Keplers Laws.docx

JINNAH UNIVERSITY FOR WOMEN

Name: Shumaila Farid Class: BS III (fall) Department: Mathematics Course Code: MAT-3067 Course Title: Classical Dynamics Assignment Topic: Kepler’s Laws, Euler’s Theorem, and Chasel’s Theorem Submitted To: Ms.Kaneez Fatima

KEPLER’S LAWS: In astronomy and classical physics, laws describing the motions of the planets in the solar system. They were derived by the German astronomer Johannes Kepler, whose analysis of the observations of the 16th-century Danish astronomer Tycho Brahe enabled him to announce his first two laws in the year 1609 and a third law nearly a decade later, in 1618. Kepler himself never numbered these laws or specially distinguished them from his other discoveries. We shall consider Kepler’s Second Law (that the planet sweeps out equal areas in equal times) first, because it has a simple physical interpretation.

Kepler’s 2nd law: Statement: The rate of sweeping out of area is proportional to the angular momentum, and equal to L/2m. Explanation: The standard approach in analyzing planetary motion is to use (r,ϴ) coordinates, where r is the distance from the origin which we take to be the center of the Sun and ϴ is the angle between the x-axis and the line from the origin to the point.

In the above figure, let us suppose the planet goes from A to B, a distance ∆s, in a short time ∆t, so its speed in orbit is ∆s/∆t. Notice that the velocity can be resolved into vector components in the radial direction ∆r/∆t ,and in the direction perpendicular to the radius r∆ϴ/∆t. The short line BC in the diagram above is perpendicular to SB (S being the center of the Sun), and therefore becoming perpendicular to SC as well in the limit of AB becoming an infinitely small distance. Consider the triangle ABS, the area of the triangle is 1 𝐴𝑟𝑒𝑎 = 𝑏𝑎𝑠𝑒 × ℎ𝑒𝑖𝑔ℎ𝑡 2 So, 1 𝑑𝐴 = (𝑟)(𝑟𝑑𝜃) 2 Where, 𝑟 = ℎ𝑒𝑖𝑔ℎ𝑡 and 𝑟𝑑𝜃 = 𝑏𝑎𝑠𝑒

And the rate at which area is swept out on the orbit is: 𝑑𝐴 𝑑𝜃 = (𝑟) (𝑟 ) 𝑑𝑡 𝑑𝑡 𝑑𝜃 ∵𝑟 = 𝑣𝜃 𝑑𝑡 𝑑𝐴 1 = 𝑟𝑣 ⟶ (1) 𝑑𝑡 2 𝜃 ∵ 𝐴𝑛𝑔𝑢𝑙𝑎𝑟 𝑚𝑜𝑚𝑒𝑛𝑡𝑢𝑚 𝐿 = 𝑚(𝑟 × 𝑣) 𝐿 = 𝑚𝑟𝑣𝜃 or 𝑟𝑣𝜃 =

𝐿 𝑚

By putting in eq.(1), we get: 𝑑𝐴 1 𝐿 = 𝑑𝑡 2 𝑚 "Equal areas in equal times" means the rate at which area is swept out on the orbit (dA/dt) is constant.

Kepler’s 1st Law: Statement: All planets move in elliptical orbits, with the sun at one focus. Explanation:

The elliptical shape of the orbit is a result of the inverse square force of gravity. The eccentricity of the ellipse is greatly exaggerated here.

We now back up to Kepler’s First Law: proof that the orbit is in fact an ellipse if the gravitational force is inverse square. As usual, we begin with Newton’s Second Law: F = ma, in vector form. The force is GMm/r2 in a radial inward direction. But what is the acceleration? Is it just d2r/dt2? Well, no, because if the planet’s moving in a circular orbit it’s still accelerating inwards at r(𝑑𝜃/𝑑𝑡)2 even though r is not changing at all. The total acceleration is the sum, so ma=F becomes: 𝑑2𝑟 𝑑𝜃 2 𝐺𝑀 − 𝑟 ( ) =− 2 2 𝑑𝑡 𝑑𝑡 𝑟 The differential equation becomes easy to solve if we make the substitution, 𝑟 ⟶ 𝑢−1 , We have 𝑑𝑟 𝑑𝑢 = −𝑢−2 𝑑𝑡 𝑑𝑡 𝑑𝑟 𝑑𝑢 𝑑𝜃 = −𝑢−2 𝑑𝑡 𝑑𝜃 𝑑𝑡 𝑑𝜃 𝐿 𝑑𝜃 𝐿𝑢2 ∵ (𝑟 )= ⟹ = 𝑑𝑡 𝑚 𝑑𝑡 𝑚 2

𝑑𝑟 1 𝐿𝑢2 𝑑𝑢 = − 2( ) 𝑑𝑡 𝑢 𝑚 𝑑𝜃 𝑑𝑟 𝐿 𝑑𝑢 =− 𝑑𝑡 𝑚 𝑑𝜃 Further we have, 𝑑2 𝑟 𝐿 𝑑 𝑑𝑢 =− ( ) 2 𝑑𝑡 𝑚 𝑑𝑡 𝑑𝜃 =−

𝐿 𝑑𝜃 𝑑 𝑑𝑢 𝑚 𝑑𝑡 𝑑𝜃 𝑑𝜃

With this identity in hand, our central equation becomes 𝐿 2 2 𝑑2𝑢 𝐿 2 3 −( ) 𝑢 − ( ) 𝑢 = −𝐺𝑀𝑢2 𝑚 𝑑𝜃 2 𝑚 Which becomes, 𝑑 2 𝑢 𝐺𝑀𝑚2 − 2 + =𝑢 𝑑𝜃 𝐿2 which has the simple solution 𝑢 = 𝐴𝑐𝑜𝑠(𝜃 + 𝛿) +

𝐺𝑀𝑚2 𝐿2

We can always define our coordinates so that 𝛿 = 0, and thus we set it to zero for the remainder of the discussion. In terms of 𝑟(𝜃), we have 1

𝑟=

𝐴𝑐𝑜𝑠𝜃 + =

𝐺𝑀𝑚2 𝐿2

1 𝐺𝑀𝑚2 𝐿2

(𝑒𝑐𝑜𝑠𝜃 + 1) 𝐴𝐿2

Note that in the line above we made the useful substitution ⟹ 𝐺𝑀𝑚2 . From this form of 𝑟, it is clear that the aphelion and perihelion (points of furthest and closest distance, respectively, to the Sun) are given by 𝑟𝑚𝑎𝑥 =

1 𝐺𝑀𝑚2 𝐿2

(1 − 𝑒)

,

𝑟𝑚𝑖𝑛 =

1 𝐺𝑀𝑚2 𝐿2

(1 + 𝑒)

The semi-major axis is given by 𝑎=

1 𝐿2 (𝑟𝑚𝑖𝑛 + 𝑟𝑚𝑎𝑥 ) = (1 − 𝑒 2 )−1 2 𝐺𝑀𝑚2

Since 𝑟𝑚𝑖𝑛 and 𝑟𝑚𝑎𝑥 are distances from the Sun, we see that the Sun is at one focus of the orbit. Thus, we have derived Kepler's first law.

Kepler’s 3rd Law: Statement: The square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit. Explanation:

We can easily derive Kepler’s 3rd Law from Newton’s laws of motion and his law of gravitation, we will first of all make two simplifying assumptions, to make the mathematics easier. First we will assume that the orbits are circular, rather than elliptical. Secondly, we will assume that the Sun is at the centre of a planet’s circular orbit. Neither of these assumptions is strictly true, but they will make the derivation much simpler. Newton’s law of gravity states that the gravitational force between two bodies of masses M and m is given by 𝐹=

𝐺𝑀𝑚 ⟶ (1) 𝑟2

Where 𝑟 is the distance between the two bodies and 𝐺 is a constant, known as Newton’s universal gravitational constant. In the case we are considering here, 𝑟 is of course the radius of a planet’s circular orbit about the Sun. When an object moves in a circle, even at a constant speed, it experiences an acceleration. This is because the velocity is always changing, as the direction of the velocity vector is always changing, even if its size is constant. From Newton’s 2nd law, 𝐹 = 𝑚𝑎, which means if there is an acceleration there must be a force causing it, and for circular motion this force is known as the centripetal force. It is given by 𝐹=

𝑚𝑣 2 ⟶ (2) 𝑟

where 𝑚 is the mass of the moving body, 𝑣 is its speed, and 𝑟 is the radius of the circular orbit. This centripetal force in this case is provided by gravity, so we can say that 𝐺𝑀𝑚 𝑚𝑣 2 = 𝑟2 𝑟 𝐺𝑀 = 𝑣 2 ⟶ (3) 𝑟 But the speed 𝑣 is given by the distance the body moves divided by the time it takes. For one full circle this is just 𝑣=

2𝜋𝑟 𝑇

where 2𝜋𝑟 is the circumference of a circle and 𝑇 is the time it takes to complete one full orbit, its period. Substituting this into equation (3) gives 𝐺𝑀 4𝜋 2 𝑟 2 = 𝑟 𝑇2

4𝜋 2 3 𝑇 = 𝑎 𝐺𝑀 2

where we have substituted 𝑎 for 𝑟 and

4𝜋 2 𝐺𝑀

is constant So, 𝑇 2 ∝ 𝑎3

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

EULER’S THEOREM Statement: In three-dimensional space, any displacement of a rigid body such that a point on the rigid body remains fixed, is equivalent to a single rotation about some axis that runs through the fixed point.

CHASLE’S THEOREM Statement: Chasles' theorem applies only to rigid body displacements and states that for such bodies, any given displacement can be treated as equivalent to a translation (T) along an axis and a rotation (R) about the same axis.

References:    

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/KeplersLaws.htm

 

https://www.quora.com/What-is-Chasles-Theorem-in-classical-mechanics

https://brilliant.org/wiki/deriving-keplers-laws/ https://www.uu.edu/dept/math/SeniorPapers/09-10/DavisEmily.pdf https://thecuriousastronomer.wordpress.com/2014/03/04/derivation-of-newtons-form-ofkeplers-3rd-law-part-1/ https://physics.stackexchange.com/questions/356829/axis-of-rotation-and-eulers-theorem-inrigid-body-dynamics