A spacecraft is launched from Earth at exactly escape velocity, then its engines cut off. Ignoring atmospheric drag, which trajectory will it follow?
AIt falls back to Earth — gravity always pulls objects back eventually
BIt enters a stable circular orbit at a very high altitude
CIt follows a parabolic trajectory and arrives at infinite distance with zero velocity
DIt follows a hyperbolic trajectory and retains kinetic energy at infinite distance
Escape velocity is defined as the minimum speed for which the total mechanical energy E = 0. With E = 0, the trajectory is a parabola, and as r → ∞ the kinetic energy approaches zero (all energy was 'used' against gravity). Option A is wrong because E = 0 is exactly the threshold of escape. Option D (hyperbola) requires E > 0, meaning the launch speed exceeded escape velocity. The parabolic trajectory is the knife-edge case — any slightly less speed produces a bound elliptical orbit.
Question 2 Multiple Choice
For a satellite in a circular orbit at radius r, how does its kinetic energy relate to its gravitational potential energy?
AKE = |PE| — kinetic and potential energy are equal in magnitude
BKE = ½|PE| — kinetic energy is half the magnitude of the potential energy
CKE = 2|PE| — kinetic energy is twice the potential energy
DKE = |PE| / r — the relationship depends on orbital radius
From the circular orbit condition GMm/r² = mv²/r, we get v² = GM/r, so KE = ½mv² = GMm/(2r). Since PE = −GMm/r, we have KE = −PE/2, or KE = ½|PE|. This is the virial theorem for gravitational systems. It also means total energy E = KE + PE = GMm/(2r) − GMm/r = −GMm/(2r) = −KE, so the total energy is always negative (bound) and equals negative of the kinetic energy.
Question 3 True / False
A comet observed to follow a hyperbolic trajectory through the inner solar system has positive total mechanical energy.
TTrue
FFalse
Answer: True
Orbit type is completely determined by the sign of total energy E = KE + PE. E < 0: bound elliptical orbit. E = 0: parabolic trajectory (barely escapes). E > 0: hyperbolic trajectory — the object has more than enough energy to escape and retains kinetic energy at infinite distance. A comet on a hyperbolic path is an unbound visitor making one pass through the solar system; it was never gravitationally captured. This also applies to interstellar objects like 'Oumuamua.
Question 4 True / False
An object launched from Earth's surface at escape velocity is expected to be aimed straight up; launching at an angle requires a higher initial speed to escape.
TTrue
FFalse
Answer: False
Gravity is a conservative force, so the work done by gravity depends only on the initial and final positions (radii), not on the path taken. For any trajectory that starts at radius R with speed v_esc = √(2GM/R), the total energy is exactly E = 0, regardless of launch direction. The object will escape no matter what angle it is launched at, as long as the speed equals v_esc. (In practice, atmosphere and terrain matter, but in idealized point-mass mechanics, direction is irrelevant to escape.)
Question 5 Short Answer
Explain why all elliptical orbits with the same semi-major axis have the same orbital period, even if their shapes (eccentricities) are very different.
Think about your answer, then reveal below.
Model answer: The total mechanical energy of an elliptical orbit is E = −GMm/(2a), where a is the semi-major axis. Since energy depends only on a, two orbits with the same a have the same energy regardless of eccentricity. Kepler's third law (T² ∝ a³) follows from this energy relationship combined with angular momentum. A circular orbit and a highly elongated ellipse with the same a have the same period because they have the same total energy — the elongated ellipse moves slowly at apogee and very fast at perigee, averaging out to the same orbital period.
This is one of the most surprising results in orbital mechanics. The 'size' of the orbit (semi-major axis) alone determines both the energy and the period — not the shape. A nearly radial orbit (very high eccentricity) and a nearly circular orbit (eccentricity ≈ 0) with the same a are energetically equivalent and have the same period. This follows because E = −GMm/(2a) is the exact energy for any conic section orbit.