3
$\begingroup$

enter image description here

A particle $A$ is initially given some velocity and is being attracted by a large object $B$ ($m_A\ll m_B$).

  • Do there exists an initial velocity so that the particle $A$ describes a circular motion and not collide with the object $B$?
  • Is there any mathematical proof for it?
$\endgroup$
5
  • $\begingroup$ @tom10 I was only interested in circular motion. $\endgroup$ – Asher2211 Apr 18 at 2:05
  • $\begingroup$ @tom10 Isn't the answer same even if replace circular orbits with non circular orbits? $\endgroup$ – Asher2211 Apr 18 at 2:06
  • $\begingroup$ Not really. Circular orbits are a very small subset of possible orbits. Most important for this question is that almost any starting condition, for example the one you show at the blue dot, will create an orbit that doesn't collide with B. Making an orbit that doesn't collide with B is easy. Making a circular orbit is hard. So if you're interesting in circular orbits you'll get one answer, and general non-colliding orbits you'll get a different answer. $\endgroup$ – tom10 Apr 18 at 2:12
  • $\begingroup$ @tom10 What about the argument I gave in the comment section mike stone's answer. Couldn't that be used for non circular orbit's? $\endgroup$ – Asher2211 Apr 18 at 2:16
  • $\begingroup$ As I understand it, I think so, but it's difficult to be clear in comments. If you're saying that if a particle is in a circular motion at one time that means it was in a circular motion at every other time, that is true (assuming no other forces are applied). That is true for other orbits as well, btw. Also, I'm not sure this is "proof" of anything, but it is a true statement. $\endgroup$ – tom10 Apr 18 at 2:22
9
$\begingroup$

Yes --- but the motion will never look like your drawing. The initial velocity must be tangential to the desired circle and be given by $mv^2/r=GmM/r^2$ .

$\endgroup$
7
  • $\begingroup$ So if a particle is at a distance $d$ it can never move in a circle of radius $r<d$. Is that what you meant? $\endgroup$ – Asher2211 Apr 17 at 19:39
  • $\begingroup$ Exactly! A spacecraft needs an orbit injection burn to circularize its orbit. $\endgroup$ – mike stone Apr 17 at 19:51
  • $\begingroup$ Is it difficult to prove it mathematically $\endgroup$ – Asher2211 Apr 17 at 21:04
  • $\begingroup$ No. Not hard assuming you know basic calculus and ODE's. Just look (and understand) up the soultion to the Kepler problem. en.wikipedia.org/wiki/Kepler_problem $\endgroup$ – mike stone Apr 17 at 21:36
  • $\begingroup$ I came to think of a logical proof. If the particle starts at $t=0$ and suppose it begins to move in a uniform circular motion at time $T$. If we go backwards in time the particle should execute circular motion (it is the reverse of going forward in time). So this implies that the particle is in circular motion from $t=0$ itself and this is only possible if velocity is tangential to the circle initially. Is this reasoning correct? $\endgroup$ – Asher2211 Apr 17 at 22:07
4
$\begingroup$

Yes, this is how satellites work. See Orbital Mechanics.

$\endgroup$
3
$\begingroup$

Yes it is possible. To see how we must first learn something about uniform circular motion. This means to move in a circle at constant speed. A particle undergoing uniform circular motion has a position $\vec r(t)$ that looks as follows. $$\vec r(t)=\pmatrix{r\cos\omega t\\r\sin\omega t}$$ Here $\omega=2\pi f$ where $f$ is the frequency. Let's take two derivatives to obtain the acceleration \begin{align} \vec v(t)=\frac{d \vec r}{dt}&=\pmatrix{-r\omega\sin\omega t\\\ \ \ r\omega\cos\omega t}\\ \vec a(t)=\frac{d \vec v}{dt}&=\pmatrix{-r\omega^2\cos\omega t\\-r\omega^2\sin\omega t}=-\omega^2\vec r(t) \end{align} If we calculate the lengths of the vectors $\vec v,\vec a$ we get $v=\omega\,r$ and $a=\omega^2\, r$. By eliminating $\omega$ we get the equation $$a=\frac{v^2}r$$ In other words, if something moves in a circle with constant speed it will accelerate with $a=v^2/r$. Or equivalently, if something accelerates with $a=v^2/r$ in the $-r$ direction it will move in a circle with constant speed.

So to get back to your problem: we want the particle to move in a circle so its acceleration must be equal to $a=v^2/r$. We also know the only force acting on the particle is gravity (this is actually not explicitly in your answer, be sure to add every relevant detail in the future because being able to ask good questions is a valuable skill!). Because the only force is gravity we know the particle accelerates with $a=\frac{Gm_B}{r^2}$. So together this gives the condition $$\frac{v^2}r=\frac{Gm_B}{r^2}$$ which allows you to solve for $v$. Now I have the following questions for you

  1. Solve for $v$. This is the speed necessary to move in a circle
  2. Is every value of $r$ allowed? For example in the function $\sqrt{r-b}$ only values of $r>b$ are allowed. If every value is allowed this means that at every radius it is possible to move in a circle if you are moving at the right speed.
  3. Convince yourself that the gravitiational force points in the right direction for circular motion. For example with a drawing.
$\endgroup$
1
$\begingroup$

I think you already made it yourself clear. When the small mass is in the depicted circular orbit and you let it make some rotations after which you reverse time then the small mass can never exit the orbit to arrive at A. And if the backward is not possible then also the forward is impossible (assuming that the small mass has no propulsion device).

$\endgroup$
0
$\begingroup$

Yes, we need to have initial velocity otherwise it would just do and collide directly with the larger mass along the line joining centers of the masses.

$\endgroup$
1
  • 1
    $\begingroup$ I am asking whether for some velocity it is possible or not. $\endgroup$ – Asher2211 Apr 17 at 18:46

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.