# Homework Help: Energy+bound orbit

1. Jul 3, 2010

### PhMichael

http://img810.imageshack.us/img810/1883/planet.jpg [Broken]

1. The problem statement, all variables and given/known data
I have a body of mass m, which is moving in a circular path around some planet. At a certain instant, the object explodes into two equal bodies. I'm given that the tangential velocity doesn't change as a result of the explosion, in addition, the system's kinetic energy increases by a factor of k (k>1). I'm asked to determine the minimal radius (measured from the binding center) of the two half-bodies.

2. The attempt at a solution

The total energy before the explosion occured is: (M=the mass of the planet)

$$E_{tot} = \frac{1}{2}mv_{\theta}^{2} - \frac{GMm}{R}$$

Now, the total energy after the explosion:

$$E_{tot} = \frac{k}{2}mv_{\theta}^{2} - \frac{GMm}{R_{min}}$$

now, we can obtain the tangential velocity from newton's II law:

$$-\frac{GMm}{R^2} = -m \frac{v_{\theta}^{2}}{R}$$

$$v_{\theta}^{2} = \frac{GM}{R}$$

equating both expressions of $$E_{tot}$$ as the total energy is conserved and using the last relation, yields:

$$-\frac{GMm}{2R} = \frac{kGMm}{2R}-\frac{GMm}{R_{min}}$$

therefore,

$$R_{min} = \frac{2R}{1+k}$$

$$R_{min} = \frac {R}{1+\sqrt{k-1}}$$

what have I done wrong?

Last edited by a moderator: May 4, 2017
2. Jul 3, 2010

### hikaru1221

I don't know what binding center is, so I cannot give you any opinion on how to solve.
In your solution, the deadly mistake you made is that you assume the total energy is conserved, while it's not. Right after the explosion, the total potential energy remains the same, but the kinetic energy increases.

So instead of approaching it from the energy angle, you may use the k factor to find the initial velocities of the parts right after explosion. I guess the binding center is somewhat related to the motions of the parts after explosion, so having the initial velocities, you may be able to find the radius.

3. Jul 3, 2010

### vela

Staff Emeritus
I don't understand how you justify using Rmin in the expression

$$E_{tot} = \frac{k}{2}mv_{\theta}^{2} - \frac{GMm}{R_{min}}$$

Immediately after the explosion, the kinetic energy has increased by a factor of k while the tangential component of the velocity is unchanged, so the two pieces must have non-zero radial components of velocity. They'll follow elliptical orbits now, and the distance of closest approach will depend on the orbits' eccentricities. The energy of an orbit doesn't depend on the eccentricity, so I don't think you can solve this problem by only considering energy.

4. Jul 3, 2010

### PhMichael

- The binding center is the planet itself.

- Ok, after the explosion there has to be a radial velocity also, right?! because:

$$E'_{k}=\frac{kmv_{\theta}^{2}}{2}=\frac{m(\sqrt{k}v_{\theta})^{2}}{2}$$

so,

$$v=\sqrt{v_{\theta}^{2}+v_{r}^{2}}=\sqrt{k}v_{\theta}$$

therefore,

$$v_{r}=\sqrt{k-1}v_{\theta}$$

so far so good, but how do I continue from here?! :blush:

5. Jul 3, 2010

### ehild

Use conservation of angular momentum.

$$-\frac{GMm}{R^2} = -m \frac{v_{\theta}^{2}}{R}$$

is valid only for circular orbit. The "R" in the expression of centripetal acceleration is the radius of curvature which is not the same as the distance from the sun if the orbit is an ellipse.

ehild

6. Jul 3, 2010

### PhMichael

thanks all.

I really followed the conservation of angular momentum approach and got the correct answer:

$$mv_{\theta}R = mv_{\theta}R_{min} + m \sqrt{k-1} v_{\theta}R_{min}$$

$$v_{\theta} = \frac{R}{1+\sqrt{k-1}}$$

this is all good, but the problem is that i just plugged in these stuff blindly without seeing the picture; in other words, why/where the angular momentum is conserved?
the angular momentum is defined as:
$$\vec{r} \times \vec{v}$$ so what i "really" should get is:
$$R_{min}\hat{r} \times (v_{\theta}\hat{\theta} + \sqrt{k-1}v_{\theta}\hat{r})$$
so, the r component of the velocity drops down and the result is: $$R_{min}v_{\theta}\hat{z}$$. :surprised

can anyone clear the picture for me?!

7. Jul 3, 2010

### ehild

The angular momentum is conserved in any central field. It means that the product of the "theta" component of the velocity and R is constant. At minimum distance, the velocity has only this "theta" component, so v*Rmin=vθ*R0 where vθ is the original "tangential" component and R0 is the original radius of the circular orbit. Both parts of the planet gained equal radial velocities (because of conservation of momentum) so equal kinetic energy. You know the total energy for one piece just after the explosion, and it is conserved along the orbit.

ehild

8. Jul 3, 2010

### PhMichael

This whole subject confuses me.
Let me put it this way, the equations to solve for Rmin are:

(1) $$mv_{\theta}R = m(v_{\theta}+v_{r})R_{min}$$

(2) $$v_{r}=v_{\theta}\sqrt{k-1}$$

Q1: now, how can I justify the right hand side of equation (1) ? I mean, OK, a radial component for the velocities of each half-body must exist as a result of the increasing kinetic energy, however, the way I should compute angular momentum in each case (even after the explosion) is by the cross product of the radius vector and the linear momentum. Doing that removes the contribution of the radial component of the velocity since it's parallel to the radius vector.

Q2: the absolute velocity of each half-body after the explosion is $$\sqrt{v_{\theta}^{2}+v_{r}^{2}}$$ ... so why isn't this expression plugged in the first equation instead of the 'regular" sum of the velocities as it appears there?!

What i've just written must be crap because the right answer comes out from these equation, so I need to settle these stuff in my mind =)

9. Jul 4, 2010

### ehild

(1) should be a vector equation and referring to the state after the explosion: initially, the velocity has both radial and azimuthal components, and only azimuthal one at Rmin.

ehild

Last edited: Jul 4, 2010
10. Jul 4, 2010

### vela

Staff Emeritus
I think you stumbled upon the answer using an incorrect method, which is why the equations probably don't make sense to you. Since $L=mv_\theta R$ is constant, you can see that $v_\theta$ must vary as R changes, so the $v_\theta$ you have in equation (1) on the lefthand side isn't equal to the $v_\theta$ you have on the righthand side.

Edit: Actually, I should say your equation (1) doesn't make sense to me so it's possible the $v_\theta$ on the two sides are, in fact, equal, but I doubt it.

Last edited: Jul 4, 2010
11. Jul 4, 2010

### PhMichael

I'm given that $$v_{\theta}$$ remains constant, so how could it not be the same in LHS and RHS of eq. (1) ?!

What really puzzles me is the use of a regular algebric sum for both $$v_{r}$$ and $$v_{\theta}$$ in eq. (1) ... why is that done this way?!

It would be extremely appreciated if someone shows me how to obtain equation (1) mathematically, especially the angular momentum after the explosion as it is given by the RHS of eq. (1).

(1) $$mv_{\theta}R = m(v_{\theta}+v_{r})R_{min}$$

12. Jul 4, 2010

### ehild

$$v_{\theta}$$ remains constant during the explosion, not after.

Equation 1. is wrong.

ehild

13. Jul 4, 2010

### vela

Staff Emeritus
The specific angular momentum is given by

$$\vec{l} = \vec{r} \times \vec{v} = r \hat{r} \times [v_r(r) \hat{r} + v_\theta(r) \hat{\theta}] = rv_\theta(r) \hat{z}$$

where the tangential and radial components of the velocity are functions of the radial distance, so conservation of angular momentum tells you

$$l = Rv_\theta(R) = R_{min} v_\theta(R_{min})$$.

Use this relationship along with conservation of energy and insights scattered in this thread to solve for Rmin.

14. Jul 4, 2010

### PhMichael

The radial velocity is found by the given data of the kinetic energy:
$$v_{r}=v_{\theta}\sqrt{k-1}$$

(1) $$Rv_{\theta} = R_{min} V_{\THETA}$$

(2) Etot(after the explosion)=Etot(at Rmin)

$$\frac{1}{2}mv_{\theta}^{2}+\frac{1}{2}mv_{r}^{2}-\frac{GMm}{R}=\frac{1}{2}mV_{\THETA}^{2}-\frac{GMm}{R_{min}}$$

is it now good?

I guess i'm too dumb to understand this =/

15. Jul 4, 2010

### vela

Staff Emeritus
Looks good so far. You just need this relationship you derived earlier:

$$v_{\theta}^{2} = \frac{GM}{R}$$

and you should be able to solve for Rmin in terms of R and k.

16. Jul 4, 2010

### PhMichael

So I solved these equations (by hand and checked it using Matlab) and got:

$$R_{min} = R \cdot \frac{-1+\sqrt{k-1}}{k-2}$$

which is an incorrect answer as it should be:

$$R_{min} = \frac {R}{1+\sqrt{k-1}}$$

what's wrong now? :(

17. Jul 4, 2010

### vela

Staff Emeritus
Well, without seeing your actual work, it's impossible to say. I worked it out here from the same equations and did get the right answer.

18. Jul 4, 2010

### PhMichael

(1) $$v_{\theta}^{2}=\frac{GM}{R}$$

(2) $$v_{r}^{2} = v_{\theta}^{2}(k-1)=\frac{GM}{R}(k-1)$$

(3) $$V^{2} = \frac{R^{2}v_{\theta}^{2}}{R_{min}^{2}} = \frac{R^{2}\frac{GM}{R}}{R_{min}^{2}}=\frac{GMR}{R _{min}^{2}$$

(4) $$\frac{1}{2}mv_{\theta}^{2}+\frac{1}{2}mv_{r}^{2}-\frac{GMm}{R}=\frac{1}{2}mV^{2}-\frac{GMm}{R_{min}}$$

I substitue the 1st 3 eqns. in eq. (4) and there pops the answer. no?

19. Jul 4, 2010

### vela

Staff Emeritus
Yes, so apparently your mistake occurs after you substitute everything into equation (4). What did you get after you did that and simplified?

20. Jul 4, 2010

### PhMichael

I get rid of "m" in eq.(4), then:

$$\frac{GM}{2R}+\frac{GM(k-1)}{2R}-\frac{GM}{R}=\frac{GMR}{2R_{min}^{2}}-\frac{GM}{R_{min}}$$

now I divide by "GM":

$$-\frac{1}{2R}+\frac{k-1}{2R}=\frac{R}{2R_{min}^{2}}-\frac{1}{R_{min}}$$

eventually,

$$R_{min}^{2}( \frac{k-2}{2R} ) +2R_{min}-R = 0$$

21. Jul 4, 2010

### vela

Staff Emeritus
You can make the algebra easier if you rewrite your first equation slightly. Combining the first two terms on the LHS and strategically introducing factors of R, you get

$$k\frac{GM}{2R}-\frac{GM}{R}=\frac{GM}{2R}\left(\frac{R}{R_{min}}\right)^{2}-\frac{GM}{R}\left(\frac{R}{R_{min}}\right)$$

Often when you recast an equation in terms of a dimensionless quantity, like R/Rmin, you'll find it is easier to work with.

Try solving that for R/Rmin.

22. Jul 4, 2010

### vela

Staff Emeritus
The reason this didn't work, at least in part, is because, going from the first equation to the second, the linear and constant terms picked up a factor of two whereas the quadratic term didn't.

23. Jul 4, 2010

### PhMichael

Thanks alot, mr. vela!!!! you are my hero =)

24. Jul 4, 2010

### kuruman

There is no mistake. Just multiply numerator and denominator by sqrt(k-1)+1 and see what you get. Radicals can be deceiving some times.

25. Jul 4, 2010

### vela

Staff Emeritus
Good catch.