Post-Test Quiz: Introduction to Orbital Mechanics

Q1 — Answer (b). LEVEL 1
Kepler's first law (law of orbits): every satellite moves in an ellipse with the Earth's center at one focus. A circle is the special case where both foci coincide. Parabolic (option d) and hyperbolic paths are escape trajectories, not bound orbits.

Q2 — Answer (c). LEVEL 1
Kepler's second law: the radius vector sweeps equal areas in equal times. To sweep the same area near the Earth (where the satellite is close) as far away, the satellite must move faster at perigee and slower at apogee.

Q3 — Answer (b). LEVEL 1
Perigee = closest point to Earth; apogee = farthest point. (The ascending node is where the orbit crosses the equator going north; the foci are geometric points of the ellipse.)

Q4 — Answer (c). LEVEL 1
Kepler's third law: T² = 4π²a³/μ — the period depends only on the semi-major axis a (the orbit's average radius), not on eccentricity, inclination, or satellite mass.

Q5 — Answer (a). LEVEL 1
ε = −μ/(2a): the specific orbital energy (energy per unit mass) depends only on the semi-major axis. This is why two satellites in orbits with the same a — one circular, one highly elliptical — have the same period and the same energy, even though their speeds and altitudes vary along the orbit.

Q6 — Answer (b). LEVEL 2
Gravity exerts no torque about the Earth's center, so angular momentum h = r×v is conserved. At perigee the radius r is smallest, so the speed v must be largest to keep h constant (this is exactly Kepler's second law).

Q7 — Answer (b). LEVEL 2
A 24-hour period only makes the satellite geosynchronous. If the orbit is inclined, the satellite's latitude oscillates between +i and −i each day while its longitude stays fixed — tracing a figure-eight (analemma) in the sky. Only a zero-inclination (equatorial) geosynchronous orbit is geostationary (appears fixed). Communication antennas could track such a path, but fixed dishes could not.

Q8 — Answer (c). LEVEL 2
In the two-body problem with no external forces, both the specific energy ε = −μ/2a and the specific angular momentum h are conserved. Speed and altitude exchange forms of energy (kinetic ↔ potential) around the ellipse, but their sum is constant.

Q9 — Answer (b). LEVEL 2
Adding speed at a point raises the energy, increasing the semi-major axis. Since the burn point is now the closest point of the new orbit, it becomes the perigee of a larger ellipse. This is precisely how a GTO is created: a burn at LEO perigee raises apogee to GEO altitude. (Above escape speed the orbit would open into a parabola, option c.)

Q10 — Answer (b). LEVEL 2
Drag removes energy, shrinking the semi-major axis (ε = −μ/2a becomes more negative). From Kepler's third law the period shortens, and since v = √(μ(2/r − 1/a)), lowering a while r decreases makes the satellite speed up as it spirals down — orbital "decay." This is the famous paradox: friction makes a satellite go faster while it falls.

Q11 — Answer (b). LEVEL 3
v = √(μ/r) = √(3.986×10⁹⁴/7.371×10⁹) = √(5.41×10⁷) ≈ 7,355 m/s ≈ 7.3 km/s
Compare with 7.6 km/s at 500 km: speed decreases as altitude increases (weaker gravity requires less centripetal force).

Q12 — Answer (b). LEVEL 3
T = 2π√(a³/μ) = 2π√((4.2164×10⁹)³/3.986×10⁹⁴) ≈ 86,160 s ≈ 24 h
This is no coincidence: the GEO altitude of 35,786 km is derived by requiring the period to equal the Earth's rotation period. 100 minutes would be a LEO period.

Q13 — Answer (c). LEVEL 3
v_p = √[μ(2/r_p − 1/a)] = √[3.986×10⁹⁴(2/7.0×10⁹ − 1/2.4582×10⁹)] ≈ 9.9 km/s
The satellite whips through perigee at ~9.9 km/s and crawls over apogee at ~1.6 km/s — a ratio equal to r_a/r_p ≈ 6, illustrating Kepler's second law dramatically.

Q14 — Answer (c). LEVEL 3
v_esc = √(2μ/r) = √(2 × 3.986×10⁹⁴/6.871×10⁹) ≈ 10.8 km/s
Note v_esc = √2 × v_circular ≈ 1.414 × 7.6 km/s. The famous 11.2 km/s applies only at the Earth's surface (r = 6,371 km).

Q15 — Answer (b). LEVEL 3
a = [μT²/4π²]⅓ = [3.986×10⁹⁴(28,800)²/39.478]⅓ ≈ 2.03×10⁹ m ≈ 20,300 km
(altitude ≈ 13,900 km — typical MEO navigation territory; GPS, with T ≈ 11 h 58 min, sits at a ≈ 26,560 km, option c.)