kepler law

Kepler’s Laws

Johann Kepler developed empirically three laws of planetary motion, based on conclusions drawn from the extensive observations of Mars by Tycho Brahe (taken around the year 1600). While they were originally defined in terms of the motion of the planets about the Sun, they apply equally to the motion of natural or artificial satellites about the Earth. Kepler‟sfirst law states that the satellite follows an elliptical path in its orbit around the Earth. The satellite does not necessarily have uniform velocity around its orbit. Kepler‟s second law states that the line joining the satellite with the centre of the Earth sweeps out equal areas in equal times. Kepler‟s third law states that the cube of the mean distance of the satellite from the Earth is proportional to the square of its period.

Kepler’s First Law

The path followed be a satellite (in our case artificial satellite) around the primary (a planet and in our case Earth) will be an ellipse.
 “The orbit of every planer is an ellipse with sun at one of the two foci. “
 An ellipse has two focal points. Let us consider F1 and F2. The centre of mass of the two body system, known as the barycentre as always cantered at one foci. Due to the great difference between the masses of the planet (Earth) and the satellite, centre of mass always coincides with the centre of Earth and hence is always at one foci.

(Note: Ellipse: A regular oval shape, traced by a point moving in a plane so that the sum of its distances from two other points (the foci) is constant. Foci: The center of interest and in our case centre of the ellipse.)
 Parameters associated with the 1st law of Kepler:

o Eccentricity (e): it defines how stretched out an ellipse is from a perfect circle.

o Semi-Major axis (a): It is the longest diameter, a line that runs through the centre and both foci, its ends being at the widest points of the shapes. This line joins the points of apogee.

o Semi-Minor axis (b): the line joining the points of perigee is called the Semi-Minor axis.
The value of e could be determined by: e = (√a2 – b2) / a

kepler law

Foci F1 and F2, Semi-major axis a and semi-minor axis b of an ellipse.

Kepler’s Second Law

  • “For equal time intervals, a satellite will sweep out equal areas in its orbital plane focussed at the barycentre”.
  •  With respect to the laws governing the planetary motion around the sun, tis law could be stated as “A line joining a planet and the sun sweeps our equal area during equal intervals of time”.
kepler second law

The areas A1 and A2 swept out in unit intervals of time.

From figure and considering the law stated above, if satellite travels distances S1 and S2 meters in 1 second, then areas A1 and A2 will be equal.
 The same area will be covered everyday regardless of where in its orbit a satellite is. As the First Keplerian law states that the satellite follows an elliptical orbit around the primary, then the satellite is at different distances from the planet at different parts of the orbit. Hence the satellite has to move faster when it is closer to the Earth so that it sweeps an equal area on the Earth.

 This could be achieved if the speed of the satellite is adjusted when it is closer to the surface of the Earth in order to make it sweep out equal areas (footprints) of the surface of the Earth.

 

Kepler’s Third Law

 The square of the periodic time of orbit is proportional to the cube of the mean distance between the two bodies.

 The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.

 This law shows the relationship between the distances of satellite from earth and their orbital period.  Example: suppose satellite Satellite-I is four times as far from Earth as Satellite-II.

Then I must traverse four times the distance of II in each orbit. Now considering the speed of I and II, suppose I travels at half the speed of II, then in order to maintain equilibrium with the reduced gravitational force (as I is four times away from Earth than what II is), then in all it will require 4 x 2 = 8 times as long for I to travel an orbit in agreement with the law which comes down to (82 = 43).

 Symbolically:

P2 α a3 (P2 is directly proportional to a3) Where P is the orbital period; a is the semi-major axis a3 = μ/n2 Where n is the mean motion of satellite in radians per second and μ is the Earth‟s geocentric gravitational constant.

μ = 3.986005 x 1014 m3/sec2 Due to Earth‟s oblateness, a new parameter called drag is taken into account. P = 2П / n Here, P is in seconds and n is in radians/ second This law also confirms the fact that there is a fixed relation between period and size.