📝 Kepler's third law formula (27 MCQs)

📖 From Calculus • 13. Vector Valued Functions • 27 questions available

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What is Kepler's third law formula?

Definition:
Kepler's third law states $T^2 = \frac{4\pi^2}{GM} a^3$ for elliptical orbits around mass $M$ .

Example:
Jupiter's moon Io has known $a$ and $T$ , allowing calculation of Jupiter's mass $M$ .

Reason:
It provides a direct method to determine celestial masses from observable orbital parameters.

Easy

Medium

Hard

📝 All Kepler's third law formula MCQs

Q1. A vector-valued function $\mathbf{r}(t)$ describes a satellite orbit where the semi-major axis is doubled while maintaining the same central mass. If the original period was $T$ , how does the magnitude of the average velocity vector over one complete orbit change?

A.It remains exactly the same because orbital shape determines speed.

B.It decreases by a factor of

\sqrt{2}

. ✅

C.It decreases by a factor of

2

D.It increases by a factor of

\sqrt{2}

due to larger path length.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Average velocity magnitude for a closed orbit is zero, but average speed relates to circumference over period. Since $T^2 \propto a^3$ , doubling $a$ increases $T$ by $2^{3/2}$ . Circumference doubles, so average speed scales as $2 / 2^{3/2} = 1/\sqrt{2}$ . Students often confuse instantaneous velocity vectors with scalar average speed in this context.

Q2. When deriving Kepler’s Third Law from Newtonian gravitation using vector calculus, which specific property of the cross product $\mathbf{r} \times \mathbf{v}$ is essential to establish that the areal velocity is constant before relating it to the period?

A.The cross product is always parallel to the position vector.

B.The time derivative of

\mathbf{r} \times \mathbf{v}

is zero for central forces. ✅

C.The magnitude of the cross product equals the total energy.

D.The cross product vanishes at perihelion and aphelion.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Conservation of angular momentum $\mathbf{L} = m(\mathbf{r} \times \mathbf{v})$ requires $d\mathbf{L}/dt = \mathbf{r} \times \mathbf{F} = 0$ for central forces. This constancy ensures equal areas are swept in equal times. Without this vector property, linking geometric area to temporal period via integration would be impossible, making option B the foundational conceptual link.

Q3. An astronomer plots $\log(T)$ versus $\log(a)$ for exoplanets and obtains a slope of 1.4 instead of the theoretical 1.5. Which systematic error in data processing most likely explains this deviation?

A.Using diameter instead of radius for planetary size measurements.

B.Failing to account for the mass of the orbiting planets relative to the star. ✅

C.Using linear regression on non-logarithmic data then converting.

D.Assuming circular orbits when eccentricities are high.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Kepler’s refined third law includes total system mass: $T^2 \propto a^3/(M+m)$ . For massive exoplanets like hot Jupiters, neglecting $m$ causes underestimation of $M$ and distorts the log-log slope. Diameter errors affect transit depth not period; eccentricity affects instantaneous speed not the period-semi-major axis relationship fundamentally.

Q4. Given two satellites with position vectors $\mathbf{r}_1(t)$ and $\mathbf{r}_2(t)$ around Earth, if $|\mathbf{r}_1|_{max} = 4|\mathbf{r}_2|_{max}$ and both have identical eccentricity, what is the ratio of their maximum orbital speeds $v_{1,max}/v_{2,max}$ ?

1/4

1/2

✅

1/\sqrt{2}

2

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Maximum speed occurs at periapsis. Using vis-viva equation combined with Kepler’s third law scaling: $v_{max} \propto \sqrt{(1+e)/a(1-e)}$ . With identical $e$ , $v_{max} \propto a^{-1/2}$ . Since $a_1 = 4a_2$ , the speed ratio is $\sqrt{a_2/a_1} = \sqrt{1/4} = 1/2$ . This integrates energy concepts with period-distance scaling.

Q5. In a simulation, a student models an orbit using $\mathbf{r}(t) = \langle a\cos(t), b\sin(t) \rangle$ and claims this satisfies Kepler’s Third Law for any $a \neq b$ . Why is this parametrization fundamentally flawed for gravitational orbits despite tracing an ellipse?

A.The parameter

t

represents eccentric anomaly, not physical time.

B.The speed is constant, violating conservation of angular momentum.

C.The foci are not at the origin unless

a=b

D.Both A and C are correct. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Gravitational orbits require the central body at a focus, not the center. The given parametrization centers the ellipse at origin. Additionally, uniform parameterization implies constant areal velocity only for circles. Real orbits use true or eccentric anomaly related nonlinearly to time via Kepler’s equation. Both geometric and temporal flaws invalidate direct application.

Q6. If a binary star system has components of equal mass $M$ orbiting their common center of mass with separation $R$ , how must the standard form $T^2 = k a^3$ be modified compared to a planet-star system where $m \ll M$ ?

A.Replace

a

with

R/2

and keep

k

unchanged.

B.Double the constant

k

because reduced mass halves effective inertia.

C.Use

a = R

and replace central mass

M

with total mass

2M

in denominator. ✅

D.No modification needed since

R

serves as semi-major axis.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: For comparable masses, Kepler’s third law uses total system mass: $T^2 = \frac{4\pi^2}{G(M_1+M_2)} a^3$ , where $a$ is the semi-major axis of relative orbit (equal to separation $R$ for circular case). The reduced mass formulation yields identical result. Option C correctly identifies both the distance measure and mass substitution required.

Q7. A graph shows orbital period squared vs. semi-major axis cubed for moons of Jupiter. One data point lies significantly above the best-fit line. Assuming measurement accuracy, what physical scenario best explains this outlier?

A.The moon has unusually high density.

B.The moon is in resonance with another moon, altering its effective period. ✅

C.The moon’s orbit is highly inclined relative to Jupiter’s equator.

D.Tidal dissipation is causing orbital decay.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Resonant interactions exchange angular momentum, effectively modifying the mean motion away from pure two-body prediction. Density affects internal structure not orbital period directly. Inclination doesn’t alter period in point-mass approximation. Tidal decay changes period gradually but wouldn’t cause static deviation above line; resonance creates persistent dynamical offset visible in such plots.

Q8. When numerically integrating $\mathbf{r}''(t) = -\mu \mathbf{r}/|\mathbf{r}|^3$ to verify Kepler’s Third Law, which numerical artifact would falsely suggest violation of $T^2 \propto a^3$ even with perfect initial conditions?

A.Using adaptive step sizing based on local truncation error.

B.Energy drift due to non-symplectic integrator over many periods. ✅

C.Round-off error in computing vector magnitudes.

D.Choosing too small a fixed time step.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Non-symplectic methods like Runge-Kutta conserve energy poorly over long integrations, causing artificial orbital expansion or contraction. This changes effective semi-major axis and period independently, breaking the expected proportionality. Symplectic integrators preserve phase-space volume and orbital elements better. Step size alone doesn’t cause systematic bias; energy drift accumulates secularly mimicking physical law violation.

Q9. Two spacecraft are in elliptical orbits with same semi-major axis but different eccentricities. Student A claims they have same period per Kepler’s Third Law. Student B argues higher eccentricity means longer path thus longer period. Who is correct and why?

A.Student B, because arc length integral depends on eccentricity.

B.Student A, because period depends solely on semi-major axis in inverse-square fields. ✅

C.Neither; period also depends on angular momentum.

D.Student A, but only approximately for small eccentricities.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Kepler’s Third Law states period depends exclusively on semi-major axis for bound orbits in $1/r^2$ potentials, regardless of eccentricity. While higher eccentricity orbits have varying speed, the time-averaged dynamics balance exactly. Path length does increase, but average speed adjusts precisely to maintain constant period. This counterintuitive result stems from virial theorem and action-angle variables.

Q10. In designing a Molniya orbit with period exactly half a sidereal day, engineers specify inclination at 63.4°. How does this inclination choice relate to satisfying Kepler’s Third Law operationally?

A.It minimizes atmospheric drag to preserve semi-major axis.

B.It nullifies secular perturbation of argument of perigee, stabilizing the orbit geometry tied to period. ✅

C.It maximizes ground coverage time at apogee independent of period.

D.It ensures the orbital plane precesses at rate matching Earth’s rotation.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: While Kepler’s Third Law fixes period via semi-major axis, real orbits experience J2 perturbations causing apsidal precession. At 63.4° critical inclination, $\dot{\omega} = 0$ , freezing perigee location. This maintains the intended ground track synchronized with the Keplerian period. Without this, drifting perigee would degrade mission performance despite correct nominal period.

Q11. If gravitational force followed $F \propto 1/r^3$ instead of $1/r^2$ , what would be the functional relationship between period $T$ and semi-major axis $a$ for stable circular orbits?

T \propto a^2

T \propto a

T

is independent of

a

D.Stable circular orbits cannot exist. ✅

💡 Difficulty: hard | ✅ Correct: D

📖 Explanation: Bertrand’s theorem proves only $1/r^2$ and harmonic potentials yield closed stable orbits. For $1/r^3$ , effective potential lacks minimum for bound states; orbits either spiral inward or escape. Thus no well-defined period-semi-major axis relation exists. This highlights uniqueness of inverse-square law underlying Kepler’s empirical discovery and distinguishes mathematical possibility from physical reality.

Q12. A student computes orbital period using $T = 2\pi \sqrt{a^3/\mu}$ but inputs $a$ in kilometers and $\mu$ in m³/s². Their answer is off by factor ~31.6. What correction resolves this?

A.Multiply result by 1000.

B.Divide result by

\sqrt{1000}

C.Convert

a

to meters before computation. ✅

D.Square the final answer.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Unit consistency is mandatory: $a$ must match length unit in $\mu$ . Since $\mu$ uses meters, $a$ in km must convert to m (multiply by 1000). Inside cube root, $(1000)^3 = 10^9$ , square root gives $10^{4.5} \approx 31623$ , explaining discrepancy. Dimensional analysis prevents such errors in applied celestial mechanics calculations.

Q13. Observations show asteroid belt objects follow $T^2 \propto a^3$ precisely, but Trojan asteroids at Jupiter’s L4/L5 points deviate slightly. What causes this apparent violation?

A.Trojans are influenced by Saturn’s gravity, breaking two-body assumption.

B.Measurement uncertainty is larger for distant Trojans.

C.Trojans orbit the Sun-Jupiter barycenter, not Sun alone, modifying effective

\mu

. ✅

D.Their orbits are chaotic and lack defined semi-major axes.

💡 Difficulty: easy | ✅ Correct: C

📖 Explanation: Trojans co-orbit with Jupiter around the system barycenter. The relevant central mass for their libration period includes Jupiter’s contribution, altering the effective gravitational parameter. While heliocentric $a$ still approximately satisfies Kepler’s law, precise dynamics require restricted three-body framework. This demonstrates domain limits of two-body Keplerian approximation in multi-body environments.

Q14. When analyzing radial velocity curves of exoplanet host stars, astronomers derive $m \sin i$ rather than true mass. How does unknown inclination $i$ affect verification of Kepler’s Third Law using observed period and derived semi-major axis?

A.It introduces systematic underestimate of

a

, breaking proportionality.

B.Period remains observable directly;

a

derived from RV assumes

\sin i = 1

, so true

a

is larger by

1/\sin i

. ✅

C.Inclination affects period measurement through projection effects.

D.Kepler’s Third Law cannot be tested without knowing

i

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Radial velocity gives $K \propto m \sin i / \sqrt{a}$ . Combined with Kepler’s law $a^3 \propto T^2$ , one solves for $m \sin i$ . Observed $T$ is inclination-independent. Derived $a$ from RV alone assumes edge-on; true $a$ scales as $1/\sin i$ . However, $T^2 \propto a^3$ still holds for true values; the law isn’t violated, just incompletely constrained.

Q15. A cube-shaped satellite tumbles in low Earth orbit. Does its irregular shape invalidate application of Kepler’s Third Law for predicting orbital period?

A.Yes, because drag depends on orientation.

B.No, because orbital period depends only on center-of-mass trajectory in vacuum. ✅

C.Only if tumbling induces significant torque coupling with gravity gradient.

D.Yes, because moment of inertia affects translational motion.

💡 Difficulty: easy | ✅ Correct: B

📖 Explanation: Kepler’s Third Law governs center-of-mass motion under central gravity, independent of attitude or shape, assuming negligible non-gravitational forces. In ideal two-body problem, extended bodies behave as point masses at COM. Atmospheric drag or gravity-gradient torques are perturbations, not violations of the fundamental law. Shape matters for decay rate, not instantaneous Keplerian period.

Q16. In a logarithmic plot of $T$ vs. $a$ for solar system planets, Mercury deviates most from the best-fit line. Beyond observational error, what relativistic effect contributes to this?

A.Frame-dragging alters Mercury’s effective semi-major axis.

B.Schwarzschild precession modifies the definition of orbital period. ✅

C.Time dilation makes Mercury’s clock run slower, affecting measured

T

D.None; deviation is purely due to solar oblateness.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: General relativity causes perihelion precession, meaning Mercury’s orbit isn’t perfectly closed. The anomalistic period (perihelion-to-perihelion) differs slightly from sidereal period used in Kepler’s law. This subtle distinction becomes measurable for Mercury due to strong field and high eccentricity. While small, it represents genuine physical departure from Newtonian prediction, distinguishable from classical perturbations.

Q17. A student argues that since $\mathbf{r}(t)$ for elliptical orbit isn’t sinusoidal, Fourier analysis can’t extract period. How would you refute this while connecting to Kepler’s Third Law?

A.Fourier series applies only to circular motion.

B.Orbital motion is periodic regardless of waveform; fundamental frequency corresponds to mean motion

n = 2\pi/T

. ✅

C.Kepler’s law replaces need for spectral analysis.

D.Elliptical orbits contain infinite harmonics masking fundamental period.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Periodicity doesn’t require sinusoidality. Any bounded Keplerian orbit repeats after time $T$ , so Fourier transform shows peak at $f = 1/T$ . Mean motion $n$ links directly to semi-major axis via $n^2 a^3 = \mu$ . Spectral methods actually validate Kepler’s law empirically by confirming single dominant frequency despite complex spatial trajectory.

Q18. During orbital transfer, a spacecraft follows Hohmann ellipse with semi-major axis $a_t = (r_1 + r_2)/2$ . Why can’t we directly apply Kepler’s Third Law to compute transfer time as full period of this ellipse?

A.Transfer uses only half the ellipse; time is

T/2

. ✅

B.Hohmann trajectories aren’t true Keplerian orbits.

C.Gravitational parameter changes during burn.

D.Eccentricity exceeds unity during transfer.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Hohmann transfer traverses exactly half an elliptical orbit between tangent points. Kepler’s Third Law gives full period $T = 2\pi\sqrt{a_t^3/\mu}$ ; actual transfer duration is $T/2$ . Misapplying full period doubles predicted time. This common error arises from conflating orbital element definition with mission segment geometry. Correct usage requires recognizing partial-orbit traversal.

Q19. If dark matter halo contributes additional mass interior to galactic orbit radius $R$ , how does observed rotation curve deviation from Keplerian expectation manifest in $T(R)$ data?

T

increases faster than

R^{3/2}

T

becomes independent of

R

at large radii.

T

decreases slower than

R^{3/2}

, implying extra enclosed mass. ✅

D.Rotation curves don’t relate to orbital periods.

💡 Difficulty: medium | ✅ Correct: C

📖 Explanation: Keplerian decline expects $v \propto R^{-1/2}$ , so $T = 2\pi R/v \propto R^{3/2}$ . Flat rotation curves ( $v \approx const$ ) imply $T \propto R$ , which grows slower than $R^{3/2}$ . This indicates mass $M(R) \propto R$ rather than constant, revealing dark matter. Period-radius relation thus diagnoses mass distribution beyond visible matter.

Q20. A computational model outputs orbital periods accurate to 0.1% but semi-major axes with 2% error. When plotting $T^2$ vs. $a^3$ , residuals show curvature. What does this indicate about error structure?

A.Random noise dominates measurements.

B.Systematic bias in

a

estimation violates power-law assumption. ✅

C.Model uses wrong gravitational constant.

D.Curvature is expected due to relativistic corrections.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: If $a$ has consistent multiplicative error $\epsilon$ , then $a^3$ error amplifies to $3\epsilon$ , creating curved residuals in log-space or quadratic deviation in linear space. Random errors scatter symmetrically; curvature implies structured miscalibration. Identifying this guides recalibration of distance metrics rather than rejecting Kepler’s law itself.

Q21. Why does Kepler’s Third Law hold exactly for test particles but only approximately for real planets in multi-planet systems?

A.Planets have finite size causing tidal deformation.

B.Mutual gravitational interactions introduce perturbations absent in two-body derivation. ✅

C.Solar radiation pressure alters effective gravity.

D.Planetary magnetic fields couple with solar wind.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Kepler’s laws derive from isolated two-body inverse-square problem. Real systems involve N-body interactions causing orbital element variations over time. While instantaneous osculating elements satisfy Kepler’s relations, long-term averaged behavior deviates. Secular resonances and mean-motion commensurabilities further complicate dynamics. Thus law remains foundational approximation, not exact description, in complex gravitational environments.

Q22. An astronaut measures local orbital period aboard ISS as 92 minutes. Ground station reports 93 minutes. Disregarding clock errors, what relativistic or kinematic effect explains discrepancy?

A.Gravitational time dilation makes ISS clocks tick faster.

B.Special relativistic time dilation slows ISS clocks relative to ground.

C.Combined GR and SR effects net to slight difference. ✅

D.Atmospheric refraction delays signal arrival.

💡 Difficulty: hard | ✅ Correct: C

📖 Explanation: ISS experiences weaker gravity (GR speeds up clocks) but high velocity (SR slows clocks). Net effect is ~microseconds/day difference, measurable with precision timing. While small, it confirms that orbital period isn’t absolute but frame-dependent. Kepler’s Third Law assumes Newtonian absolute time; relativistic corrections become relevant for high-precision applications like GPS, illustrating theory’s domain boundaries.

Q23. When fitting exoplanet transit data, assuming circular orbit yields period $P_c$ . Allowing eccentricity gives $P_e \approx P_c$ but different duration. Why doesn’t eccentricity significantly alter derived period despite changing transit geometry?

A.Transit timing depends on mean motion, which is fixed by semi-major axis via Kepler’s Third Law. ✅

B.Eccentricity only affects ingress/egress, not mid-transit epoch spacing.

C.Data quality insufficient to detect period change.

D.Circular assumption biases period estimate upward.

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Mean motion $n = 2\pi/P$ depends solely on $a$ per Kepler’s Third Law. Eccentricity redistributes time spent near star but preserves average angular rate. Transit intervals remain governed by $n$ , so period extraction is robust. Duration changes reflect velocity variation at conjunction, not period alteration. This decoupling enables reliable period determination even with unknown eccentricity.

Q24. A student derives $T^2 \propto a^3$ using dimensional analysis with $G, M, a$ . They obtain correct exponent but miss constant $4\pi^2$ . Why can’t dimensional analysis recover this factor?

A.Dimensional analysis ignores geometric constants arising from integration. ✅

\pi

is dimensionless and thus invisible to unit balancing.

C.The derivation assumed circular orbit, hiding elliptical geometry.

D.Constants depend on choice of units.

💡 Difficulty: hard | ✅ Correct: A

📖 Explanation: Dimensional analysis determines functional form up to dimensionless constants. Factors like $4\pi^2$ emerge from solving differential equations or integrating over orbital geometry, involving transcendental numbers unrelated to physical dimensions. While powerful for scaling laws, it cannot capture numerical coefficients rooted in mathematical structure. Full derivation via calculus or action principles is necessary for complete expression.

Q25. In a debris field around Earth, smaller fragments exhibit slightly shorter periods than predicted by Kepler’s Third Law for their altitude. What non-gravitational force likely causes this?

A.Electrostatic repulsion from charged debris.

B.Atmospheric drag reducing semi-major axis over time. ✅

C.Solar radiation pressure increasing effective gravity.

D.Magnetic Lorentz force opposing motion.

💡 Difficulty: medium | ✅ Correct: B

📖 Explanation: Drag removes orbital energy, decreasing semi-major axis and thus period below Keplerian prediction for initial altitude. Smaller fragments have higher area-to-mass ratio, enhancing drag susceptibility. Electrostatic and magnetic forces are typically negligible at LEO altitudes. Radiation pressure pushes outward, increasing period. Observed period shortening uniquely implicates dissipative drag as culprit violating pure two-body assumption.

Q26. Comparing analytical solution of Kepler’s equation $M = E - e\sin E$ with numerical root-finding, why might iterative methods fail near $e \approx 1$ despite satisfying Kepler’s Third Law theoretically?

A.High eccentricity violates conservation of angular momentum.

B.Newton-Raphson converges slowly when derivative approaches zero near apoapsis. ✅

C.Kepler’s Third Law breaks down for parabolic orbits.

D.Numerical precision limits prevent solving transcendental equations.

💡 Difficulty: hard | ✅ Correct: B

📖 Explanation: Near $e=1$ , mean anomaly $M$ changes very slowly with eccentric anomaly $E$ near apoapsis, making $dM/dE \approx 0$ . Newton-Raphson step $\Delta E = -f/f'$ blows up. This numerical instability doesn’t reflect physical law failure but algorithmic limitation. Alternative methods like bisection or universal variables handle near-parabolic cases robustly while preserving Keplerian relationships.

Q27. If a planet’s mass were suddenly doubled while keeping semi-major axis fixed, how would its orbital period change according to the generalized Kepler’s Third Law?

A.Period decreases by factor

\sqrt{2}

. ✅

B.Period remains unchanged because central mass dominates.

C.Period increases due to stronger self-gravity.

D.Period decreases by factor

2

💡 Difficulty: easy | ✅ Correct: A

📖 Explanation: Generalized law: $T^2 = \frac{4\pi^2}{G(M+m)} a^3$ . Doubling planet mass $m$ increases denominator, reducing $T$ . For $m \ll M$ , effect is tiny, but principle holds. Common misconception assumes only central mass matters; however, two-body dynamics depend on total mass. This question tests understanding beyond test-particle approximation.

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