Solve for trajectory (if you can)

First thing we do is that we switch to polar co-ordinates $\overrightarrow { r } =rcos\theta \hat { i } +rsin\theta \hat { j }$

Also the velocity will have two components :

1)Tangential velocity = ${ v }_{ t }$

2)Radial velocity= ${ v }_{ r }$

Then we note that since the force is central hence angular momentum for the particle is conserved. So we write :

$m{ v }_{ 0 }{ r }_{ 0 }=m{v}_{t}r$ (i)

Also by conservation of total energy we have K.E+P.E.=Constant

$\frac { 1 }{ 2 } m{ v }_{ 0 }^{ 2 }-\frac { 8 }{ { r }_{ 0 } } =\frac { 1 }{ 2 } m{ v }^{ 2 }-\frac { 8 }{ r }$ (ii)

Also ${ v }_{ t }^{ 2 }+{ v }_{ r }^{ 2 }={ v }^{ 2 }$ (iii)

Using (iii) in (ii) we get

$\frac { 1 }{ 2 } m{ v }_{ 0 }^{ 2 }-\frac { 8 }{ { r }_{ 0 } } =\frac { 1 }{ 2 } m({ v }_{ t }^{ 2 }+{ v }_{ r }^{ 2 })-\frac { 8 }{ r }$

Using (i) to get

$\frac { 1 }{ 2 } m{ v }_{ 0 }^{ 2 }-\frac { 8 }{ { r }_{ 0 } } =\frac { 1 }{ 2 } m{ v }_{ r }^{ 2 }+\frac { m{ v }_{ 0 }^{ 2 }{ r }_{ 0 }^{ 2 } }{ 2{ r }^{ 2 } } -\frac { 8 }{ r }$

Put the values to get :

$\frac { dr }{ dt } ={ v }_{ r }=\sqrt { \frac { 16 }{ r } -\frac { 4 }{ { r }^{ 2 } } -7 }$ (iv)

Also ${ v }_{ t }=\omega r$ hence in (i) we have

$\omega { r }^{ 2 }={ r }_{ 0 }{ v }_{ 0 }$

$\Rightarrow \omega =\frac { 2 }{ { r }^{ 2 } } =\frac { d\theta }{ dt }$ (v)

Dividing eq. (iv) by (v) we have

$\frac { dr }{ d\theta } =\frac { { r }^{ 2 } }{ 2 } \sqrt { \frac { 16 }{ r } -\frac { 4 }{ { r }^{ 2 } } -7 }$

$\Rightarrow \int { \frac { 2dr }{ r\sqrt { 16r-7{ r }^{ 2 }-4 } } =\int { d\theta } }$

$C-\tan ^{ -1 }{ (\frac { 2(2r-1) }{ \sqrt { 16r-7{ r }^{ 2 }-4 } } ) } =\theta$

Finding the constant of integration and putting it we have :

$\cot ^{ -1 }{ (\frac { 2(2r-1) }{ \sqrt { 16r-7{ r }^{ 2 }-4 } } ) } =\theta$

Solving for $\theta$ we have

$r=\frac { 2 }{ 4-3cos\theta }$

Polar equation of a general ellipse is given by $r=\frac { l }{ 1-ecos\theta }$

e=eccentricity and l=latus rectum

Comparing it with our original equation we have :

$\boxed { e=\frac { 3 }{ 4 } }$

Clearly, the potential energy at a point is $\dfrac{-8}{r}$

Let $p$ and $q$ be the major and minor axes , then clearly,

$p(1+e) = r_{0} = 2$

Also, let $r = p(1-e)$ be the other extreme end of the ellipse, and $v$ be the velocity at that point.

Then by conservation of angular momentum,

$vr = v_{0}r_{0} \Rightarrow v = \dfrac{v_{0}r_{0}}{r}$

Also, by conservation of mechanical energy,

$\dfrac{1}{2} mv^2 -8/r = \dfrac{1}{2} mv_{0}^2 - 8/r_{0}$

Substitute the values of $m , v_{0} , r_{0}$ and solve to get $r = \dfrac{2}{7}$

Clearly, $p(1+e) = 2, \text{ and } p(1-e) = \dfrac{2}{7}$

Solve to get $\boxed{e = \dfrac{3}{4}}$

Preamble: in cartesian coordinates $(O,\,x,\,y)\,$ :

$\vec {r}=x\hat {\imath}+y\hat {\jmath};$

in polar coordinates $(O,\,r,\,\theta)\,$ :

$\vec {r}=r\hat {e_r},\,\hat {e_r}={\cos\theta}\hat {\imath}+{\sin\theta}\hat {\jmath},\,\hat {e_\theta}={\frac {d}{d\theta}}{\hat {e_r}}=-{\sin\theta}\hat {\imath}+{\cos\theta}\hat {\jmath}$

The system is conservative, the potential energy by the central force, with $V(\infty)=0\,$ :

$\vec {F}=-{\frac 8 {r^2}}\hat {e_r},\,$ is: $V=-\frac 8 r$

so there are two constant of the motion, the total mechanical energy and the angular momentum:

$E={\frac 1 2}mv^2-\frac 8 r={\frac 1 2}m{v_0}^2-\frac 8 {r_0}=E_0$

$\vec L=m\vec v\times\vec r=\vec {L_0}$

the cinematic values in polar coordinates are in this case:

$v_{r}=\dot r,\,v_{\theta}=r{\dot\theta},\,a_r={\ddot r}-r{\dot\theta}^2,\,a_{\theta}=r{\ddot\theta}+2\dot r\dot\theta=0$

so:

$\vec {L}=mr^2{\dot\theta}\hat {k},\,L=mr^2{\dot\theta}=mr_0v_0=L_0,\,$

and it can be writed after some algebraic tricks:

$v^2={v_r}^2+{v_{\theta}}^2=({\frac {L_0} m})^2{\left(({\frac d {d\theta}}{\frac 1 r})^2+\frac 1 {r^2}\right)}$

$a_{\theta}=0,\,a_r=-({\frac {L_0} {mr}})^2{\left({\frac {d^2}{d{\theta}^2}}{\frac 1 r}+\frac 1 r\right)}$

we can write the simple equation of motion;

$a_r=-\frac 8 {r^2},$

${\left({\frac {d^2}{d{\theta}^2}}{\frac 1 r}+\frac 1 r\right)}=\frac {8m^2}{{L_0}^2}$

but now i choose the method that use the conservation of the constant of the motion $E_0,\,L_0,\,$ thus:

$E_0={\frac 1 2}m({\frac {L_0} m})^2{\left(({\frac d {d\theta}}{\frac 1 r})^2+\frac 1 {r^2}\right)}-\frac 8 r$

that can be rewrited:

${d\theta}=\frac {dr}{r^2{\sqrt{{\frac {2m}{L^2}}(E_0+\frac 8 r)-\frac 1 {r^2}}}}$

${d\theta}={\frac {L_0} {\sqrt {2m}}}\frac {dr}{r^2{\sqrt{E_0+{\frac 8 r}-(\frac {L_0} {\sqrt{2m}})^2{\frac 1 {r^2}}}}}$

changing the variable $r\,$ in:

$u-u_0=\frac {L_0}{r\sqrt{2m}}\,$ so that in the square root there will be a difference of squares:

$(a+u)(a-u)=(a^2-u^2)\,$ which implies:

$u_0=-\frac {8m} {L_0\sqrt {2m}},\,a^2=E_0+\frac {32m}{L^2}\,$ we obtain:

${d\theta}=-{\frac {d(u/a)}{\sqrt{1-(u/a)^2}}}\,$

so integrating:

$\theta-{\theta}_i=-\arccos(\frac u a);\,\frac u a =cos(\theta-{\theta}_i)\,$

reintroducing $r\,$ :

$r=\frac p {1+e\cos(\theta-{\theta}_i)}$

$p={\frac {{L_0}^2} 8} > 0,\,e=\sqrt{1+\frac {E_0{L_0}^2}{32m}}\ge 0,\,E_0\ge -\frac {32m}{{L_0}^2}$

substituting values for $r_0,\,v_0,\,m,\,$ we get:

$E_0=-{\frac 7 2},\,L_0=2,\,p={\frac 1 2},\,e={\frac 3 4}$

so the eccentricity is : $e={\frac a b}={\frac 3 4}$

$a\times b=3\times 4=\boxed {12}$

the equation of the trajectory is:

$r={\frac 2 {4+3\cos(\theta-{\theta}_i)}}$

$t_0=0,\,r=r_0=2,\,\theta={\theta}_0=0,\,$ so:

$\cos(-{\theta}_i)=-1,\,{\theta}_i=\pi,\,$ so the equation of the trajectory became:

$r={\frac 2 {4-3\cos(\theta)}}$

using the equation of motion:

${\left({\frac {d^2}{d{\theta}^2}}{\frac 1 r}+\frac 1 r\right)}=\frac {8m^2}{{L_0}^2}$

imposing $w={\frac 1 r},\,c=\frac {8m^2}{{L_0}^2},\,$ we get:

${\ddot w}+w=c,\,$ with general solution: $\frac 1 r = a_1\cos\theta + a_2\sin\theta + c$

$t=0,\,\theta = {\theta}_0,\,$ imposing: $\frac {a_1}{\sqrt{{a_1}^2+{a_2}^2}} =\cos{\theta}_0,\,\frac {a_2}{\sqrt{{a_1}^2+{a_2}^2}} =\sin{\theta}_0,\,\sqrt{{a_1}^2+{a_2}^2}=A,\,$

we get:

$\frac 1 r = A\cos(\theta - {\theta}_0) + c,\,$ imposing: $c=\frac 1 p ,\,A=\frac e p ,\,$

$r=\frac p {1+e\cos(\theta - {\theta}_0)},\,$ the same equation of the trajectory.

I solved this problem numerically, using Euler's numerical method to find the particle as it moves along the force field. Here is an image that I made of the trajectory using 2000 data points from python simulated for 2.5 seconds and scatter plot:

As you can see, the particle does trace out an elliptical path. Calculating the value of $e = \Large \frac{ \sqrt{a^2-b^2} }{a}$ , with the data, you roughly get $0.75$ , which is equal to $\boxed{ \frac{3}{4}}$

Here is my python code below, which has timestep of $10^-3$ . Plot the data points in Excel.

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import math

time = 0 #initial time              
deltaT = 10**-3 #timestep

r  = [2,0] #initial position vector
rDot = [0,1] #initial velocity vector   

Rx = [] #x values   
Ry = [] #y values

while time <=2.5: #simulating to 2.5 seconds
    Rx.append(r[0]) 
    Ry.append(r[1])

    rDotDot = [-8/((math.hypot(r[0],r[1]))**3)*r[0], -8/((math.hypot(r[0],r[1]))**3)*r[1]] #Force field calculation 

    rDot[0] = rDot[0] + rDotDot[0]*deltaT #numerically integrating to get x velocity
    rDot[1] = rDot[1] + rDotDot[1]*deltaT #numerically integrating to get y velocity

    r[0] = r[0] + rDot[0]*deltaT #numerically integrating xdot to get x position
    r[1] = r[1] + rDot[1]*deltaT #numerically integrating ydot to get y position

    time = time + deltaT #updating value of time


print("x values: ")

print("")

for i in Rx:
    print(i)

print("")

print("y values: ")

print("")

for j in Ry:
    print(j)

If we know that the trajectory is an ellipse as given in the question and given that it is an inverse square force,it must be directed towards one of the focii.

Therefore one of the focii is the origin.

further by shifting origin, x = ae + acos(wt)

                                                      y  =  bsin(wt)

at t=0

Vy = 1 => wb=1......(i)

acceleraton in x direction = 2 => w^2 * a = 2............(ii)

also distance of particle at t=0 from focs is a(1+e) = 2.........(iii)

Solving these three equation for e gives e = 3/4. Is it correct?