Animated Solution for Physics - Gravitation: The ratio of earth's orbital angular momentum (about the sun) to its mass is 4.4×1015 m2/s. The area enclosed by earth's orbit is approximately ......... m2.
Visualized Solution
Visualizing Kepler's Second Law
Consider Earth of mass m orbiting the Sun in an elliptical path.
The radius vector r sweeps out an area dA in an infinitesimal time interval dt.
Areal Velocity and Angular Momentum
The rate at which area is swept, known as areal velocity, is given by:
dtdA=2mL
where L is the angular momentum of Earth about the Sun, and m is Earth's mass.
Integrating to Find Total Area
To find the total area A enclosed by Earth's orbit over one complete revolution (period T):
A=∫0TdtdAdt=∫0T2mLdt
Simplifying the Area Equation
Since L and m are constant:
A=2mL∫0Tdt=2mLT
Identifying Given Parameters
We are given:
mL=4.4×1015 m2/s
T=365 days
Converting Time Period to Seconds
Convert T to seconds:
T=365×24×3600 s
T=3.1536×107 s
Substituting Values into the Area Formula
Substitute the values of mL and T:
A=21(mL)T
A=21×(4.4×1015)×(3.1536×107)
Calculating the Final Area
Perform the multiplication:
A=2.2×1015×3.1536×107
A≈6.94×1022 m2
The Way Forward
Since gravity is a central force, angular momentum is conserved in any central force field.
Thus, Kepler's second law holds true for all elliptical orbits, regardless of eccentricity.
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The Sigma Insight: Kepler's Laws of Planetary Motion
Solution Diagram
The Cosmic Dance of Kepler's Laws
Imagine standing in the silent, vast expanse of the cosmos, watching our blue planet Earth glide effortlessly around the brilliant Sun.
For thousands of years, humanity looked up at the night sky, wondering what rules governed this majestic, silent dance.
It was Johannes Kepler who, in the early 17th century, shattered the ancient belief in perfect circular orbits and revealed the true elliptical nature of planetary motion.
Kepler's Second Law, often called the Law of Equal Areas, states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.
This means that when Earth is closer to the Sun (at perihelion), it moves faster, and when it is farther away (at aphelion), it slows down.
But what is the deep, underlying physics that enforces this cosmic speed limit?
The answer lies in one of the most fundamental principles of the universe: the Conservation of Angular Momentum.
The Mathematical Bridge
Areal Velocity
To translate Kepler's geometric observation into the language of modern physics, we define a quantity called areal velocity, denoted as dtdA.
This represents the rate at which area is swept out by the planet's radius vector over time.
Let's derive this relationship from first principles to see its elegant connection to angular momentum.
Consider a planet of mass m moving with velocity v at a distance r from the Sun.
In an infinitesimally small time interval dt, the planet undergoes a small displacement dr=vdt.
The area dA of the tiny triangle swept out by the radius vector is given by:
dA=21∣r×dr∣
Substituting dr=vdt, we get:
dA=21∣r×v∣dt
Now, let's divide both sides by dt to find the rate of area sweep:
dtdA=21∣r×v∣
Recall that the orbital angular momentum L of the planet about the Sun is defined as:
L=r×p=m(r×v)
This means the magnitude of the angular momentum is:
L=m∣r×v∣
By substituting this back into our areal velocity equation, we discover a breathtakingly simple relation:
dtdA=2mL
This is the mathematical heart of Kepler's Second Law!
Since gravity is a central force directed entirely along the line connecting the planet to the Sun, it exerts zero torque on the planet.
With no external torque, the angular momentum L remains absolutely constant.
And since the mass m of the planet is also constant, the areal velocity dtdA must be constant throughout the entire orbit.
Integrating Over Time
Now that we have established that the rate of area sweep is constant, how do we find the total area A enclosed by Earth's entire orbit?
We simply integrate this constant rate over one complete orbital period T, which corresponds to one Earth year:
A=∫0TdtdAdt
Since dtdA=2mL is constant, we can pull it out of the integral:
A=2mL∫0Tdt
Integrating dt from 0 to T simply gives the total time period T:
A=2mLT
We can rewrite this as:
A=21(mL)T
This formula is incredibly powerful because it tells us that we do not need to know the semi-major axis, the semi-minor axis, or even the eccentricity of the orbit to find its area.
All we need is the ratio of the angular momentum to the mass, mL, and the total time period T.
The Final Calculation
Let's plug in the values provided in the problem to find the numerical value of this cosmic area.
We are given the ratio of Earth's orbital angular momentum to its mass:
mL=4.4×1015 m2/s
The time period T is one Earth year, which is approximately 365 days.
To keep our units consistent, we must convert this time period into seconds:
T=365 days×24 hours/day×3600 seconds/hour
T=365×86400 s=3.1536×107 s
Now, let's substitute these values into our master equation:
A=21×(4.4×1015)×(3.1536×107)
First, simplify the fraction:
A=2.2×1015×3.1536×107
Now, multiply the decimal coefficients:
2.2×3.1536=6.93792
Next, combine the exponents of ten:
1015×107=1022
Putting it all together, we get:
A≈6.94×1022 m2
This is the total area enclosed by Earth's orbit—an astronomical figure of nearly seventy billion trillion square meters!
Why This Matters
This problem beautifully demonstrates the power of conservation laws in physics.
By focusing on a conserved quantity—angular momentum—we bypassed the need for complex geometric calculations involving ellipses.
Whether the orbit is nearly circular (like Earth's actual orbit) or highly elongated (like a comet's), the relationship between areal velocity and angular momentum remains completely unchanged.
The next time you look up at the night sky, remember that the silent, sweeping motion of the planets is governed by these elegant, unchanging laws of conservation, keeping the universe in perfect, predictable harmony.