Kepler's Laws:
Kepler formulated following three laws concerning the motion of the planets around the sun.
(1) Kepler's first law of elliptical orbits:
All planets move in an elliptical orbits, with the sun at its foucs.Proof: The eccentricity (ε) of the orbit of a particle , moving under attractive central force is given by:
ε²=1+ 2EJ÷mC² E= - mC²÷2J²(1-ε²)
Where , J = Angular momentum of the particle of reduced mass ‛m’.
E = Total energy of a particle.
C = Constant= GMm
(in case of gravity force).
M= Mass of the body about which the particle moves.
In the case of a planet revolving round the sun , the sun is at one foucs which is takes as the centre of the coordinate system of the ellipse. If ε <1 then the total energy of the system ( planet and the sun ) is negative and the planet remains bound to the sun .
As the planet is bound to the sun and cannot escape from it, the planet moves around the sun , in closed elliptical orbit. This proves Kepler's first law of elliptical orbits.
(2) Kepler's second law for Areal velocity:
The speed of a planet along its orbit varies in such a way that the radius vector drawn from a sun to a planet sweeps out equal areas of the orbit in equal interval of times.
In short, the Areal velocity of the planet around the sun is Constant.
As shown in figure , the planet moves from the position a to b and from the position c to d in the same time. Then two shaded areas swept out by a planet in equal times are equal.
Proof: Suppose a planet of mass ‛m’ is moving in an elliptic orbit around the sun as shown in figure . Let S be the centre of the sun.
Let‛r’ be the radius vector of the planet with respect to sun. Suppose the planet moves from P to P' in a small time dt.
The area swept out by the radius vector is the area SPP'.
If ‛dt’ is infinitesimally small then PP' is a straight line = r dθ
Now , SPP' is a triangle.
.•. Area of ∆ SPP'=dA = 1 r . rdθ = 1 r²dθ
2 2
This is the area swept out in time dt.
.•. Rate at which the area is swept is Called Areal Velocity of the planet.
.•. Areal Velocity , dA = 1 × r² dθ ........(1)
dt 2 dt
Now , the Angular momentum of the planet , J = mvr = mwr² is a constant under a central force .
.•. ω = J/ mr² or dθ/dt = J/mr²
Hence , eq.(1) becomes:
dA = J/2m = constant .........(2)
dt
Thus , the Areal velocity of the planet around the sun is Constant. This proves the Kepler's second law that the radius vector joining the sun to the planet sweeps out equal areas in equal intervals of time.
(3) Kepler's third law of period:
The square of the time period of any planet about the sun is proportional to the cube of the planet's mean distance from the sun i.e semi-major axis of the elliptical orbit of the planet.The time period(T) of the planet(i.e time taken by it to complete one revolution around the sun is given by:
T= Area of the ellipse
Areal velocity of the radius vector
= πab
J / 2m
Where , a and b are the semi- major and the semi-minor axes of the elliptical orbit of the planet respectively.
.•. T² = 4π²m²a²b²
J² ......(1)
Now, if l is the letus rectum of the ellipse, then we have :
l=b²/a .•. b²= la
Hence, eq. (1) becomes:
T²= 4π²m²a³l
J²
But, 4π²m²l = Constant
J²
.•. T² ∝ a³ ....(2)
Hence, the square of the time period of the planet is proportional to the cube of semi- major axis of the ellipse . This is Kepler's Third Law of Period.
