Kepler’s Laws Of Planetary Motion

Kepler’s Laws Of Planetary Motion

Edited By Vishal kumar | Updated on Sep 25, 2024 06:26 PM IST

Kepler’s Laws of Planetary Motion, formulated by the German astronomer Johannes Kepler in the early 17th century, revolutionized our understanding of the solar system by describing how planets orbit the Sun. These laws not only laid the foundation for modern astronomy but also provided insight into the predictable and harmonious nature of celestial movements. In real life, the principles behind Kepler's laws are observed in various scenarios, such as satellite orbits around Earth, which rely on the same gravitational forces that govern planetary motion. Even the rhythm of seasons and the timing of solar eclipses can be traced back to the precise mechanics described by Kepler, demonstrating how these ancient laws continue to shape our daily experiences.

Kepler's Law

Kepler gives three empirical laws which govern the motion of the planets which are known as Kepler’s laws of planetary motion.

As we know planets are large natural bodies rotating around a star in definite orbits.

So, Kepler's laws are

(a) First Law (Law of Ellipses)
Each planet orbits the Sun in an elliptical path, with the Sun located at one of the two foci of the ellipse. This law explains why the distance between a planet and the Sun changes as the planet moves along its orbit.

(b) Second Law (Law of Equal Areas)
A line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time. This means that a planet moves faster when it is closer to the Sun and slower when it is farther away, ensuring that the area covered over a given time period is the same.

The below figure

Area of velocity =dAdt
dAdt=12(r)(Vdt)dt=12rV
Where
dAdtAreal velocity
dA small area traced
Kepler's 2nd law is Similar to the Law of conservation of momentum
As dAdt=L2m
where
L=mvr Angular momentum

(c) The Law of Periods

The square of the orbital period of a planet (the time it takes to complete one orbit around the Sun) is directly proportional to the cube of the semi-major axis of its orbit (the average distance from the Sun)

The below figure


AB=AF+FB2a=r1+r2a=r1+r12
Where
a= semi-major Axis
r1= The shortest distance of the planet from the sun (perigee)
r2= The largest distance of the planet from the sun (apogee)
So if T= Time period of revolution
Then according to Kepler's 3rd law.

T2αa3 or T2α(r1+r22)3

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Solved Examples Based on Kepler’s Laws of Planetary Motion

Example 1: A planet moving around the sun sweeps area A1 in 2 days A2 in 3 days and A3 in 6 days. Then the relation between A1 ,A2 and A3 is

1) 3A1 = 2 A2 = A3

2) 2 A1 = 3 A2 = 6 A3

3) 3 A1 = 2 A2 = 6 A3

4) 6 A1 = 3 A2 = 2 A3

Solution:

According to Kepler's second law, the average velocity of the planet around the sun remains constant.

A1t1=A2t2=A3t3A12=A23=A36=3A1=2A2=A3

Hence, the answer is the option (1).

Example 2: If the angular momentum of a planet of mass m, moving around the Sun in a circular orbit is L, about the centre of the Sun, its areal velocity is

1) Lm
2) 4Lm
3) L2m
4) 2Lm

Solution:

Kepler's 2nd law

dAdt=12(r)(Vdt)dt=12rV

dAdtAreal velocity
dA small area traced
wherein
Similar to the Law of conservation of momentum

dAdt=L2m

L Angular momentum
it is also known as the Law of Area
So from the question

dA=12r2dθdAdt=12r2dθdtdAdt=12r2ωL=mr2ωdAdt=L2m

Hence, the answer is the option (3).

Example 3: Figure shows elliptical path abcd of a planet around the sun S such that the area of triangle csa is 1/4 the area of the ellipse. (See figure) With db as the semimajor axis, and ca as the semiminor axis. If t1 is the time taken for the planet to go over path abc and t2 for the path taken over cda then :

1) t1 = t2

2) t1 = 2t2

3) t1 = 3t2

4) t1 = 4t2

Solution:

Kepler's 3rd law


T2αa3
From fig.

AB=AF+FB2a=r1+r2a=r1+r12a= semi major Axis r1= Perigee

wherein
Known as the law of periods

r2= apogee T2α(r1+r22)3r1+r2=2at1t2=A/2+A/4A/2A/4=3A/4A/4=3t1=3t2

Hence, the answer is the option (3).

Example 4: India’s Mangalyan was sent to Mars by launching it into a transfer orbit EOM around the sun. It leaves the earth at E and meets Mars at M. If the semi-major axis of Earth’s orbit is ae = 1.5\times1011 m, that of Mar’s orbit am = 2.28\times1011 m, taken Kepler’s laws give the estimate of time for Mangalyan to reach Mars from Earth to be close to : (In days)

1) 260

2) 320

3) 500

4) 220

Solution:

Kepler's 3rd law


T2αa3
From fig.

a=r1+r12

a= semi major Axis
r1= Perigee

r2= apogee
For EOM

rm=1.5+2.282=1.89TmTe=(1.891.5)32⇒>Tm=Te(1.891.5)32

time for Mangalyan to reach Mars from Earth=t= =Tm2

So t=3652×1.41=257.3 days

Hence, the answer is the option (1).

Example 5: The time period of a satellite of Earth is 5 hours. If the separation between the earth and the satellite is increased to 4 times the previous value, the new time period will become ( In hours)

1) 40

2) 80

3) 10

4) 20

Solution:

Kepler's 3rd law


T2αa3
From fig.

AB=AF+FB2a=r1+r2a=r1+r12a= semi major Axis r1= Perigee

wherein

r2= apogee T2α(r1+r22)3r1+r2=2aT2r3(T1T2)2=(r1r2)3=(14)3=164T1T2=18T2=8T1=40hrs

Hence, the answer is the option (1).

Summary

Kepler's Laws of Planetary Motion provide a framework for understanding the elliptical orbits of planets, their varying speeds, and the relationship between their orbital periods and distances from the Sun. These laws are essential in predicting planetary motion, designing satellite orbits, and solving real-world problems, such as estimating travel times for space missions. By applying these principles, we gain insight into the predictable and harmonious patterns that govern celestial mechanics.

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