Question

Motion in a Force Field a mathematical model for the position x(t) of a body moving rectilinearly on the x-axis in an inverse-square force field is given by

$\frac{{\mathbf{d}}^{\mathbf{2}}\mathbf{x}}{\mathbf{d}{\mathbf{t}}^{\mathbf{2}}}\mathbf{=}\mathbf{-}\frac{{\mathbf{k}}^{\mathbf{2}}}{{\mathbf{x}}^{\mathbf{2}}}$

Suppose that at t = 0 the body starts from rest from the position $\mathit{x}\mathbf{=}{\mathit{x}}_{\mathbf{0}}\mathbf{,}{\mathit{x}}_{\mathbf{0}}\mathbf{>}\mathbf{0}$. Show that the velocity of the body at time t is given by ${\mathit{v}}^{\mathbf{2}}\mathbf{=}\mathbf{2}{\mathit{k}}^{\mathbf{2}}\mathbf{(}\mathbf{1}\mathbf{/}\mathit{x}\mathbf{-}\mathbf{1}\mathbf{/}{\mathit{x}}_{\mathbf{0}}\mathbf{)}$. Use the last expression and a CAS to carry out the integration to express time t in terms of x.

Short answer

The solution of the above question is

$$\begin{gathered} t=-\frac{1}{2k\sqrt{x}} \\ \left(\sqrt{x-2}x+\sqrt{x}\ln \frac{\sqrt{\frac{x-2}{x}}-1}{\sqrt{\frac{x-2}{x}}+1}\right)+c_2 \end{gathered}$$

Step-by-step solution

Motion in a force field:

We consider the motion of a particle in a random isotropic force field. Assuming that the force field arises from a Poisson field in , d ≥ 4, and the initial velocity of the particle is sufficiently large, we describe the asymptotic behaviour of the particle

Concept of Motion in a force field:

In this question we will apply the concept Motion in a force field;

$$\begin{gathered} \frac{d^{2}x}{d{t}^2}=-\frac{k^2}{x^2} \\ x(v=0)=x_0 \\ \frac{dx}{dt}=v \\ \frac{d^{2}x}{d{t}^2}=\frac{dv}{dt} \\ =v\frac{dv}{dx} \\ v\frac{dv}{dx}=-\frac{k^2}{x^2} \\ \int vdv=-k^2\int \frac{1}{x^2}dx \\ {v}^2=2k^2\frac{1}{x}+h \\ v=\frac{dx}{dt} \\ {\left(\frac{dx}{dt}\right)}^2=-2k^2\left(\frac{1}{x}-\frac{1}{x_0}\right) \\ -\frac{dx}{dt}=\sqrt{2}k\sqrt{\frac{x_0}{x_{0}x}} \\ \int \left(\sqrt{\frac{x-x_0}{x_{0}x}}dx\right)=-\sqrt{2}k\int dt \end{gathered}$$

Hence, the final answer is$t=-\frac{1}{2k\sqrt{x}}\left(\sqrt{x-2}x+\sqrt{x}\ln \frac{\sqrt{\frac{x-2}{x}}-1}{\sqrt{\frac{x-2}{x}}+1}\right)+c_2$