Question

A particle is given some speed $v_0$ at a height h = R above the surface of earth such that it revolves around earth. The minimum value of speed $v_0$ required is $\sqrt{\frac{\alpha GM}{\beta R}}$. Find the value of $(\alpha + \beta)$

(Assume the mass of the earth is M and its radius is R)

Answer 1

4

Answer 2

The particle is at a height $h = R$ above the surface of the Earth. The radius of the Earth is $R$. So, the distance of the particle from the center of the Earth is $r = R + h = R + R = 2R$.

We assume that the speed $v_0$ is given horizontally (tangentially to a circle around the Earth). For the particle to revolve around the Earth in a closed orbit without hitting the Earth's surface, the perigee distance of the orbit must be greater than or equal to the radius of the Earth, $R$.

When a particle is at a distance $r_a = 2R$ from the center of the Earth and has a tangential speed $v_0$, this point is the apogee of the elliptical orbit if $v_0 < v_{orbit} = \sqrt{\frac{GM}{2R}}$.

The total energy of the particle is $E = \frac{1}{2}mv_0^2 - \frac{GMm}{r_a} = \frac{1}{2}mv_0^2 - \frac{GMm}{2R}$. The angular momentum of the particle is $L = mr_a v_0 = m(2R)v_0$.

For an elliptical orbit, the semi-major axis is given by $a = -\frac{GMm}{2E} = -\frac{GMm}{2(\frac{1}{2}mv_0^2 - \frac{GMm}{2R})} = \frac{GM}{ \frac{GM}{R} - v_0^2}$. The distances of apogee ($r_a$) and perigee ($r_p$) are related to the semi-major axis ($a$) and eccentricity ($e$) by $r_a = a(1+e)$ and $r_p = a(1-e)$. Also, $r_a + r_p = 2a$. Given $r_a = 2R$, we have $2R + r_p = 2a$. $r_p = 2a - 2R = 2 \left( \frac{GM}{ \frac{GM}{R} - v_0^2 } \right) - 2R$.

For the orbit to not intersect the Earth's surface, the perigee distance must be greater than or equal to the radius of the Earth, i.e., $r_p \ge R$. $2 \left( \frac{GM}{ \frac{GM}{R} - v_0^2 } \right) - 2R \ge R$ $2 \left( \frac{GM}{ \frac{GM}{R} - v_0^2 } \right) \ge 3R$ $\frac{2GM}{ \frac{GM}{R} - v_0^2 } \ge 3R$ $2GM \ge 3R \left( \frac{GM}{R} - v_0^2 \right)$ $2GM \ge 3GM - 3Rv_0^2$ $3Rv_0^2 \ge 3GM - 2GM$ $3Rv_0^2 \ge GM$ $v_0^2 \ge \frac{GM}{3R}$ $v_0 \ge \sqrt{\frac{GM}{3R}}$

The minimum value of speed $v_0$ required for the particle to revolve around the Earth (in a closed orbit) is $v_{0,min} = \sqrt{\frac{GM}{3R}}$. The given form of the minimum speed is $\sqrt{\frac{\alpha GM}{\beta R}}$. Comparing the two expressions, we have: $\sqrt{\frac{GM}{3R}} = \sqrt{\frac{\alpha GM}{\beta R}}$ $\frac{GM}{3R} = \frac{\alpha GM}{\beta R}$ $\frac{1}{3} = \frac{\alpha}{\beta}$

We need to find integer values for $\alpha$ and $\beta$ that satisfy this equation. The simplest integer values are $\alpha = 1$ and $\beta = 3$. We are asked to find the value of $(\alpha + \beta)$. $\alpha + \beta = 1 + 3 = 4$.