Question

If a particle is thrown with a speed of v at an angle theta with its orbit around the earth while revolving around in a radius R0 R=Radius of Earth, M mass of earth M0 mass of partile , analyse its elliptical motion mathematicall

Answer 1

The elliptical orbit is described by the equation $r(\phi) = \frac{a(1 - e^2)}{1 + e \cos(\phi - \phi_0)}$, where $a = \frac{G M R_0}{2 G M - R_0 v^2}$ is the semi-major axis, $e = \sqrt{1 + \frac{(v^2 - 2 G M/R_0) R_0^2 v^2 \cos^2\theta}{(G M)^2}}$ is the eccentricity, and $\phi_0$ is the angle of periapsis. The initial position is at $r=R_0$ and angle $\phi=0$, with initial radial velocity $v_r = v \sin\theta$ and tangential velocity $v_t = v \cos\theta$. The eccentricity $e$ must be between 0 and 1 for an ellipse, which requires $v^2 < 2GM/R_0$.

Answer 2

The motion of a particle under the gravitational force of the Earth is governed by the laws of central forces. The orbit of the particle is a conic section, which is an ellipse in this case. The parameters of the elliptical orbit (semi-major axis $a$ and eccentricity $e$) are determined by the initial conditions: the initial position (radius $R_0$) and the initial velocity (speed $v$ and angle $\theta$ with the local tangent).

Let the initial position be at radius $r = R_0$. Let the initial velocity be $\vec{v}$ with magnitude $v$. The angle between the initial velocity vector and the local tangent is $\theta$. The angle between the initial velocity vector and the initial radius vector is $\alpha = 90^\circ - \theta$ (assuming $\theta$ is measured towards the center) or $\alpha = 90^\circ + \theta$ (assuming $\theta$ is measured away from the center). Let's assume the angle between the velocity vector and the radius vector is $\alpha$. Then the radial component of velocity is $v_r = v \cos\alpha$ and the tangential component is $v_t = v \sin\alpha$. The problem states the angle is $\theta$ with the orbit (tangent). So, the tangential component is $v_t = v \cos\theta$ and the radial component is $v_r = v \sin\theta$ (assuming $\theta$ is the angle between $\vec{v}$ and the tangent, and the radial component is positive outward).

The total mechanical energy $E$ and the angular momentum $\vec{L}$ are conserved.

Initial total energy $E = \frac{1}{2} M_0 v^2 - \frac{G M M_0}{R_0}$.

Initial angular momentum magnitude $L = |\vec{r}_0 \times M_0 \vec{v}_0| = R_0 M_0 v_t = R_0 M_0 v \cos\theta$.

For an elliptical orbit, the total energy $E$ must be negative:

$E < 0 \implies \frac{1}{2} M_0 v^2 < \frac{G M M_0}{R_0} \implies v^2 < \frac{2 G M}{R_0}$. This means the initial speed $v$ must be less than the escape velocity at $R_0$.

The parameters of the ellipse are related to $E$ and $L$:

Semi-major axis $a = -\frac{G M M_0}{2 E} = -\frac{G M M_0}{2 (\frac{1}{2} M_0 v^2 - \frac{G M M_0}{R_0})} = \frac{G M}{\frac{2 G M}{R_0} - v^2} = \frac{G M R_0}{2 G M - R_0 v^2}$.

Eccentricity $e$ is given by $L^2 = G M M_0^2 a (1 - e^2)$.

$(R_0 M_0 v \cos\theta)^2 = G M M_0^2 \frac{G M R_0}{2 G M - R_0 v^2} (1 - e^2)$

$R_0^2 v^2 \cos^2\theta = \frac{(G M)^2 R_0}{2 G M - R_0 v^2} (1 - e^2)$

$1 - e^2 = \frac{R_0^2 v^2 \cos^2\theta (2 G M - R_0 v^2)}{(G M)^2 R_0} = \frac{R_0 v^2 \cos^2\theta (2 G M - R_0 v^2)}{(G M)^2}$

$e^2 = 1 - \frac{R_0 v^2 \cos^2\theta (2 G M - R_0 v^2)}{(G M)^2}$.

Alternatively, $e^2 = 1 + \frac{2 E L^2}{(G M M_0)^2} = 1 + \frac{2 (\frac{1}{2} M_0 v^2 - \frac{G M M_0}{R_0}) (R_0 M_0 v \cos\theta)^2}{(G M M_0)^2}$

$e^2 = 1 + \frac{M_0 (\frac{1}{2} v^2 - \frac{G M}{R_0}) 2 R_0^2 M_0^2 v^2 \cos^2\theta}{G^2 M^2 M_0^2} = 1 + \frac{(v^2 - \frac{2 G M}{R_0}) R_0^2 v^2 \cos^2\theta}{G^2 M^2}$

$e^2 = 1 + \frac{v^4 R_0^2 \cos^2\theta}{G^2 M^2} - \frac{2 G M}{R_0} \frac{R_0^2 v^2 \cos^2\theta}{G^2 M^2} = 1 + \frac{v^4 R_0^2 \cos^2\theta}{G^2 M^2} - \frac{2 R_0 v^2 \cos^2\theta}{G M}$.

The equation of the elliptical orbit in polar coordinates with the Earth at the origin is:

$r(\phi) = \frac{a(1 - e^2)}{1 + e \cos(\phi - \phi_0)}$

where $\phi$ is the angle from the direction of periapsis (closest approach).

Let's set the initial position at $\phi = 0$. So $r(0) = R_0$.

$R_0 = \frac{a(1 - e^2)}{1 + e \cos(-\phi_0)} = \frac{a(1 - e^2)}{1 + e \cos\phi_0}$.

We also need to determine $\phi_0$. The radial velocity at the initial point is $v_r = v \sin\theta$.

The radial velocity in polar coordinates is $\dot{r} = \frac{d}{dt} \left( \frac{a(1 - e^2)}{1 + e \cos(\phi - \phi_0)} \right) = a(1 - e^2) \frac{-1}{(1 + e \cos(\phi - \phi_0))^2} (-e \sin(\phi - \phi_0)) \dot{\phi}$

$\dot{r} = \frac{a(1 - e^2) e \sin(\phi - \phi_0)}{(1 + e \cos(\phi - \phi_0))^2} \dot{\phi}$.

Using $r = \frac{a(1 - e^2)}{1 + e \cos(\phi - \phi_0)} \implies (1 + e \cos(\phi - \phi_0)) = \frac{a(1 - e^2)}{r}$.

$\dot{r} = \frac{r^2}{a(1 - e^2)} e \sin(\phi - \phi_0) \dot{\phi}$.

Using conservation of angular momentum, $r^2 \dot{\phi} = h = L/M_0 = R_0 v \cos\theta$.

$\dot{r} = \frac{h}{a(1 - e^2)} e \sin(\phi - \phi_0)$.

At the initial point ($\phi=0, r=R_0, \dot{r}_0 = v \sin\theta$):

$v \sin\theta = \frac{R_0 v \cos\theta}{a(1 - e^2)} e \sin(-\phi_0) = -\frac{R_0 v \cos\theta}{a(1 - e^2)} e \sin\phi_0$.

$\frac{v \sin\theta a(1 - e^2)}{R_0 v \cos\theta} = -e \sin\phi_0$

$\frac{\tan\theta a(1 - e^2)}{R_0} = -e \sin\phi_0$.

From $R_0 = \frac{a(1 - e^2)}{1 + e \cos\phi_0}$, we have $1 + e \cos\phi_0 = \frac{a(1 - e^2)}{R_0}$.

$e \cos\phi_0 = \frac{a(1 - e^2)}{R_0} - 1$.

We have two equations for $e$ and $\phi_0$:

$e \cos\phi_0 = \frac{a(1 - e^2)}{R_0} - 1$

$e \sin\phi_0 = -\frac{a(1 - e^2)}{R_0} \tan\theta$

Squaring and adding gives $e^2 = (\frac{a(1 - e^2)}{R_0} - 1)^2 + (\frac{a(1 - e^2)}{R_0})^2 \tan^2\theta$.

This is consistent with the earlier expression for $e^2$ derived from $E$ and $L$.

The mathematical analysis involves calculating the parameters of the ellipse ($a, e, \phi_0$) from the initial conditions ($R_0, v, \theta$) using the conservation of energy and angular momentum. The equation of the orbit in polar coordinates describes the path of the particle. The position and velocity of the particle at any point on the orbit can be derived from the orbit equation.

The semi-major axis $a = \frac{G M R_0}{2 G M - R_0 v^2}$.

The eccentricity $e = \sqrt{1 + \left(\frac{v^2}{G M} - \frac{2}{R_0}\right) \frac{R_0^2 v^2 \cos^2\theta}{1}}$. (Using $e^2 = 1 + \frac{2 E h^2}{(G M)^2}$ where $h = L/M_0$)

$h = R_0 v \cos\theta$. $E/M_0 = \frac{1}{2} v^2 - \frac{G M}{R_0}$.

$e^2 = 1 + \frac{2 (\frac{1}{2} v^2 - \frac{G M}{R_0}) (R_0 v \cos\theta)^2}{(G M)^2} = 1 + \frac{(v^2 - \frac{2 G M}{R_0}) R_0^2 v^2 \cos^2\theta}{(G M)^2}$.

The angle of periapsis $\phi_0$ can be found from $\tan\phi_0 = \frac{e \sin\phi_0}{e \cos\phi_0} = \frac{-\frac{a(1 - e^2)}{R_0} \tan\theta}{\frac{a(1 - e^2)}{R_0} - 1} = \frac{- \tan\theta}{1 - \frac{R_0}{a(1 - e^2)}}$.

We know $R_0 = \frac{a(1 - e^2)}{1 + e \cos\phi_0}$, so $\frac{R_0}{a(1 - e^2)} = \frac{1}{1 + e \cos\phi_0}$.

$\tan\phi_0 = \frac{-\tan\theta}{1 - \frac{1}{1 + e \cos\phi_0}} = \frac{-\tan\theta}{\frac{1 + e \cos\phi_0 - 1}{1 + e \cos\phi_0}} = \frac{-\tan\theta (1 + e \cos\phi_0)}{e \cos\phi_0}$.

This doesn't simplify nicely. Let's use the initial velocity components.

Initial radial velocity $\dot{r}_0 = v \sin\theta$. Initial tangential velocity $v_{t0} = R_0 \dot{\phi}_0 = v \cos\theta$.

The equation of the orbit is $r(\phi)$.

$\frac{dr}{d\phi} = \frac{dr}{dt} \frac{dt}{d\phi} = \frac{\dot{r}}{\dot{\phi}}$.

At the initial point $\phi=0$: $(\frac{dr}{d\phi})_{\phi=0} = \frac{\dot{r}_0}{\dot{\phi}_0} = \frac{v \sin\theta}{v \cos\theta / R_0} = R_0 \tan\theta$.

From $r(\phi) = \frac{C}{1 + e \cos(\phi - \phi_0)}$, where $C = a(1 - e^2)$.

$\frac{dr}{d\phi} = C \frac{-1}{(1 + e \cos(\phi - \phi_0))^2} (-e \sin(\phi - \phi_0)) = \frac{C e \sin(\phi - \phi_0)}{(1 + e \cos(\phi - \phi_0))^2}$.

At $\phi = 0$: $(\frac{dr}{d\phi})_{\phi=0} = \frac{C e \sin(-\phi_0)}{(1 + e \cos(-\phi_0))^2} = \frac{-C e \sin\phi_0}{(1 + e \cos\phi_0)^2}$.

We know $R_0 = \frac{C}{1 + e \cos\phi_0}$, so $1 + e \cos\phi_0 = C/R_0$.

$(\frac{dr}{d\phi})_{\phi=0} = \frac{-C e \sin\phi_0}{(C/R_0)^2} = \frac{-C e \sin\phi_0 R_0^2}{C^2} = -\frac{e \sin\phi_0 R_0^2}{C}$.

So, $R_0 \tan\theta = -\frac{e \sin\phi_0 R_0^2}{a(1 - e^2)}$.

$\tan\theta = -\frac{e \sin\phi_0 R_0}{a(1 - e^2)}$.

We also have $R_0 = \frac{a(1 - e^2)}{1 + e \cos\phi_0}$, so $\frac{R_0}{a(1 - e^2)} = \frac{1}{1 + e \cos\phi_0}$.

$\tan\theta = -e \sin\phi_0 \frac{1}{1 + e \cos\phi_0}$.

$e \sin\phi_0 = -\tan\theta (1 + e \cos\phi_0)$.

This is the same as the previous equation for $e \sin\phi_0$.

The state of the elliptical orbit is fully described by the parameters $a$ and $e$, and the orientation $\phi_0$. These are calculated from the initial conditions.