A Strange Central Force

First note that if the radial component of radial force is represented by $F(r) = - \frac{1}{r^{(n+1)}}, \quad n = 2/3$ then the potential energy is $U(r) = - \frac{1}{nr^n}$

At the initial point, radial velocity is zero. At the point of closest approach also, the radial velocity must be zero.

Then by conservation of energy, $\frac{1}{2} m v_{t0}^2 - U(r_0) = \frac{1}{2} m v_{t}^2 - U(r),$ where $v_t$ and $r$ tangential speed and radial distance from center at the point of closest approach.

Next note that, by conservation of angular momentum $m r_0 v_{t0} = m r v_t \quad \Rightarrow v_t = v_{t0} \frac{r_0}{r}.$

Thus, $T = \frac{1}{2} m v_{t}^2 = \frac{1}{2} m v_{t0}^2 \left( \frac{r_0}{r}\right)^2 = T_0 \left( \frac{r_0}{r}\right)^2.$ Similarly, factoring out $U(r)$ we get $U(r) = U(r_0) \left( \frac{r_0}{r}\right)^n$

Dividing the equation by $T_0$ , and defining $u \equiv \left( \frac{r_0}{r}\right)^n$ We obtain a cubic equation of the form $u^3 - a u + b = 0, \quad, a = \frac{U_0}{T_0} = 161.583, \quad b = a -1,$ whose solutions are $1, 12.182$ . Thus $r = \frac{(r_0 = 10)}{12.182^{(1/n = 1.5)}} = 0.2352$

A small 10 kg sphere is attracted to a central point at the origin by a force of F=-r^(5/3)r where r is measured in meters. At time t=0 the sphere is at r=10 m with a tangential velocity of 0.02 m/s and no radial velocity. What is the closest distance the sphere comes to the origin in meters? Let r 0 be the initial radial vector and v 0 be the initial velocity vector. Since this system is very similar to that of celestial bodies, it makes sense that the motion of this particle is kind of elliptical. So in the moment of time when the particle is as close as possible to the origin, the velocity vector is perpendicular to the radial vector. Let v be the velocity vector of the particle. The angular momentum of the particle is defined to be r mv, where m is the mass of the particle. Since the only force acting on the particle is radial, we have that the angular momentum of the system about the origin is constant, thus r_0 v 0=r*v. When t=0 and when the particle is as close to the origin as possible, we have that the radial vectors are perpendicular to the respective velocity vectors, so this is equivalent to saying that the products of the magnitudes of these vectors are equal; r 0 v_0=r v=(10m)(0.02m/s)=0.2 m^2/s. Now consider the mechanical energy of the system. Clearly it is not constant, but since this setup is similar to the motion of celestial bodies, we can define a pseudo-gravitational potential energy U. We shall define it like so: U 1-U 2=int {r 1}^{r 2} -r^(-5/3)dr = (-3/2)r 2^(-2/3)+(3/2)r 1^(-2/3) U=-1.5r^(-2/3)=(-3/2)r^(-2/3). Taking this into account, we have that the mechanical energy of the system is conserved, thus (-3/2)r 0^(-2/3)+(1/2)mv_0^2=(-3/2)r^(-2/3)+(1/2)mv^2. The LHS of this equation can be calculated as 0.321. Therefore 0.321=(-3/2)r^(-2/3)+5v^2. We already have that rv=0.2, so we plug in v=0.2/r to get: 0.321=(-3/2)r^(-2/3)+0.2r^(-2). We now must solve for r. Substituting x=r^(-2/3) yields the cubic equation (0.2)x^3-(1.5)x-0.321=0. We can then use the cubic root calculator here: http://easycalculation.com/algebra/cubic-equation.php The roots of this cubic are 2.624, -2.840, and 0.215. The only value of x that gives a value of r that makes sense is x=2.624; thus r=2.624^(-3/2)=0.235.

The potential energy associated with the central force is given by:

$U(r) = -\int F \ \mathrm{d}r = -\int -r^{-5/3} \ \mathrm{d}r = -\frac{3}{2}r^{-2/3}$

For this motion, both the total energy $E_{tot}$ and angular momentum $L$ of the sphere is conserved. We will calculate the two values based on the initial condition given.

$E_{tot} = \frac{1}{2}mv^2 +U(r) = \frac{1}{2}(10)(0.02)^2 - \frac{3}{2}(10^{-2/3})=-0.3212 \ \mathrm{J}$ $L=mvr = 10 \times 0.02 \times 10 = 2 \ \mathrm{kgm^2s^{-1}}$

The total energy of the sphere at any point is given by $E_{tot} = \frac{1}{2}m\dot{r}^2 + \frac{1}{2}mr^2\dot{\phi}^2 +U(r)$ At the turning points (closest or furthest distance), we have $\dot{r}=0$ , so we obtain the following equation $\frac{1}{2}mr^2\dot{\phi}^2 +U(r)=-0.3212$

There are two unknowns $r$ and $\dot{\phi}$ , but we can express $\dot{\phi}$ in terms of $r$ using the conservation of angular momentum. We have $L=mr^2\dot{\phi} = 2$ , so $\dot{\phi} = \frac{2}{mr^2} = \frac{0.2}{r^2}$ . Substituting this into the equation gives us

$\frac{1}{2}mr^2\left(\frac{0.2}{r^2}\right)^2 +U(r)=-0.3212$

Simplifying, we get

$0.2r^{-2} - \frac{3}{2}r^{-2/3} =-0.3212$

This is an equation in $r$ so $r$ can be solved easily. Let $k=r^{-2}$ , we would have $0.2k+0.3212 = \frac{3}{2}k^{1/3}$ . Then, cubing both sides and simplifying, we have

$0.008k^3+0.038544k^2-3.3131k+0.03314=0$

This is a cubic equation, and we also know that $k=(10)^{-2}=0.01$ is a solution because $r=10 \ \mathrm{m}$ is a turning point as given. Thus we obtain the other two solutions $k=18.078$ and $k=-22.90$ (rejected since it is negative).

Thus the other turning point occurs when $r^{-2} = 18.078$ or $r=(18.078)^{-1/2}=0.235 \ \mathrm{m}$ . This is indeed the closest distance, so the answer is $\fbox{0.235}$

When the distance from origin is minimum , velocity must be perpendicular to $\vec{r}$ .

Let the closest distance be $r$ and velocity at that instant be $v$

Since , the force is always imparted towards origin , we get that angular impulse about origin = 0.

Hence, we can conserve angular momentum about origin.

$\Rightarrow mvr = m(0.02)(10) \Rightarrow v = \frac{0.2}{r}$

Clearly , potential energy in this field = $-\frac{3}{2}r^{-\frac{2}{3}}$

Using conservation of energy,

$\frac{1}{2} 10 v^2 - \frac{3}{2}r^{-\frac{2}{3}} = \frac{1}{2} 10 (0.02)^2 - \frac{3}{2}10^{-\frac{2}{3}}$

Putting $v = \frac{0.2}{r}$ in this above equation ,we get $r = \fbox{0.235}$

Conservation of angular momentum: V.r {min} = v.r = 0.02 \times 10 = 0.2 (v is the tangential velocity at the distance r) when the mass reach the closest to the origin, the radical velocity is obviously zero, therefore the total velocity equals to the tangential velocity. Conservation of energy : \frac{mV^{2}}{2} = \frac{m\times 0.02^{2}}{2} + \frac{3}{2}(\frac{1}{r {min}^{2/3}} -\frac{1}{10^{2/3}}) (the expression of work done by the central force can be calculated by basic integrate) substitute V=\frac{0.2}{r {min}}, we got an equation containing only a variable r {min} Ultimately r_{min} = 0.235 m