⭐derivativon of third law of Kepler (law of periods)✌️☺️
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The square of the time period of revolution of a planet is proportional to the cube of the semi major axis of the ellipse traced out by the planet. this law are known as Kepler third law of periods
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Consider earth is revolving around the sun on an elliptical path . So the force due to masses btw them is
Fg = GMm/r^2 ------eq.1
Fg = Force due to gravity
M = mass of sun
m = mass of earth
As the centripetal force also acts on it
Fc = mv^2/r ( but v = rw) (w=omega)
Fc = mr^2w^2/r
Fc = mrw^2 -------eq.2
From eq 1 & 2-------
GMm/r^2 = mrw^2 (but w=2#/T) #=pie
GM/r^2 = r(2#/T)^2
GM/r^2 = 4r#^2/T^2
GMT^2 = 4r^3#^2
T^2 = 4#2r^3/GM
Here 4#^2/GM = K (constant)
So from here
T^2 is directly proportional to r^3
And this is Kepler's law of time period.
Fg = GMm/r^2 ------eq.1
Fg = Force due to gravity
M = mass of sun
m = mass of earth
As the centripetal force also acts on it
Fc = mv^2/r ( but v = rw) (w=omega)
Fc = mr^2w^2/r
Fc = mrw^2 -------eq.2
From eq 1 & 2-------
GMm/r^2 = mrw^2 (but w=2#/T) #=pie
GM/r^2 = r(2#/T)^2
GM/r^2 = 4r#^2/T^2
GMT^2 = 4r^3#^2
T^2 = 4#2r^3/GM
Here 4#^2/GM = K (constant)
So from here
T^2 is directly proportional to r^3
And this is Kepler's law of time period.
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