Question

A particle is projected from the surface of the earth (of radius $R_e$ and mass $M_e$) with speed equal to $\sqrt{\frac{GM_e}{R_e}}$, at certain angle from local horizontal as shown in the figure such that the angle subtended by arc between launching and landing site at earth's centre is $\beta=60^\circ$. Calculate the maximum separation of the particle from the centre of the earth. Consider earth to be uniformly dense and air resistance to be absent.

Answer 1

$R_e\left(1+\frac{\sqrt{3}}{2}\right)$

Answer 2

The problem describes the motion of a particle projected from the Earth's surface. This is a problem of two-body motion under a central gravitational force, where the trajectory is an ellipse with the Earth's center at one focus.

1. Calculate the total mechanical energy (E) of the particle: The particle is projected from the surface of the Earth ($r = R_e$) with speed $v = \sqrt{\frac{GM_e}{R_e}}$. The total mechanical energy $E$ is the sum of kinetic energy ($K$) and potential energy ($U$): $E = K + U = \frac{1}{2}mv^2 - \frac{GM_e m}{r}$ Substitute the given values: $E = \frac{1}{2}m\left(\sqrt{\frac{GM_e}{R_e}}\right)^2 - \frac{GM_e m}{R_e}$ $E = \frac{1}{2}m\frac{GM_e}{R_e} - \frac{GM_e m}{R_e}$ $E = \frac{GM_e m}{2R_e} - \frac{GM_e m}{R_e} = -\frac{GM_e m}{2R_e}$

2. Relate energy to the semi-major axis (a) of the elliptical orbit: For an elliptical orbit, the total mechanical energy is given by: $E = -\frac{GM_e m}{2a}$ Comparing this with the calculated energy: $-\frac{GM_e m}{2a} = -\frac{GM_e m}{2R_e}$ This implies $a = R_e$. So, the semi-major axis of the elliptical orbit is equal to the radius of the Earth.

3. Determine the eccentricity (e) of the orbit: The launching point (P) and landing point (Q) are both on the surface of the Earth, so their distance from the Earth's center (focus O) is $R_e$. The angle subtended by the arc between launching and landing site at the Earth's center is $\beta = 60^\circ$. Let P and Q be the launching and landing points, respectively. So, $\angle POQ = 60^\circ$. For an ellipse, the general equation for the distance $r$ from the focus is: $r = \frac{a(1-e^2)}{1+e \cos \nu}$ where $\nu$ is the true anomaly (angle measured from the perigee). Since $OP = OQ = R_e$, and $a=R_e$, we have: $R_e = \frac{R_e(1-e^2)}{1+e \cos \nu_P}$ and $R_e = \frac{R_e(1-e^2)}{1+e \cos \nu_Q}$ This implies $1+e \cos \nu_P = 1-e^2$ and $1+e \cos \nu_Q = 1-e^2$. So, $e \cos \nu_P = -e^2$ and $e \cos \nu_Q = -e^2$. Since the orbit is not circular (it lands at a different point, implying $e \ne 0$), we can divide by $e$: $\cos \nu_P = -e$ and $\cos \nu_Q = -e$. This means $\nu_P$ and $\nu_Q$ are angles such that their cosine is $-e$. Given that $\beta = |\nu_P - \nu_Q| = 60^\circ$. For $\cos \nu_P = \cos \nu_Q = -e$, the angles $\nu_P$ and $\nu_Q$ must be symmetric with respect to the major axis (perigee direction). So, if the perigee is at $\nu=0$, then $P$ is at $\nu = \nu_0$ and $Q$ is at $\nu = -\nu_0$. The angular separation is $2\nu_0 = \beta = 60^\circ$, which means $\nu_0 = 30^\circ$. Therefore, $\cos(30^\circ) = -e$. $\frac{\sqrt{3}}{2} = -e$. This gives $e = -\frac{\sqrt{3}}{2}$, which is not possible as eccentricity $e$ must be non-negative ($0 \le e < 1$ for an ellipse).

   This indicates that the perigee is not between the launching and landing points in terms of angular position. Instead, the major axis must be such that both points P and Q are on one side of the major axis or the other side. The true anomaly $\nu$ is measured from the perigee. Let the launching point be at true anomaly $\nu_1$ and the landing point be at $\nu_2$. We have $\cos \nu_1 = -e$ and $\cos \nu_2 = -e$. This means $\nu_1$ and $\nu_2$ must be such that $\cos \nu = -e$. Since $e$ must be positive, $-e$ is negative. So $\nu_1$ and $\nu_2$ must be in the second or third quadrant. So, $\nu_1 = \pi - \theta$ and $\nu_2 = \pi + \theta$ for some angle $\theta$, or vice versa. The angular separation between these points is $\beta = |\nu_1 - \nu_2| = 60^\circ$. Let's assume the perigee is at $\nu=0$. Then the two points P and Q are at true anomalies $\nu_P$ and $\nu_Q$ such that $\cos \nu_P = \cos \nu_Q = -e$. This implies $\nu_Q = 2\pi - \nu_P$ (or $\nu_Q = -\nu_P$). The angle between the position vectors $OP$ and $OQ$ is $\beta = 60^\circ$. So, $\nu_P - \nu_Q = 60^\circ$ (or $360^\circ - 60^\circ$). This means that one point is at $\nu_P$ and the other is at $\nu_P + 60^\circ$. So, $\cos \nu_P = -e$ and $\cos (\nu_P + 60^\circ) = -e$. This implies $\cos \nu_P = \cos (\nu_P + 60^\circ)$. For this equality to hold, either $\nu_P = \pm (\nu_P + 60^\circ) + 2k\pi$. Case 1: $\nu_P = \nu_P + 60^\circ + 2k\pi \implies 0 = 60^\circ + 2k\pi$, which is not possible. Case 2: $\nu_P = -(\nu_P + 60^\circ) + 2k\pi$ $2\nu_P = -60^\circ + 2k\pi$ $\nu_P = -30^\circ + k\pi$. Since $\nu_P$ is a true anomaly, it usually ranges from $0$ to $2\pi$. If $k=1$, $\nu_P = -30^\circ + 180^\circ = 150^\circ$. Then $\cos(150^\circ) = -e$. $-\frac{\sqrt{3}}{2} = -e \implies e = \frac{\sqrt{3}}{2}$. This is a valid eccentricity ($0 < e < 1$).

4. Calculate the maximum separation from the center of the Earth: The maximum separation is the apogee distance, $r_{max}$. $r_{max} = a(1+e)$. Substitute $a=R_e$ and $e=\frac{\sqrt{3}}{2}$: $r_{max} = R_e \left(1 + \frac{\sqrt{3}}{2}\right)$ $r_{max} = R_e \left(\frac{2+\sqrt{3}}{2}\right)$

The maximum separation of the particle from the centre of the earth is $R_e \left(1 + \frac{\sqrt{3}}{2}\right)$.

Explanation of the solution:

1. Calculate the total mechanical energy of the particle using its initial speed and position.
2. Relate this total energy to the semi-major axis ($a$) of the elliptical orbit, finding $a = R_e$.
3. Use the fact that the launching and landing points are at radius $R_e$ and are separated by an angle of $60^\circ$. Apply the polar equation of an ellipse $r = \frac{a(1-e^2)}{1+e \cos \nu}$.
4. From $r=R_e$ and $a=R_e$, deduce that $\cos \nu = -e$ for both points.
5. Since the two points are separated by $60^\circ$ and have the same $\cos \nu$ value, their true anomalies must be $\nu$ and $\nu+60^\circ$ (or $\nu$ and $\nu-60^\circ$) where $\cos \nu = \cos(\nu \pm 60^\circ)$. This implies $\nu = 150^\circ$ (or $210^\circ$).
6. Calculate the eccentricity $e = -\cos(150^\circ) = \sqrt{3}/2$.
7. The maximum separation is the apogee distance $r_{max} = a(1+e)$. Substitute $a=R_e$ and $e=\sqrt{3}/2$ to get the result.