Question

Consider a spacecraft in an elliptical orbit around the earth. At the lowest point or perigee, of its orbit it is $300 km$ above the earth’s surface at the highest point or apogee, it is $3000 km$ above the earth’s surface.

(a) What is the period of the spacecraft’s orbit ?

(b) Find the ratio of the spacecraft’s speed at perigee to its speed at apogee.

(c) Find the speed at perigee and the speed at apogee.

(d) It is desired to have the spacecraft escape from the earth completely. If the spacecraft’s rockets are fired at perigee, by how much would the speed have to be increased to achieve this? What if the rockets were fired at apogee ? Which point in the orbit is the most efficient to use? (Let the radius of earth is $R = 6400 km$ )

Accepted Answer

The angular momentum of the spacecraft will be conserved. The motion of the spacecraft is shown in figure below,

The radius of the perigee is given by $3000 km$ and radius of apogee will be $r_{a} = R + h_{a} = 6400 + 3000 = 9400 km$ . Let the mass of the spacecraft be $m$ , then the average gravitational force is balanced by the average centrifugal force, i.e.,

$\frac{G M m}{a^{2}} = \frac{m v^{2}}{a}$ ; where $v$ is average speed of spacecraft, $M$ is mass of earth and $a = (\frac{r_{a} + r_{p}}{2})$ is average radius.

$v = \sqrt{\frac{G M}{a}} . . . (1)$

(a) The time period of the spacecraft is given by,

$T = \frac{2 π a}{v}$

$T = \frac{2 π a}{\sqrt{\frac{G M}{a}}} = \frac{2 π a^{3 / 2}}{R \sqrt{g}}$ ; since acceleration due to gravity will be $g = \frac{G M}{R^{2}}$ .

$T = \frac{2 \times π \times {(\frac{9400 \times 10^{3} + 6700 \times 10^{3}}{2})}^{3 / 2}}{6400 \times 10^{3} \times \sqrt{9.8}}$

$T = 7.16 \times 10^{3} s$

(b). The angular momentum of the spacecraft at perigee and apogee will be equal, i.e.,

$m v_{p} r_{p} = m v_{a} r_{a}$

Hence, the ratio of the spacecraft’s speed at perigee to its speed at apogee will be,

$\frac{v_{p}}{v_{a}} = \frac{r_{a}}{r_{p}} = \frac{9400}{6700} = \frac{94}{67} = 1.4$

(c) Apply conservation of mechanical energy at perigee and apogee, we get,

$- \frac{G M m}{r_{p}} + \frac{1}{2} m v_{p}^{2} = - \frac{G M m}{r_{a}} + \frac{1}{2} m v_{a}^{2}$

$v_{p}^{2} - v_{a}^{2} = 2 G M (\frac{1}{r_{p}} - \frac{1}{r_{a}}) . . . (2)$

And we have the ratio of velocity, i.e.,

$\frac{v_{p}}{v_{a}} = \frac{r_{a}}{r_{p}} . . . (3)$

From equation $(2)$ and $(3)$ , we get,

$v_{a} = \sqrt{\frac{G M r_{p}}{r_{a} (r_{a} + r_{p})}} = \sqrt{\frac{R^{2} g r_{p}}{r_{a} (r_{a} + r_{p})}}$

$v_{a} = \sqrt{\frac{{(6400 \times 10^{3})}^{2} \times 9.8 \times 6700 \times 10^{3}}{9400 \times 10^{3} (6700 \times 10^{3} + 9400 \times 10^{3})}} = 5.95 km s^{- 1}$

And $v_{p} = 1.4 v_{a} = 1.4 \times 5.95 = 8.33 km s^{- 1}$

(d) Let the change in velocity is $∆ v$ . From conservation of energy, the gravitational potential energy of the spacecraft will be converted into the kinetic energy of the escape velocity,

$\frac{1}{2} m {(v_{p} + ∆ v)}^{2} = \frac{G M m}{r_{p}}$

$v_{p} + ∆ v = \sqrt{\frac{2 G M}{r_{p}}} = \sqrt{\frac{2 R^{2} g}{r_{p}}}$

$∆ v = \sqrt{\frac{2 R^{2} g}{r_{p}}} - v_{p} = \sqrt{\frac{2 \times {(6400 \times 10^{3})}^{2} 9.8}{6700 \times 10^{3}}} - 8.33 \times \times 10^{3}$

$∆ v = 3.09 km s^{- 1}$ .

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