0
$\begingroup$

Imagine a $x$-$y$ plane, which induces an acceleration field obeyed by the equation $ a= -\frac{C}{z^2} \vec{k}$ where $z$ is the perpendicular distance from the plane to the ball and $C$ is just some constant. Now a ball is thrown from a height of 1 meter from the plane, find trajectory of the ball for an oblique projection.

This is a hypothetical question why my friend made-up. I could solve for an exact solution of differential equation probably but I know I cannot approximate the acceleration as something like $-mg$ because the denominator is starting from $z=0$. If we are on earth, the gravitational law takes the form $ \frac{ GmM}{(R+x)^2}$ and because of the form of the denominator, we can use taylor expansion approximation and solve for acceleration close to the surface.

However we can't do that because we can not taylor expand $ -\frac{1}{x^2}$ around $x=0$, because neither the function nor it's derivatives are defined. Now, how would I find the trajectory considering all of that?

Also, another observation which I made is horizontal component once it's projected would be effected since there is no horizontal force.

On deeper analysis, I realize this is a problem with any point like particle. The reason we don't encounter it with earth, is because the 'bulk' of the earth puts a limit on how close we can get to the earth, so how would we deal with these singularities super dense objects (i.e: point masses?)

Edit:

Is the radial two-body problem solvable?

This stack question goes into detail about solving the differential equation and interpreting solutions to it. I thought this was good, if someone could show how can we interpret the mathematical solution of the differential equation (maybe with solving differential equation) and apply it in this context then it'd be great. I am not so well versed with solving second order differential equations or interpreting the solution , so I hope someone more knowledgeable can help in this aspect

$\endgroup$
5
  • $\begingroup$ A plane doesn’t produce this kind of field. $\endgroup$
    – G. Smith
    Jul 11 '20 at 17:52
  • $\begingroup$ It's a purely hypothetical question with consequences in real life as well $\endgroup$
    – 666User666
    Jul 11 '20 at 19:47
  • $\begingroup$ If you are interested in how two point masses fall directly toward each other as the force between them becomes infinite, there are numerous questions and answers on this site addressing that calculation. $\endgroup$
    – G. Smith
    Jul 11 '20 at 20:25
  • $\begingroup$ I guess that would be a start for trying this question. Can you suggest some? $\endgroup$
    – 666User666
    Jul 11 '20 at 20:26
  • $\begingroup$ physics.stackexchange.com/questions/515402/… $\endgroup$
    – G. Smith
    Jul 11 '20 at 20:32
2
$\begingroup$

Consider radial infall of two point masses, with the equation of motion

$$\ddot{r}=-\frac{1}{r^2}$$

where we ignore constants like $G$, masses, etc. (Note that the equation in the question you linked to was missing a minus sign. Gravity is attractive.)

The trick to solving this second-order differential equation is to convert it into a first-order one by multiplying both sides by $\dot{r}$:

$$\dot{r}\ddot{r}=-\frac{\dot{r}}{r^2}$$

$$\frac{d}{dt}\left(\frac12\dot{r}^2\right)=\frac{d}{dt}\left(\frac1r\right)$$

$$\frac12\dot{r}^2-\frac1r=\text{const}$$

You can recognize the final equation as simply the conservation of kinetic energy plus gravitational potential energy. The latter is negative because the force is attractive.

You can solve this simpler equation by first separating the variables $r$ and $t$ and then integrating both sides. An especially easy case to start with is when the constant is zero, corresponding to the two masses being released from rest at infinite separation.

Even without solving the equation, you can see that as the separation approaches zero, the infall speed approaches infinity.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.