To understand how a planet moving in a circular orbit around the sun obeys Kepler's Third Law of planetary motion, we need to delve into the relationship between the orbital radius and the orbital period. Kepler's Third Law states that the square of the orbital period (T) of a planet is directly proportional to the cube of the semi-major axis (r) of its orbit. Mathematically, this is expressed as T² ∝ r³. Let's break this down step by step.
Understanding the Circular Motion of a Planet
When a planet orbits the sun in a circular path, it experiences centripetal acceleration, which is necessary to keep it in that circular motion. The gravitational force between the planet and the sun provides this centripetal force. The gravitational force can be described by Newton's law of gravitation:
F = G * (m1 * m2) / r²
Here, F is the gravitational force, G is the gravitational constant, m1 is the mass of the sun, m2 is the mass of the planet, and r is the distance between the center of the planet and the sun.
Relating Gravitational Force to Centripetal Force
For a planet in circular motion, the required centripetal force (Fc) is given by:
Fc = (m * v²) / r
where m is the mass of the planet and v is its linear speed. Setting the gravitational force equal to the centripetal force gives us:
G * (m1 * m2) / r² = (m2 * v²) / r
Notice that the mass of the planet (m2) cancels out from both sides:
G * m1 / r = v² / r
Deriving Kepler's Third Law
From the equation above, we can rearrange it to find the linear speed:
v² = G * m1 / r
Now, the orbital period T (the time it takes for one complete orbit) can be expressed in terms of the linear speed and the circumference of the orbit:
T = 2πr / v
Substituting the expression for v from our earlier equation:
T = 2πr / √(G * m1 / r)
This simplifies to:
T = 2π√(r³ / G * m1)
From this equation, we can see that T² is proportional to r³:
T² ∝ r³
This confirms Kepler's Third Law, as it shows that the square of the period of orbit is directly proportional to the cube of the average distance from the sun.
Calculating Pluto's Orbital Period
Now, let’s find the length of a year on Pluto, which is, on average, 40 times farther from the sun than Earth. If we denote the average distance of Earth from the sun as r₁ and that of Pluto as r₂, we have:
r₂ = 40 * r₁
According to Kepler's Third Law:
T₂² / T₁² = r₂³ / r₁³
Substituting r₂:
T₂² / T₁² = (40 * r₁)³ / r₁³
This simplifies to:
T₂² / T₁² = 40³
Calculating 40³ gives us 64000. Therefore:
T₂² = 64000 * T₁²
Assuming Earth's orbital period (T₁) is 1 year, we find:
T₂² = 64000
Taking the square root gives:
T₂ = √64000 ≈ 253.98 years
Thus, the length of a year on Pluto is approximately 254 Earth years. This lengthy orbital period is a fascinating aspect of Pluto's distant and elliptical orbit around the sun.
