta}$ from $E$. Then you get
$$E=\frac{m}{2} \dot{r}^2 + \frac{L^2}{m r^2}-\frac{G M m}{r}.$$
In this you substitute
$$r=\frac{1}{u(\theta)}$$
for the trajectory, expressed in polar coordinates, thinking for the angle as a function of time, $\theta(t)$. Then you take the time derivative, which leads to a differential equation for $u(\phi)$.

It is not so easy to get the trajectory as function of time. This leads to Kepler's equation, which is only solvable in implicit form!

I have done the derivation again as you suggested.

I did the substitution, but I think I did something wrong the differential equation looks very complicated.

Attachments

• gravitation3.pdf
207.8 KB · Views: 277
Ok, I guess it's time to solve the problem, but I think you already did a good job!

We start from your equation (3)
$$\frac{\dot{r}^2}{2}+\frac{h^2}{2 r^2} - \frac{G M}{r}=\text{const}.$$
Here $h=L/m$.

First we substitute $r=1/u[\theta(t)]$. Using the chain rule this gives
$$\dot{r}=-\frac{u'}{u^2} \dot{\theta},$$
where the prime refers to differentiation wrt. $\theta$ and the dot to differentiation wrt. time.

Now use the angular-momentum conservation (it's also on your paper):
$$h=\frac{L}{m}=r^2 \dot{\theta}.$$
Subsitute this into the equation above, you get
$$\dot{r}=-\frac{u'}{u^2} \frac{h}{r^2}=-h u'.$$
Put this into the energy-conservation equation above
$$\frac{h^2 u'^2}{2} + \frac{h^2}{2} u^2-G M u=\text{const}.$$
The time derivative of this gives
$$h^2 u' u'' \dot{\theta}+h^2 u u' \dot{\theta}-G M u' \dot{\theta}=0.$$
Using again the angular-momentum conservation equation gives after some algebra
$$u''=\frac{G M}{h^2}-u.$$
This is the equation of an harmonic oscillator subject to an external constant force. The general solution obviously is
$$u(\theta)=\frac{G M}{h^2}+A \cos \theta + B \sin \theta.$$
Now it is the usual convention to choose the polar coordinate system such that $\vartheta=0$ denotes the perihel, i.e., the point, where the planet comes next to the sun. At this point $u=1/r$ should be maximal, i.e., $u'(0)=0 \; \Rightarrow \; B=0$ and (assuming $A>0$) yields
$$u(\theta)=\frac{G M}{h^2} + A \cos \theta.$$
This gives
$$r(\theta)= \frac{1}{G M/h^2 + A \cos \theta}.$$
Put this a bit in a more familiar shape:
$$r(\theta)=\frac{p}{1+\epsilon \cos \theta},$$
where
$$p=\frac{h^2}{G M}, \quad \epsilon=\frac{h^2 A}{G M}.$$
One can show that, if the total energy is negative (bound motion), then $\epsilon<1$. This means that the planet moves along an ellipse with the sun in one of its foci. That's Kepler's first law.

1.Switch to polar coordinates
2.Eliminate the angular coordinate by the conservation of angular momentum
3.Use above mentioned trick u=1/r to get a r(θ) solution.

If you try to solve for r(t) you'll get a non elementary integral, feed it into mathematica if you're still interested.

Hi,

I realized what I needed to do this morning. I have attached the derivation again for completeness till we get the final differential equation. Now I would like to solve this differential equation with initial conditions: position and speed. I will try to figure it out myself and if I can't do it I will ask for help.

I have also derived a differential equation at the end using the time derivative expressions for energy and angular momentum. This DE is in d2/dt2(r) and r. I wonder whether this DE is useful for anything.

Thanks a lot for the support so far, this forum rocks!

Attachments

• gravitation4.pdf
210.2 KB · Views: 302
Hi,

I've done it! Please see the attached pdf. I have a formula for the trajectory as a function of theta and the initial conditions. I rechecked it with my father, there shouldn't be any errors inside.

Thanks for everyone's help!

Attachments

• gravitation4.pdf
214.4 KB · Views: 288
Hello,

I started plugging in numbers and I started to get weird graphs as you can see at the end of my attached file.

What is wrong in my derivation of the Coefficients for the solution of the differential equation?

Attachments

• gravitation4.pdf
320.1 KB · Views: 267

1. What is the path calculation of a point mass in a gravitational field?

The path calculation of a point mass in a gravitational field is the process of determining the trajectory or path that a point mass will follow in the presence of a gravitational force. This calculation takes into account the mass of the point and the strength of the gravitational field.

2. How is the path of a point mass affected by the strength of a gravitational field?

The path of a point mass is affected by the strength of a gravitational field in that the stronger the gravitational field, the greater the force acting on the point mass and the more curved its path will be.

3. What is the difference between path calculation in a uniform and non-uniform gravitational field?

In a uniform gravitational field, the strength of the gravitational force remains constant, resulting in a straight-line path for the point mass. In a non-uniform gravitational field, the strength of the gravitational force varies, causing the path of the point mass to curve.

4. How is the path calculation of a point mass related to Newton's laws of motion?

The path calculation of a point mass is related to Newton's laws of motion in that it follows Newton's second law, which states that the acceleration of an object is directly proportional to the net force acting on it. In the case of a point mass in a gravitational field, the net force is the gravitational force, resulting in a curved path.

5. What factors besides mass and gravitational field strength can affect the path calculation of a point mass?

Other factors that can affect the path calculation of a point mass include the initial velocity and angle of the point mass, as well as the presence of other forces acting on the point mass, such as air resistance or friction. These factors can alter the trajectory of the point mass and must be taken into account in the calculation.

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