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Kepler’s laws of planetary motion1. The orbit of a planet is an ellipse with the Sun at its focus.
2. The line segment joining a planet and the Sun sweeps out equal areas duringequal intervals of time.
3. The square of the orbital period of a planet is proportional to the cube ofthe semi-major axis of its orbit.
𝜏2 ∝ 𝑎3
2a
Newton’s law of gravitationAny two bodies in the universe attract each other with a force thatis directly proportional to the product of their masses and inverselyproportional to the square of the distance between them.
𝐹 = −𝐺𝑚1𝑚2
𝑟2 𝑟
𝑚2
𝑚1
− 𝐹
𝐹
𝑟
The force between two bodies is called central if:
1. It acts along the line joining the bodies,
2. Its magnitude depends only on the distance between the bodies.
In general, a central force has the form 𝐹 = 𝑓(𝑟) 𝑟, where 𝑟 is theseparation vector between the bodies, with magnitude 𝑟 and directionalong the unit vector 𝑟.
Looks familiar?
For Gravitation, 𝐹 = −𝐺𝑚1𝑚2
𝑟2= −
𝑘
𝑟2
Potential 𝑉 = 𝑟∞𝑓 𝑟 d𝑟 = −
𝑘
𝑟
Central force
𝑚2
𝑚1
− 𝐹
𝐹
𝑟
Working out the Central Force Problem
Step 1
Reducing it to a one-body problem
Concerned only with the relative motionof the bodies
No external force: Centre of mass (c.m.)is fixed
Measure positions from c.m.
𝑟1 =𝑚2
𝑚1+𝑚2 𝑟, 𝑟2 =
𝑚2
𝑚1+𝑚2 𝑟
Thus, if one is known, we can find the other.
𝑚1
𝑚2 − 𝐹
𝐹
𝑟
c.m.
𝑟1
𝑟2
Let’s find the equation of motion for the mass m1.
• From Newton’s law (of motion this time),
• But here 𝑟1 =𝑚2
𝑚1+𝑚2 𝑟. Thus,
• Here, is a constant, with the dimensions ofmass. We will call it m, the reduced mass of thesystem.
• Now the equation of motion is simply that of aparticle of mass m, attracted towards the originwith a force of magnitude f(r). If we solve it tofind how 𝑟1 changes with time, Our job is done.
• If one mass is much less than other (say m1 ismuch less than m2), the reduced mass is almostequal to the smaller. Example: Sun and planet.
𝑚2
𝑚1
− 𝐹
𝐹
𝑟
c.m.
𝑟1
𝑟2
𝑟
𝐹
Still working out (yep, it’s tedious)Step 2
Making it 2D
For 3D problems, we need 3 coordinates ((x,y,z), (r,θ,φ), whatever).
But central force motion is always in a plane.
Thus, 2 coordinates 𝑟, 𝜃 will do.
𝑟
𝐹m
𝜃
𝑣
X
Y
Getting closer…Step 3
Getting the equations
Force (radially towards centre) = Mass × (radial acceleration + centripetal acceleration)
…(1)
Radial force ⇒ Torque about origin = 0 ⇒ Angular momentum = constant
…(2)
Substituting (2) in (1), we get
Which “can be” solved to get r(t). Once we know that, (2) gives us 𝜃(𝑡).
And closer still…Step 4
Solving them
Can be solved by numerical methods
(e.g. Consider very small time intervals, then find position and velocity after each interval, assuming rate of change to be almost constant)
Can be solved analytically for some forces, but notall
From energy conservation,
Now,
Hence,
Integrating, we get …(A)
(A) Gives t as a function of r. From it we “can get” r as a function of t. Then we integrate the equation to get 𝜃 as a function of t.
We also get the orbit equation from
Technically, we have solved the central force problem. But our work is not over yet…
Solving the differential equation
Kepler’s second lawFrom angular momentum conservation,
Therefore, if A is the area swept out by the radius vector in time t,
Thus, we find that Kepler’s second law holds not only for gravitationalforce, but for any central force, even repulsive.
𝑑𝜃
r𝑟𝑑𝜃
Source: Geophysics.ou.edu
Kepler’s first law
We have previously got the equation
For gravitation, 𝑉 𝑟 = −𝑘
𝑟.
Substituting u=1/r, we get after a few tedious steps.
Therefore, the equation of the orbit is
Is this an ellipse?
Conic sectionWe have
A fixed point S (the focus),
A fixed straight line D (thedirectrix).
A point P moves such that itsdistance from S (PS) and itsdistance from D (PM) are in afixed ratio e (the eccentricity)
Then P traces out a curve knownas a conic section.
The shape of the curve dependson e.
D
S
PM
PS
PM= 𝑒
We take S to be the origin and D tobe parallel to the y-axis on thenegative side.
Then, PS=r,
PM=d+rcos𝜃
Thus,𝑟
𝑑+𝑟 cos 𝜃= 𝑒
i.e.,1
𝑟=
1−𝑒 cos 𝜃
𝑑𝑒= 𝐶(1 − 𝑒 cos 𝜃)
For arbitrary orientation ofdirectrix,
1
𝑟= 𝐶(1 − 𝑒 cos(𝜃 − 𝜃′))
Yes, same as the planetary orbitequation.
D
S
PM
PS
PM= 𝑒
X
Y
r
𝜃
d 𝑟 cos 𝜃
Eccentricity and shape of conic
e=0 → Circle (sometimesconsidered special case ofellipse)
0<e<1 → Ellipse (flattenedcircle)
e=1 → Parabola (open)
e>1 → Hyperbola (withtwo arms)
Source: Wikipedia
Back to Kepler; First law
Equation of Kepler orbit:
Equation of conic section: 1
𝑟= 𝐶(1 − 𝑒 cos(𝜃 − 𝜃′))
Therefore,
Thus, is not possible.
(circle, special case of elliptical orbits)
(elliptical orbit of planets; Kepler’s 1st law)
(parabola, e.g. some comets)
(hyperbola, e.g. encounter between stars)
Kepler’s third law
• Can be derived from first and second laws
• Let area of orbit be A, time period be τ
• From first law, orbit is an ellipse with area 𝐴 = 𝜋𝑎 3 2 1 − 𝑒2 = 𝑗𝑎 3 2
•𝑑𝐴
𝑑𝑡=
𝐴
𝜏=
𝑗𝑎 3 2
𝜏= constant = 𝑢
• So 𝜏2 = (𝑗
𝑢)2𝑎3 (Kepler’s third law)
Source: burro.astr.cwru.edu
How Newton did it
We just got Newton’s law ⇒ Kepler’s laws.
But Newton had gone the other way round!
Kepler’s 2nd law ⇒ Acceleration of planet is radial
Kepler’s 1st (or 3rd) law ⇒ Acceleration of planet, 𝑎 ∝1
𝑟2
Therefore, Force 𝐹 ∝𝑚1
𝑟2(m1 = mass of planet)
By symmetry, 𝐹 ∝𝑚2
𝑟2(m2 = mass of sun)
Therefore, 𝐹 =𝐺𝑚1𝑚2
𝑟2
𝑚2
𝑚1
− 𝐹
𝐹
𝑟
What if?
What if gravity did not act the way it does?Would there be orbits?
What would be their shape?
Can they be circular?
Will they be closed?
Effective potential
We have seen the equation
i.e.,
This is the equation for a particle moving in one dimension under an effective potential
We can not get the motion from this, but it gives some idea of the range of values r can take.
Effective potential=Actual potential+Potential due to centrifugal force
Conditions for “good” orbitsWe can plot effective potential for any central force.
But will the orbits be of the same kind?
We can find that (not proved here)
Stable circular orbit of radius r0 exists only for f(r) satisfyingIf f(r) = −krn, then n > −3
Closed orbit exists only for f(r) = −kr (Hooke’s law force) and f(r) = −kr −2
(inverse square force)
Hooke’s law force blows up as r increases, but we expect gravitationalforce to decrease with distance
This also implies that gravity is an inverse-square force
More than two bodies
Okay, so we get what to do for 2 bodies.
Our solar system has 1 star, 9 planets (or 8, for Pluto-haters), satellites,dwarf planets, asteroids,…
For more three or more bodies, the situation gets much morecomplicated
Motion can not be found analytically except for some special cases.
Numerical simulation is possible, but difficult.
For planets, elliptical orbit is perturbed by influence of other planets
Discovery of Neptune
Lagrange pointsWe consider special case of 3-body problem where one mass is much smaller than others (e.g. asteroid in Sun-Jupiter system, artificial satellite in earth-moon system, etc.)
Third mass has negligible effect on other two
Two masses are revolving around each other
We consider forces on third mass
Lagrange points: positions where it rotates with same angular speed, i.e. is at rest in rotating frame
Lagrange points (contd.)• There are 5 Lagrange points
• L1, L2 and L3 are collinear with the two masses
• L4 and L5 form the third corners of the equatorial triangle with the masses at two vertices
• At these points, gravity equals centrifugal force, forming equilibrium
• L1, L2 and L3 are in unstable equilibrium, L4 and L5 are stable.
• Asteroids orbiting around L4 and L5: Trojans
• Artificial satellites can orbit L1, L2 and L3
in planes perpendicular to the line
Source: Classical Mechanics (Goldstein, Poole and Safko)
What is left
Many body simulations
Tidal effects
Tidal locking
Precession of planetary axis
General relativity