A spacecraft is in an elliptical orbit. At which point does the vis-viva equation predict the highest orbital speed?
AApogee — the highest point, where gravitational potential energy is greatest
BPerigee — the closest point, where r is smallest and 2/r is largest
CThe midpoint of the ellipse, where kinetic and potential energy are equal
DSpeed is constant throughout an elliptical orbit by conservation of energy
The vis-viva equation v² = GM(2/r − 1/a) shows speed depends directly on 2/r. At perigee (closest point), r is minimum, so 2/r is maximum, giving maximum speed. At apogee, r is maximum, so 2/r is minimum, giving minimum speed. Option D is the classic error: energy (not speed) is conserved. Conservation of energy is exactly why speed varies — as the spacecraft moves closer, gravitational potential energy decreases and kinetic energy (speed) increases.
Question 2 Multiple Choice
Using v² = GM(2/r − 1/a), what is the orbital speed of a satellite in a perfectly circular orbit of radius r?
Av = √(2GM/r) — the escape velocity formula
Bv = √(GM/r)
Cv = √(GM/2r)
Dv = √(2GM/a)
For a circular orbit, the semi-major axis equals the orbital radius: a = r. Substituting: v² = GM(2/r − 1/r) = GM(1/r), so v = √(GM/r). Option A is the escape velocity formula, which corresponds to a → ∞ (parabolic orbit). The factor-of-√2 difference between escape speed and circular orbital speed (√(2GM/r) vs √(GM/r)) means escape speed is exactly √2 ≈ 1.41 times circular orbital speed at any given radius.
Question 3 True / False
The vis-viva equation is an independent law of orbital mechanics, separate from and in addition to conservation of energy.
TTrue
FFalse
Answer: False
The vis-viva equation is simply conservation of energy rearranged. Starting from ½v² − GM/r = −GM/(2a) (total specific orbital energy), solving for v² gives v² = GM(2/r − 1/a). No new physics is introduced — only algebra applied to the energy conservation equation. Understanding vis-viva as a restatement of energy conservation is the key insight: any change in orbital geometry (r and a) changes speed in a predictable way because energy is conserved throughout the orbit.
Question 4 True / False
For an escape trajectory (parabolic orbit), the semi-major axis a approaches infinity, so the term 1/a vanishes and vis-viva gives escape speed v_esc = √(2GM/r).
TTrue
FFalse
Answer: True
As the orbit grows larger (more elongated), a → ∞ and total orbital energy −GM/(2a) → 0 (just barely bound). At this limit, vis-viva gives v² = GM(2/r − 0) = 2GM/r, so v_esc = √(2GM/r). The formula interpolates smoothly: circular orbit (a = r) gives v = √(GM/r); larger ellipses (a > r) give higher speed at any given r; parabolic orbit (a → ∞) gives v = √(2GM/r) = √2 × circular speed.
Question 5 Short Answer
Explain using the vis-viva equation why an orbiting body moves faster at perihelion than at aphelion. What does this imply about how a comet on a highly elliptical orbit spends its time?
Think about your answer, then reveal below.
Model answer: The vis-viva equation v² = GM(2/r − 1/a) shows that speed increases as r decreases, since 2/r grows when r shrinks. At perihelion r is smallest, giving maximum speed; at aphelion r is largest, giving minimum speed. For a highly elliptical orbit, the speed at perihelion is enormously faster than at aphelion. A comet therefore moves through the inner solar system rapidly but crawls through the outer solar system — spending most of its orbital period near aphelion, where it moves slowest.
This is why Halley's Comet spends roughly 70 of its 75-year period far from the Sun. The comet's fast, visible passage through the inner solar system occupies only a few years; the remaining decades are spent slowly traversing the outer solar system near aphelion. The same physics governs spacecraft: a Hohmann transfer ellipse has its fastest point at closest approach, which is why burn timing and placement are critical for efficient orbital mechanics. The vis-viva equation makes this speed-distance relationship immediately quantitative.