Concept Card: Kepler’s Laws of Planetary Motion & Orbital Dynamics
1. Historical Context and Empirical Foundations:
Between 1609 and 1619, German astronomer Johannes Kepler deduced three empirical laws of planetary motion based on decades of precision astronomical observations compiled by Tycho Brahe. Decades later, Sir Isaac Newton demonstrated that all three of Kepler’s laws are direct mathematical deductions from his Universal Law of Gravitation ($F \propto 1/r^2$) and Newton’s Laws of Motion.
2. Kepler’s First Law (The Law of Orbits)
Statement: Every planet revolves around the Sun in an elliptical orbit, with the Sun situated at one of the two foci of the ellipse.
Geometry of the Planetary Ellipse:
- An ellipse is defined as the locus of a point whose sum of distances from two fixed points (foci $S$ and $S’$) is a constant equal to the major axis length:
$$r_1 + r_2 = 2a$$ - Semi-Major Axis ($a$): Half the length of the longest diameter of the ellipse.
- Semi-Minor Axis ($b$): Related to $a$ and orbital eccentricity $e$ by:
$$\mathbf{b = a\sqrt{1 – e^2}}$$ - Orbital Eccentricity ($e$): A dimensionless measure of orbital elongation ($0 \le e < 1$).
- $e = 0$: Perfect circle ($b = a$, both foci coincide at the center).
- $0 < e < 1$: Elliptical orbit (most planetary orbits have small $e$, e.g., Earth has $e \approx 0.0167$, whereas comets possess high $e \approx 0.97$).
- Perihelion Distance ($r_p$): The closest point of approach to the Sun:
$$\mathbf{r_p = a(1 – e)}$$ - Aphelion Distance ($r_a$): The farthest point from the Sun:
$$\mathbf{r_a = a(1 + e)}$$ - Relationship between $a$, $r_p$, and $r_a$:
$$\mathbf{a = \frac{r_p + r_a}{2}}, \quad \mathbf{e = \frac{r_a – r_p}{r_a + r_p}}$$
3. Kepler’s Second Law (The Law of Areas) & Areal Velocity
Statement: The radius vector drawn from the Sun to a planet sweeps out equal areas in equal intervals of time. That is, the areal velocity of a planet is constant throughout its orbit:
$$\mathbf{\frac{dA}{dt} = \text{constant}}$$
Rigorous Derivation from Angular Momentum Conservation:
Consider a planet of mass $m$ at position vector $\vec{r}$ moving with velocity $\vec{v}$. In an infinitesimal time interval $dt$, the planet undergoes a vector displacement $d\vec{r} = \vec{v} dt$.
The infinitesimal area $dA$ swept out by the radius vector is the area of a triangle with sides $\vec{r}$ and $d\vec{r}$:
$$d\vec{A} = \frac{1}{2} (\vec{r} \times d\vec{r}) = \frac{1}{2} (\vec{r} \times \vec{v}) dt$$
Dividing by $dt$, the areal velocity is:
$$\frac{d\vec{A}}{dt} = \frac{1}{2} (\vec{r} \times \vec{v}) = \frac{1}{2m} (\vec{r} \times m\vec{v}) = \mathbf{\frac{\vec{L}}{2m}}$$
where $\vec{L} = \vec{r} \times m\vec{v}$ is the orbital angular momentum of the planet about the Sun.
Conservation of Angular Momentum Under a Central Force:
The gravitational force exerted by the Sun on the planet is a strictly central force:
$$\vec{F}_g = -\frac{G M m}{r^2} \hat{r}$$
The net torque $\vec{\tau}$ acting on the planet about the Sun is:
$$\vec{\tau} = \vec{r} \times \vec{F}_g = \vec{r} \times \left(-\frac{G M m}{r^2} \hat{r}\right) = \vec{0}$$
Since $\vec{\tau} = \frac{d\vec{L}}{dt} = \vec{0}$, orbital angular momentum $\vec{L}$ is strictly conserved in magnitude and direction.
Fundamental Theorem:
Kepler’s Second Law is equivalent to the conservation of orbital angular momentum. Consequently:
- The orbital motion of any planet is strictly confined to a single fixed plane perpendicular to $\vec{L}$.
- The areal velocity is constant: $\mathbf{\frac{dA}{dt} = \frac{L}{2m} = \text{constant}}$.
Velocity Extremes at Perihelion and Aphelion:
At both perihelion and aphelion, the velocity vector $\vec{v}$ is strictly perpendicular to the radius vector $\vec{r}$ ($\sin 90^\circ = 1$):
$$L = m r_p v_p = m r_a v_a = \text{constant}$$
$$\mathbf{r_p v_p = r_a v_a \implies \frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{1 + e}{1 – e}}$$
– A planet moves with maximum speed at perihelion ($v_p = v_{\text{max}}$).
– A planet moves with minimum speed at aphelion ($v_a = v_{\text{min}}$).
4. Kepler’s Third Law (The Law of Periods / Harmonic Law)
Statement: The square of the orbital period $T$ of revolution of any planet is directly proportional to the cube of the semi-major axis $a$ of its elliptical orbit:
$$\mathbf{T^2 \propto a^3 \implies \frac{T^2}{a^3} = \text{constant}}$$
Derivation for an Elliptical Orbit:
The total area of an ellipse is:
$$A_{\text{total}} = \pi a b = \pi a \left(a\sqrt{1 – e^2}\right) = \pi a^2 \sqrt{1 – e^2}$$
Since the areal velocity $\frac{dA}{dt} = \frac{L}{2m}$ is constant, the orbital period $T$ is:
$$T = \frac{A_{\text{total}}}{dA/dt} = \frac{\pi a^2 \sqrt{1 – e^2}}{L / (2m)} = \frac{2\pi m a^2 \sqrt{1 – e^2}}{L}$$
From celestial mechanics, the conserved angular momentum for an inverse-square field is:
$$L = m \sqrt{G M a (1 – e^2)}$$
Substituting $L$:
$$T = \frac{2\pi m a^2 \sqrt{1 – e^2}}{m \sqrt{G M a (1 – e^2)}} = \frac{2\pi a^2}{\sqrt{G M a}} = \frac{2\pi}{\sqrt{G M}} a^{3/2}$$
Squaring both sides:
Kepler’s Third Law Equation:
$$\mathbf{T^2 = \left(\frac{4\pi^2}{G M}\right) a^3}$$
where $M$ is the mass of the central body (the Sun). The constant of proportionality $\frac{4\pi^2}{G M}$ is independent of the mass $m$ of the orbiting planet (assuming $m \ll M$).
5. Energy Dynamics & The Vis-Viva Equation in Elliptical Orbits
- Total Mechanical Energy ($E$):
$$E = K + U = \frac{1}{2} m v^2 – \frac{G M m}{r} = -\mathbf{\frac{G M m}{2a}}$$
Crucial Fact: The total orbital energy depends solely on the semi-major axis $a$, and is completely independent of the eccentricity $e$! - The Vis-Viva Equation (Speed at Any Distance $r$):
Equating $\frac{1}{2} m v^2 – \frac{G M m}{r} = -\frac{G M m}{2a}$:
$$\mathbf{v(r) = \sqrt{G M \left(\frac{2}{r} – \frac{1}{a}\right)}}$$- At perihelion ($r = a(1-e)$): $\mathbf{v_p = \sqrt{\frac{G M}{a} \left(\frac{1+e}{1-e}\right)}}$
- At aphelion ($r = a(1+e)$): $\mathbf{v_a = \sqrt{\frac{G M}{a} \left(\frac{1-e}{1+e}\right)}}$
- At distance $r = a$ (ends of semi-minor axis): $\mathbf{v = \sqrt{\frac{G M}{a}} = \sqrt{v_p v_a}}$
6. High-Yield JEE Traps & Insights:
- Trap 1 (Central Force Scope vs. Inverse-Square Law):
- Kepler’s 2nd Law (Law of Areas) holds for ANY central force ($F \propto r^n$, for any exponent $n$).
- Kepler’s 1st Law (Elliptical Orbits) and 3rd Law ($T^2 \propto a^3$) hold strictly for inverse-square forces ($F \propto 1/r^2$). If $F \propto 1/r^n$, $T^2 \propto r^{n+1}$.
- Trap 2 (Semi-Major Axis Misidentification): Never confuse semi-major axis $a$ with perihelion $r_p$ or aphelion $r_a$. Always write $a = \frac{r_p + r_a}{2}$.
- Trap 3 (Non-Negligible Satellite Mass): If the orbiting body has significant mass $m$ relative to $M$, Kepler’s Third Law becomes $T^2 = \frac{4\pi^2}{G(M + m)} a^3$ (using reduced mass).
Solved Examples
Example 1 (Direct Ratio Application – Orbital Geometry and Speed Extremes):
An asteroid revolves around the Sun in an elliptical orbit such that its perihelion distance is $r_p = 1.0\text{ AU}$ and its aphelion distance is $r_a = 3.0\text{ AU}$.
(a) Calculate the semi-major axis $a$ and eccentricity $e$ of the orbit.
(b) Determine the ratio of its orbital speed at perihelion to that at aphelion ($v_p / v_a$).
(c) Find the orbital time period $T$ of the asteroid in Earth years.
(d) If the asteroid’s speed at perihelion is $v_p = 42.0\text{ km/s}$, find its speed at aphelion $v_a$.
Solution:
(a) Semi-Major Axis and Eccentricity:
$$a = \frac{r_p + r_a}{2} = \frac{1.0 + 3.0}{2} = \mathbf{2.0\text{ AU}}$$
$$e = \frac{r_a – r_p}{r_a + r_p} = \frac{3.0 – 1.0}{3.0 + 1.0} = \frac{2.0}{4.0} = \mathbf{0.50}$$
(b) Ratio of Speed at Extremes:
By conservation of angular momentum about the Sun:
$$m r_p v_p = m r_a v_a \implies \frac{v_p}{v_a} = \frac{r_a}{r_p} = \frac{3.0\text{ AU}}{1.0\text{ AU}} = \mathbf{3.0}$$
(c) Orbital Period via Kepler’s Third Law:
Using Earth’s orbit as a reference ($T_E = 1.0\text{ yr}, a_E = 1.0\text{ AU}$):
$$\left(\frac{T}{T_E}\right)^2 = \left(\frac{a}{a_E}\right)^3 \implies T^2 = (2.0)^3 = 8.0$$
$$\mathbf{T = \sqrt{8.0} = 2\sqrt{2}\text{ years} \approx 2.83\text{ years}}$$
(d) Speed at Aphelion:
$$v_a = \frac{v_p}{3.0} = \frac{42.0\text{ km/s}}{3.0} = \mathbf{14.0\text{ km/s}}$$
Example 2 (Calculus & Time Partition in Highly Eccentric Orbits):
Halley’s Comet revolves around the Sun in a highly eccentric elliptical orbit with semi-major axis $a = 18.0\text{ AU}$ and eccentricity $e = 0.967$.
(a) Determine the perihelion distance $r_p$ and aphelion distance $r_a$.
(b) Calculate the total orbital period $T$ of Halley’s Comet in Earth years.
(c) Explain qualitatively using Kepler’s Second Law why the comet spends more than $95\%$ of its lifetime beyond the orbit of Jupiter ($r_J \approx 5.2\text{ AU}$).
Solution:
(a) Perihelion and Aphelion Distances:
$$r_p = a(1 – e) = 18.0(1 – 0.967) = 18.0(0.033) = \mathbf{0.594\text{ AU}} \quad (\text{inside the orbit of Venus!})$$
$$r_a = a(1 + e) = 18.0(1 + 0.967) = 18.0(1.967) = \mathbf{35.406\text{ AU}} \quad (\text{beyond Neptune and Pluto!})$$
(b) Orbital Period:
$$T^2 = a^3 = (18.0)^3 = 5832 \implies \mathbf{T = \sqrt{5832} \approx 76.37\text{ years}}$$
(Halley’s Comet returns approximately every 76 years!).
(c) Areal Velocity & Time Distribution:
According to Kepler’s Second Law:
$$\frac{dA}{dt} = \frac{L}{2m} = \text{constant} \implies dt = \frac{2m}{L} dA$$
The time spent traversing any segment of the orbit is directly proportional to the area $dA$ swept by the radius vector from the Sun.
Because the Sun is at one focus, the area of the orbit outside $r > 5.2\text{ AU}$ encompasses the vast majority of the total ellipse area ($\approx 95\%$). Furthermore, because $v(r) \propto 1/r$, the comet crawls sluggishly near aphelion ($v_a \approx 0.9\text{ km/s}$) and whips rapidly past perihelion ($v_p \approx 54.5\text{ km/s}$), spending the overwhelming majority of its 76-year period in the cold, distant outer Solar System.
Example 3 (Multi-Concept Linkage – Hohmann Transfer Orbit & The Vis-Viva Equation):
A telecommunications satellite of mass $m$ in a low Earth circular parking orbit of radius $R_1 = 7000\text{ km}$ needs to be transferred to a geostationary circular orbit of radius $R_2 = 42000\text{ km} = 6 R_1$ via an elliptical Hohmann transfer orbit.
(a) Determine the semi-major axis $a_{\text{trans}}$ of the elliptical transfer orbit.
(b) Find the speed of the satellite in the initial circular orbit $v_{c1}$.
(c) Using the Vis-Viva equation, derive the required speed $v_{\text{trans}, 1}$ at perigee of the transfer orbit, and the velocity boost $\Delta v_1$ required from the rocket thrusters.
(d) Calculate the time required for the satellite to journey from $R_1$ to $R_2$.
Solution:
(a) Semi-Major Axis of Transfer Ellipse:
The perigee of the transfer ellipse is at $R_1$ and its apogee is at $R_2$:
$$2 a_{\text{trans}} = R_1 + R_2 = R_1 + 6 R_1 = 7 R_1$$
$$\mathbf{a_{\text{trans}} = \frac{7}{2} R_1 = 3.5 R_1 = 24500\text{ km}}$$
(b) Speed in Initial Circular Orbit:
$$v_{c1} = \sqrt{\frac{G M}{R_1}}$$
(c) Vis-Viva Equation at Perigee:
$$v_{\text{trans}, 1} = \sqrt{G M \left(\frac{2}{R_1} – \frac{1}{a_{\text{trans}}}\right)} = \sqrt{G M \left(\frac{2}{R_1} – \frac{1}{3.5 R_1}\right)} = \sqrt{\frac{G M}{R_1} \left(2 – \frac{2}{7}\right)} = \sqrt{\frac{12}{7} \frac{G M}{R_1}}$$
$$\mathbf{v_{\text{trans}, 1} = \sqrt{\frac{12}{7}} v_{c1} \approx 1.309 v_{c1}}$$
The velocity boost needed at perigee is:
$$\mathbf{\Delta v_1 = v_{\text{trans}, 1} – v_{c1} = \left(\sqrt{\frac{12}{7}} – 1\right) v_{c1} \approx 0.309 v_{c1}}$$
(d) Transfer Time:
The transfer duration is exactly half of the orbital period of the transfer ellipse:
$$t_{\text{transfer}} = \frac{T_{\text{trans}}}{2} = \frac{1}{2} \left[ 2\pi \sqrt{\frac{a_{\text{trans}}^3}{G M}} \right] = \mathbf{\pi \sqrt{\frac{(3.5 R_1)^3}{G M}}}$$
With $T_1 = 2\pi \sqrt{\frac{R_1^3}{G M}} \approx 90\text{ minutes}$:
$$t_{\text{transfer}} = \frac{T_1}{2} (3.5)^{3/2} \approx (45\text{ min}) (6.548) \approx \mathbf{294.6\text{ minutes} \approx 4.91\text{ hours}}$$
Example 4 (Theoretical Generalization – Arbitrary Central Force Law $F \propto 1/r^n$):
Suppose the gravitational force between two masses varied according to a modified power law:
$$F(r) = \frac{k}{r^n}$$
where $k$ is a constant and $n$ is a real number.
(a) For a circular orbit of radius $R$, derive the generalized relationship between the orbital period $T$ and orbital radius $R$.
(b) Determine whether Kepler’s Second Law (Law of Areas) remains valid under this modified force law.
(c) Deduce Kepler’s Third Law for $n = 2$ (Newtonian gravity) and $n = 3$.
Solution:
(a) Generalized Orbital Law for Circular Orbits:
For a particle of mass $m$ revolving in a circular orbit of radius $R$:
$$\frac{m v^2}{R} = F(R) = \frac{k}{R^n} \implies v^2 = \frac{k}{m R^{n-1}} \implies v = \sqrt{\frac{k}{m}} R^{-(n-1)/2}$$
The orbital period is:
$$T = \frac{2\pi R}{v} = \frac{2\pi R}{\sqrt{\frac{k}{m}} R^{-(n-1)/2}} = 2\pi \sqrt{\frac{m}{k}} R \cdot R^{(n-1)/2} = 2\pi \sqrt{\frac{m}{k}} R^{\frac{n+1}{2}}$$
Squaring both sides:
$$\mathbf{T^2 \propto R^{n+1}}$$
(b) Validity of Kepler’s Second Law:
Since $\vec{F}(r) = -\frac{k}{r^n} \hat{r}$ remains purely radial, the torque about the center is:
$$\vec{\tau} = \vec{r} \times \vec{F} = \vec{0}$$
Angular momentum $\vec{L}$ is conserved regardless of the exponent $n$.
Therefore, Kepler’s Second Law ($\frac{dA}{dt} = \frac{L}{2m} = \text{constant}$) remains strictly valid for ANY value of $n$!
(c) Special Cases:
– For Newtonian inverse-square gravity ($n = 2$): $T^2 \propto R^{2+1} = \mathbf{R^3}$ (Kepler’s Third Law!).
– For inverse-cube force ($n = 3$): $T^2 \propto R^{3+1} = \mathbf{R^4} \implies T \propto R^2$.
Practice Worksheet: Kepler’s Laws & Orbital Dynamics
Section A: Single Correct Questions (Q1 to Q5)
Q1. If the distance of a planet from the Sun were to become 4 times its present distance, the duration of its year (orbital period) would become:
(A) $2\text{ times}$
(B) $4\text{ times}$
(C) $8\text{ times}$
(D) $16\text{ times}$
Q2. A planet moves in an elliptical orbit about the Sun with speeds $v_p$ at perihelion and $v_a$ at aphelion. At a position where the distance from the Sun equals the semi-major axis $a$, the speed of the planet is:
(A) $\frac{v_p + v_a}{2}$
(B) $\sqrt{v_p v_a}$
(C) $\frac{2 v_p v_a}{v_p + v_a}$
(D) $\sqrt{\frac{v_p^2 + v_a^2}{2}}$
Q3. A planet of mass $m$ moves in an elliptical orbit of semi-major axis $a$ and semi-minor axis $b$. If the areal velocity of the planet is denoted by $\frac{dA}{dt}$, the orbital period $T$ of the planet is given by:
(A) $\frac{2\pi a b}{dA/dt}$
(B) $\frac{\pi a b}{dA/dt}$
(C) $\frac{\pi a^2}{dA/dt}$
(D) $\frac{4\pi a b}{dA/dt}$
Q4. The orbital period of an exoplanet orbiting a distant star is found to be 8 Earth years. Assuming the central star has the same mass as our Sun, the semi-major axis of the exoplanet’s orbit is:
(A) $2\text{ AU}$
(B) $4\text{ AU}$
(C) $8\text{ AU}$
(D) $16\text{ AU}$
Q5. If a satellite is orbiting in an elliptical path, the physical quantity that remains strictly constant throughout the entire orbit is:
(A) Linear speed
(B) Kinetic energy
(C) Linear momentum
(D) Angular momentum about the focus
Section B: One or More than One Correct Questions (Q6 to Q8)
Q6. Which of the following statements regarding Kepler’s Second Law is/are correct?
(A) It is a direct consequence of the conservation of orbital angular momentum.
(B) It is valid for any central force, even if the force does not follow the inverse-square law.
(C) It implies that the speed of the planet is maximum when it is closest to the Sun.
(D) It implies that the acceleration vector of the planet is always perpendicular to its velocity vector.
Q7. For a planet of mass $m$ moving in an elliptical orbit of semi-major axis $a$ around a star of mass $M$, which of the following expressions is/are correct?
(A) Total mechanical energy is $E = -\frac{G M m}{2a}$.
(B) The total mechanical energy depends only on $a$ and is independent of the orbital eccentricity $e$.
(C) The speed of the planet at any distance $r$ satisfies $v^2 = G M \left(\frac{2}{r} – \frac{1}{a}\right)$.
(D) The speed of the planet at perihelion is $v_p = \sqrt{\frac{G M}{a} \left(\frac{1+e}{1-e}\right)}$.
Q8. A comet revolves in an elliptical orbit around the Sun. As the comet moves from aphelion to perihelion:
(A) Its gravitational potential energy decreases.
(B) Its kinetic energy increases.
(C) Its orbital angular momentum about the Sun remains constant.
(D) The work done by the gravitational force on the comet is positive.
Section C: Numerical Value Type Questions (Q9 to Q10)
Q9. A planet orbits the Sun in an ellipse. Its distance at perihelion is $r_1$, and its distance at aphelion is $r_2 = 3 r_1$. If the linear speed of the planet at perihelion is $v_1 = 30\text{ km/s}$, calculate its speed $v_2$ at aphelion in $\text{km/s}$.
Q10. A space probe is placed into an elliptical orbit around Mars. The maximum distance of the probe from the center of Mars is $r_{\text{max}} = 5 R_M$, and its minimum distance is $r_{\text{min}} = 3 R_M$, where $R_M$ is the radius of Mars. The semi-major axis of the orbit can be written as $a = n R_M$. Find the integer value of $n$.
Comprehensive Solutions and Answer Key
Answer Key
| Question | Correct Answer | Core Physics Concept |
|---|---|---|
| Q1 | (C) | Kepler’s Third Law: $T’ = (4)^{3/2} T = 8 T$ |
| Q2 | (B) | Vis-Viva Relation at $r = a$: $v = \sqrt{G M/a} = \sqrt{v_p v_a}$ |
| Q3 | (B) | Total Period in Terms of Areal Velocity: $T = \frac{\pi a b}{dA/dt}$ |
| Q4 | (B) | Harmonic Law Inversion: $a = T^{2/3} = 8^{2/3} = 4\text{ AU}$ |
| Q5 | (D) | Conservation of Angular Momentum Under Central Force |
| Q6 | (A, B, C) | Properties and Foundations of Kepler’s Second Law |
| Q7 | (A, B, C, D) | Vis-Viva Equation and Invariant Total Orbital Energy |
| Q8 | (A, B, C, D) | Energy and Momentum Evolution from Aphelion to Perihelion |
| Q9 | 10 | Angular Momentum Conservation: $v_2 = v_1(r_1/r_2) = 10\text{ km/s}$ |
| Q10 | 4 | Semi-Major Axis Definition: $a = \frac{5 + 3}{2} R_M = 4 R_M$ |
Detailed Explanations
Solution 1:
By Kepler’s Third Law:
$$T^2 \propto a^3 \implies T \propto a^{3/2}$$
If the distance $a’ = 4a$:
$$T’ = (4)^{3/2} T = (\sqrt{4})^3 T = 2^3 T = \mathbf{8 T}$$
Correct Answer: (C)
Solution 2:
From the Vis-Viva equation:
$$v^2 = G M \left(\frac{2}{r} – \frac{1}{a}\right)$$
At $r = a$ (the ends of the semi-minor axis):
$$v^2 = G M \left(\frac{2}{a} – \frac{1}{a}\right) = \frac{G M}{a}$$
Now evaluate the product $v_p v_a$:
$$v_p = \sqrt{\frac{G M}{a} \left(\frac{1+e}{1-e}\right)}, \quad v_a = \sqrt{\frac{G M}{a} \left(\frac{1-e}{1+e}\right)}$$
$$v_p v_a = \sqrt{\frac{G M}{a} \left(\frac{1+e}{1-e}\right) \cdot \frac{G M}{a} \left(\frac{1-e}{1+e}\right)} = \frac{G M}{a}$$
Therefore:
$$v = \mathbf{\sqrt{v_p v_a}}$$
Correct Answer: (B)
Solution 3:
The total area of an ellipse of semi-major axis $a$ and semi-minor axis $b$ is:
$$A_{\text{total}} = \pi a b$$
Since the areal velocity $\frac{dA}{dt}$ is constant throughout the motion:
$$A_{\text{total}} = \left(\frac{dA}{dt}\right) T \implies \mathbf{T = \frac{\pi a b}{dA/dt}}$$
Correct Answer: (B)
Solution 4:
Using Kepler’s Third Law with Earth reference units:
$$T^2 = a^3$$
Given $T = 8\text{ years}$:
$$a^3 = 8^2 = 64 \implies a = (64)^{1/3} = \mathbf{4\text{ AU}}$$
Correct Answer: (B)
Solution 5:
In an elliptical orbit under a central gravitational force:
– Speed $v$ varies continuously ($v_{\text{max}}$ at perihelion, $v_{\text{min}}$ at aphelion).
– Kinetic energy $K = \frac{1}{2}mv^2$ and linear momentum $\vec{p} = m\vec{v}$ change continuously.
– The net torque about the focus (the Sun) is $\vec{\tau} = \vec{r} \times \vec{F}_g = \vec{0}$, so the angular momentum about the focus is strictly conserved.
Correct Answer: (D)
Solution 6:
– (A) True: Areal velocity $\frac{dA}{dt} = \frac{L}{2m}$ is constant because $L$ is conserved.
– (B) True: For any central force $\vec{F} = f(r)\hat{r}$, torque $\vec{\tau} = \vec{r} \times \vec{F} = \vec{0}$, so $L$ is conserved.
– (C) True: $r v = \text{constant}$ at apsides; closest distance means highest speed.
– (D) False: In an elliptical orbit, the gravitational acceleration is directed toward the focus, which is NOT perpendicular to the velocity vector except at perihelion and aphelion.
Correct Answer: (A, B, C)
Solution 7:
– (A) True: Total orbital energy is $E = -\frac{G M m}{2a}$.
– (B) True: Total energy depends only on $a$, independent of eccentricity $e$.
– (C) True: Vis-Viva equation $v^2 = G M(2/r – 1/a)$.
– (D) True: At perihelion $r = a(1-e) \implies v_p = \sqrt{\frac{GM}{a}\frac{1+e}{1-e}}$.
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 8:
As the comet moves from aphelion ($r_a$) to perihelion ($r_p$):
– $r$ decreases $\implies U = -G M m / r$ becomes more negative, so $U$ decreases. (A is true).
– By energy conservation, $K$ increases as $U$ decreases. (B is true).
– Angular momentum about the Sun is invariant under a central force. (C is true).
– Work done by gravity is $W_g = -\Delta U > 0$ (positive). (D is true).
All statements (A, B, C, D) are correct.
Correct Answer: (A, B, C, D)
Solution 9:
By conservation of angular momentum about the Sun:
$$m r_1 v_1 = m r_2 v_2 \implies v_2 = v_1 \left(\frac{r_1}{r_2}\right)$$
Given $r_2 = 3 r_1$ and $v_1 = 30\text{ km/s}$:
$$v_2 = 30\text{ km/s} \left(\frac{1}{3}\right) = \mathbf{10\text{ km/s}}$$
Correct Answer: 10
Solution 10:
The semi-major axis $a$ is the arithmetic mean of the maximum and minimum focal distances:
$$a = \frac{r_{\text{max}} + r_{\text{min}}}{2} = \frac{5 R_M + 3 R_M}{2} = \frac{8 R_M}{2} = \mathbf{4 R_M}$$
Comparing with $a = n R_M$:
$$n = 4$$
Correct Answer: 4