Orbital Mechanics — Kepler & Newton
Orbital Mechanics — Kepler & Newton
Spacecraft trajectories are governed by Newton's law of gravitation and Kepler's laws of planetary motion.
$$F = \frac{G M m}{r^2}$$
where $G = 6.674 \times 10^{-11}$ N·m²/kg², $M$ = central body mass, $m$ = satellite mass, $r$ = distance from center.
$$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{\mu}{r}}$$
where $\mu = GM$ is the gravitational parameter. For Earth: $\mu = 3.986 \times 10^{14}$ m³/s².
$$T = 2\pi\sqrt{\frac{a^3}{\mu}}$$
where $a$ = semi-major axis and $T$ = orbital period.
Find the orbital velocity and period for a satellite in a circular orbit at altitude $h = 400$ km above Earth ($R_E = 6{,}371$ km).
$r = R_E + h = 6{,}771$ km $= 6.771 \times 10^6$ m
$$v = \sqrt{\frac{3.986 \times 10^{14}}{6.771 \times 10^6}} = 7{,}672 \text{ m/s} \approx 7.67 \text{ km/s}$$
$$T = \frac{2\pi r}{v} = \frac{2\pi \times 6.771 \times 10^6}{7{,}672} = 5{,}542 \text{ s} \approx 92.4 \text{ min}$$
Find the altitude of a geostationary orbit ($T = 24$ hours).
$T = 86{,}400$ s. From Kepler's third law:
$$a = \left(\frac{\mu T^2}{4\pi^2}\right)^{1/3} = \left(\frac{3.986 \times 10^{14} \times 86{,}400^2}{4\pi^2}\right)^{1/3} = 42{,}164 \text{ km}$$
Altitude: $h = 42{,}164 - 6{,}371 = 35{,}793$ km
What escape velocity is needed from Earth's surface?
$$v_{\text{esc}} = \sqrt{\frac{2\mu}{R_E}} = \sqrt{\frac{2 \times 3.986 \times 10^{14}}{6.371 \times 10^6}} = 11{,}186 \text{ m/s} \approx 11.2 \text{ km/s}$$
Practice Problems
Show Answer Key
1. $r = 6{,}571$ km; $v = \sqrt{3.986 \times 10^{14}/6.571 \times 10^6} = 7{,}788$ m/s
2. $r = 6{,}779$ km; $T = 2\pi\sqrt{(6.779\times10^6)^3/3.986\times10^{14}} = 5{,}554$ s $\approx 92.6$ min
3. $r = 1{,}837$ km; $v = \sqrt{4.905\times10^{12}/1.837\times10^6} = 1{,}634$ m/s
4. $T = 43{,}200$ s; $a = (\mu T^2/4\pi^2)^{1/3} = 26{,}560$ km
5. $v = \sqrt{2 \times 4.283\times10^{13}/3.39\times10^6} = 5{,}027$ m/s
6. $a = (r_p + r_a)/2 = (6{,}600 + 42{,}000)/2 = 24{,}300$ km
7. $T = 2\pi\sqrt{a^3/\mu} = 37{,}729$ s $\approx 10.5$ hours
8. $v_p = \sqrt{3.986\times10^{14}(2/6.6\times10^6 - 1/2.43\times10^7)} = 9{,}876$ m/s
9. $v_{\text{circ}} = 7{,}784$ m/s; $a_{\text{transfer}} = (6{,}571 + 42{,}157)/2 = 24{,}364$ km; $v_{\text{transfer}} = 10{,}152$ m/s at perigee; $\Delta v_1 \approx 2{,}368$ m/s
10. $v \propto 1/\sqrt{r}$ — velocity decreases with altitude
11. $T \propto r^{3/2}$; factor $= 2^{3/2} = 2\sqrt{2} \approx 2.83$
12. $g = \mu/r^2 = 3.986\times10^{14}/(6.771\times10^6)^2 = 8.69$ m/s²