CBSE Class 11-science Answered

derive keplers 2nd and 3rd laws?

Asked by Sanmit Ratnaparkhi | 12 Sep, 2015, 02:52: AM

Expert Answer

Kepler's second law:

Consider that a planet of mass 'm' revolving around the Sun of mass 'M' in a circular orbit of radius 'r'. Let 'v' be its orbital velocity. Suppose that at any isntant the planet is at point A in its orbit and after an in small time dt, it reaches point B. As such, the circular path of the planet between points A and B may be considered as straight.

If dA is small area swept by the line joining the planet to the Sun in time dt, then

dA=area of ΔABS=

begin mathsize 14px style 1 half cross times base cross times perpendicular end style

or dA=

begin mathsize 14px style 1 half cross times AB cross times AS end style

.... (1)

If dθ is the angular displacement of the planet in time dt i.e., when it moves from point A to B, then AB=r dθ.

Substituting AB with rdθ and AS with 'r' in equation(1), we get,

begin mathsize 14px style dA equals 1 half cross times straight r space dθ cross times straight r dA equals 1 half straight r squared space dθ end style

Dividing the above equation by dt, we get,

begin mathsize 14px style dA over dt equals 1 half straight r squared dθ over dt or dA over dt equals 1 half straight r squared straight omega space space space space space space space space space space... space left parenthesis 2 right parenthesis end style

where ω=

begin mathsize 14px style dθ over dt end style

is the angular speed of the planet in its orbit and

begin mathsize 14px style dA over dt end style

is the areal velocity of the planet.

Multiplying and divind the R.H.S of the equation(2) by 'm' i.e., the mass of the planet, we get,

begin mathsize 14px style dA over dt equals fraction numerator 1 over denominator 2 space straight m end fraction cross times straight m space straight r squared space straight omega end style

Since, m r² ω = L, the angular momentum of the planet about the axis through the Sun, we have,

begin mathsize 14px style dA over dt equals fraction numerator straight L over denominator 2 space straight m end fraction end style

... (3)

As no external torque acts on the planet during its orbital motion, its angular momentum (L) must remain constant. Since both L amd m are constant, the equation(3) becomes,

begin mathsize 14px style dA over dt equals constant end style

... (4)

Hence, when a planet moves around the Sun, its areal velocity remains constant. It proves Kepler's second law of planetary motion.

Kepler's third law:

The centripetal force required for making the planet to revolve in the circular orbit is provided by the gravitational pull of the Sun i.e.,

begin mathsize 14px style fraction numerator straight m space straight v squared over denominator straight r end fraction equals fraction numerator straight G space straight M space straight m over denominator straight r squared end fraction end style

begin mathsize 14px style straight v equals square root of fraction numerator straight G space straight M over denominator straight r end fraction end root end style

.... (1)

The period of revolution of the planet around the Sun is,

begin mathsize 14px style straight T equals fraction numerator circumference space of space the space orbit over denominator orbital space velocity end fraction equals fraction numerator 2 πr over denominator square root of straight G space straight M divided by straight r end root end fraction end style

begin mathsize 14px style straight T squared equals fraction numerator 4 straight pi squared over denominator straight G space straight M end fraction straight r cubed end style

.... (2)

Since,

Syntax error from line 1 column 94 to line 1 column 135. Unexpected 'mathvariant'.

is constant,

begin mathsize 14px style therefore straight T squared equals constant cross times straight r cubed end style

begin mathsize 14px style straight T squared proportional to straight r cubed end style

Hence, the square of the period of revolution of a planet around the Sun is directly proportional to the cube of the radius of its circular orbit. It proves Kepler's third law of planetary motion.

Answered by Faiza Lambe | 12 Sep, 2015, 09:59: PM