Kepler's Laws of Planetary Motion

Table of Contents
1. Kepler's Laws of Planetary Motion
2. Kepler's First Law - The Law of Orbits
3. Kepler's Second Law - The Law of Areas
4. Kepler's Third Law - The Law of Periods
5. Examples and Worked Problems
6. Summary
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Kepler's Laws of Planetary Motion

The motions of planets in the sky fascinated astronomers for centuries. In the early 17th century Johannes Kepler, using precise measurements made by Tycho Brahe, formulated three empirical laws that describe planetary motion. These laws are kinematical statements about the shapes of planetary orbits, the variation of orbital speed, and the relation between orbital size and period. The three laws are stated and derived below, with examples and applications relevant to school and competitive examinations.

Kepler's First Law - The Law of Orbits

Kepler's First Law states:

All planets move around the Sun in elliptical orbits with the Sun at one focus of the ellipse, not at its centre.

An ellipse is a conic section defined by two foci. Important parameters of an ellipse are the semi-major axis a, semi-minor axis b, and eccentricity e, where 0 ≤ e < 1 for an ellipse. The eccentricity is given by e = c/a, where c is the distance from the centre to each focus. For small eccentricity (e ≈ 0) the ellipse is nearly circular; planets in the Solar System have small eccentricities so their orbits are nearly circular but precisely elliptical.

When the Sun is placed at a focus and polar coordinates (r, θ) are used with origin at that focus, the radial distance r of a planet from the Sun is given by the standard polar equation of an ellipse:

r = a(1 - e²) / (1 + e cos θ)

Kepler's Second Law - The Law of Areas

Kepler's Second Law states:

The line joining the Sun and a planet sweeps out equal areas in equal times.

This law implies that a planet moves faster when it is nearer to the Sun and slower when it is farther away. The nearest point in the orbit to the Sun is called the perihelion and the farthest point is the aphelion. Conservation of angular momentum explains Kepler's Second Law.

Derivation using conservation of angular momentum

Consider a planet of mass m moving with speed v at a distance r from the Sun. Let θ be the angle between the position vector and the velocity vector. The angular momentum about the Sun is

L = m r v sinθ

In a small time interval dt the planet moves a small distance dl = v dt and the position vector sweeps out an area equal to the area of triangle with base dl and height r sinθ. The elemental area dA swept out in time dt is

dA = 1/2 × r × (v dt) × sin(π - θ) = 1/2 r v sinθ dt

Thus the rate at which area is swept out is

dA/dt = 1/2 r v sinθ

From the expression for angular momentum

L = m r v sinθ

we obtain

dA/dt = L / (2m)

Since angular momentum L is conserved (no external torque about the Sun), L/(2m) is constant, so dA/dt is constant. This proves Kepler's Second Law.

Kepler's Third Law - The Law of Periods

Kepler's Third Law states:

The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit.

In symbols, for a planet with orbital period T and semi-major axis a,

T² ∝ a³

The constant of proportionality is the same for all planets orbiting the same central mass (the Sun), so for two planets

T₁² / a₁³ = T₂² / a₂³ = constant

A straightforward derivation for circular orbits uses Newton's law of gravitation and centripetal force.

Derivation for circular orbits

For a planet of negligible mass orbiting the Sun of mass M_s at radius r (circular orbit):

Gravitational force provides centripetal acceleration:

G M_s m / r² = m v² / r

Therefore

v² = G M_s / r

The orbital period is T = 2πr / v, so

T = 2π r / √(G M_s / r) = 2π √(r³ / (G M_s))

Squaring,

T² = (4π² / (G M_s)) r³

This shows T² ∝ r³. For elliptical orbits the same form holds if r is replaced by the semi-major axis a:

T² = (4π² / (G M_s)) a³

Examples and Worked Problems

The moon revolves around the earth 13 times per year. If the ratio of the distance of the earth from the sun to the distance of the moon from the earth is 392, find the ratio of mass of the sun to the mass of the earth.

Sol.

The time period of Earth around the Sun (Tₑ) and the time period of Moon around Earth (Tₘ) are related by Kepler's third law for the same type of central force. For circular approximation:

Tₑ² = (4π² / (G M_s)) rₑ³

Tₘ² = (4π² / (G M_e)) rₘ³

Divide the two expressions:

(Tₑ / Tₘ)² = (M_e / M_s) × (rₑ / rₘ)³

Rearrange:

M_s / M_e = (rₑ / rₘ)³ × (Tₘ / Tₑ)²

Given the moon revolves 13 times per year, Tₘ = (1 year) / 13, so Tₘ / Tₑ = 1/13. Given rₑ / rₘ = 392. Substitute:

M_s / M_e = 392³ × (1/13)²

A satellite revolves around a planet in an elliptical orbit. Its maximum and minimum distances from the planet are 1.5 × 10⁷ m and 0.5 × 10⁷ m respectively. If the speed of the satellite at the farthest point is 5 × 10³ m/s, calculate the speed at the nearest point.

Sol.

At apogee (farthest point) r₁ = 1.5 × 10⁷ m and speed v₁ = 5 × 10³ m/s.

At perigee (nearest point) r₂ = 0.5 × 10⁷ m and speed v₂ is to be found.

Angular momentum conservation about the focus (central body) gives:

m v₁ r₁ = m v₂ r₂

Therefore

v₂ = v₁ (r₁ / r₂)

Substitute values:

v₂ = 5 × 10³ × (1.5 × 10⁷ / 0.5 × 10⁷) = 5 × 10³ × 3 = 1.5 × 10⁴ m/s

Imagine a light planet revolving around a very massive star in a circular orbit of radius r with a period of revolution T. On what power of r will the square of the time period depend if the gravitational force of attraction between the planet and the star is proportional to r^-5/2?

Sol.

Assume the gravitational force F ∝ r^-5/2, i.e., F = k / r^5/2 for some constant k.

For circular motion, centripetal force = gravitational force:

m v² / r = k / r^5/2

So

v² = (k / m) × r^-3/2

Period T = 2πr / v, hence

T = 2π r / √( (k/m) r^-3/2 ) = 2π √(r² / (k/m) r^-3/2)

Simplify powers of r:

T ∝ r^7/4

Therefore

A satellite is revolving round the earth in a circular orbit of radius a with velocity v₀. A particle is projected from the satellite in forward direction with relative velocity

Sol.

Initial orbital speed of the satellite is

v₀ = √(G M / a)

If the particle is given an additional forward velocity v (relative to the satellite), the resultant speed immediately after projection is

v_R = v₀ + v

The particle will move in an elliptical orbit with perigee at r_min = a (the launch point) and apogee r_max = r. Use conservation of angular momentum and energy between perigee and apogee.

Angular momentum conservation about Earth's centre:

m v₁ r = m (v₀ + v) a

Energy conservation (total mechanical energy constant):

1/2 m v₁² - G M m / r = 1/2 m (v₀ + v)² - G M m / a

Solve these two equations to find r. The algebra leads to a quadratic in r:

3 r² - 8 a r + 5 a² = 0

The solutions are r = a and r = (5/3) a. Thus the minimum distance is a and the maximum distance is (5/3) a.

A sky lab of mass 2 × 10³ kg is first launched from the surface of earth in a circular orbit of radius 2 R (from the centre of earth) and then it is shifted from this circular orbit to another circular orbit of radius 3 R. Calculate the minimum energy required (a) to place the lab in the first orbit (b) to shift the lab from first orbit to the second orbit. Given R = 6400 km and g = 10 m/s².

Sol.

(a) Total mechanical energy on Earth's surface (take zero of potential at infinity):

E_S = KE + PE = 0 + (- G M m / R)

Total energy in circular orbit of radius 2R:

E_1 = - G M m / (4 R)

Energy required to place the lab from surface to orbit at radius 2R is ΔE = E_1 - E_S.

Using g = G M / R² and substituting numerical values yields the required energy.

(b) Total energy in circular orbit of radius 3R is

E_2 = - G M m / (6 R)

Energy required to transfer from orbit 2R to 3R is ΔE = E_2 - E_1.

A satellite is revolving around a planet of mass M in an elliptic orbit of semi-major axis a. Show that the orbital speed of the satellite when it is at a distance r from the focus will be given by :

Sol.

For an ellipse of semi-major axis a the total specific mechanical energy (energy per unit mass) of the satellite is constant and equals

ε = - G M / (2 a)

At any position r with speed v the specific mechanical energy equals kinetic plus potential energy per unit mass:

ε = 1/2 v² - G M / r

Equate the two expressions for ε:

1/2 v² - G M / r = - G M / (2 a)

Solve for v²:

v² = G M (2 / r - 1 / a)

Therefore the orbital speed at distance r is

v = √[ G M (2 / r - 1 / a) ]

Summary

Kepler's three laws give a complete kinematical description of planetary motion around a central mass: elliptical orbits with the Sun at one focus (First Law), equal areas in equal times due to conservation of angular momentum (Second Law), and the period-size relation T² ∝ a³ (Third Law). These laws follow naturally from Newtonian gravitation and conservation laws and are useful for solving a wide variety of orbital mechanics problems.