Proving Elliptical Trajectory Acceleration Vector Passes Through Focus

  • Thread starter Thread starter aim1732
  • Start date Start date
  • Tags Tags
    Trajectory
aim1732
Messages
428
Reaction score
2
If a particle's trajectory is defined by the law x=acos[pt] and y=bsin[pt] where t is parameter of time then we have to prove that it's acceleration vector passes through the focus of the conic----ellipse in this case as can be clearly seen.
If we write out the position vector in the vector notation and differentiate twice we get a=-p2r and this clearly is directed towards the centre of the axes system and not the focus.Any ideas?
 
Physics news on Phys.org
aim1732 said:
If we write out the position vector in the vector notation and differentiate twice we get a=-p2r and this clearly is directed towards the centre of the axes system and not the focus.
Correct.

Any ideas?
Ideas regarding what? I assume you are trying to prove Kepler's first law. The given equation does not describe a planet's motion. You need to find the right equation.
 
have to prove that it's acceleration vector passes through the focus of the conic----ellipse in this case as can be clearly seen.

I meant this.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top