what is the Kepler 1 ,2and 3 Laws
Answers
Answer:
Kepler’s Three Law:
Kepler’s Law of Orbits – The Planets move around the sun in elliptical orbits with the sun at one of the focii.
Kepler’s Law of Areas – The line joining a planet to the Sun sweeps out equal areas in equal interval of time.
Kepler’s Law of Periods – The square of the time period of the planet is directly proportional to the cube of the semimajor axis of its orbit.
Kepler’s 1st Law of Orbits:
This law is popularly known as the law of orbits. The orbit of any planet is an ellipse around the Sun with Sun at one of the two foci of an ellipse. We know that planets revolve around the Sun in a circular orbit. But according to Kepler, he said that it is true that planets revolve around the Sun, but not in a circular orbit but it revolves around an ellipse. In an ellipse, we have two focus. Sun is located at one of the foci of the ellipse.
Kepler’s 2nd Law of Areas:
This law is known as the law of areas. The line joining a planet to the Sun sweeps out equal areas in equal interval of time. The rate of change of area with time will be constant. We can see in the above figure, the Sun is located at the focus and the planets revolve around the Sun.
Assume that the planet starts revolving from point P1 and travels to P2 in a clockwise direction. So it revolves from point P1 to P2, as it moves the area swept from P1 to P2 is Δt. Now the planet moves future from P3 to P4 and the area covered is Δt.
As the area traveled by the planet from P1 to P2 and P3 to P4 is equal, therefore this law is known as the Law of Area. That is the aerial velocity of the planets remains constant. When a planet is nearer to the Sun it moves fastest as compared to the planet far away from the Sun.
Kepler’s 3rd Law of Periods:
This law is known as the law of Periods. The square of the time period of the planet is directly proportional to the cube of the semimajor axis of its orbit.
T² ∝ a³
That means the time ‘ T ‘ is directly proportional to the cube of the semi major axis i.e. ‘a’. Let us derive the equation of Kepler’s 3rd law. Let us suppose,
m = mass of the planet
M = mass of the Sun
v = velocity in the orbit
So, there has to be a force of gravitation between the Sun and the planet.
F = GmMr²
Since it is moving in an elliptical orbit, there has to be a centripetal force.
Fc = mv²r²
Now, F = Fc
⇒ GMr = v²
Also, v = circumferencetime = 2πrt
Combining the above equations, we get
⇒ GMr = 4π²r²T²
T² = 4π2r3)GM
⇒ T² ∝ r³
Explanation: