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Title: Minimum energy problemPost by: Fu-Kwun Hwang on February 22, 2010, 02:36:55 pm
Assume a partciel A is moving in a confined circular orbit (with radius a), another particle B is located away from the center (at r=b).
And there is a gravitation field between particles $\vec{g}=\frac{k}{r^2}\hat{r}$ where $\hat{r}$ is the unit vector between those two particles. What is the minimum initial velocity for particle a to circular the orbit? Because the field is $\vec{g}=\frac{k}{r^2}\hat{r}$ so the potential energy is $V(r)=\frac{-k}{r}$ From consevation of energy $\frac{1}{2}mv^2-\frac{mk}{a-b}=\frac{1}{2}mu^2-\frac{mk}{a+b}$ $v^2-u^2\ge \frac{2k}{a-b}-\frac{2k}{a+b}=\frac{4kb}{a^2-b^2}$ So the minimum velocity is $v=\sqrt{\frac{4kb}{a^2-b^2}}$ What if we want the particle always touch the inner surface at r=a (i.e. the Normal force provided by the circular orbit is always pointing into the center of the circle) Do you know how to solve it? The following is the simulation for you to play with. You can drag particle B ( to change b) Mode: 1. N out the normal force is always pointing away from the center of the circle and v is the minimum velocity to reach another end.2. N inthe normal force is always pointing into the center of the circle and v is the minimum velocity3. You can change cst to change the velocity ratio. The red arrow is the velocity of the particle a (You can drag the arrow to change velocity) The blue arrow is the gravitation force between particle A and B. The magenta arrow is the required Centripetal force. Another black arrow is the normal force supplied by the circular orbit. The kinetic energy/potential energy and total energy as function of angle are drawn as red/blue and green curves. |