Prove of Kepler third law of planetary motion ?
Answers
The square of the orbital period of a planet is directly proportional to the cube of the semi-major axis of its orbit.
This captures the relationship between the distance of planets from the Sun, and their orbital periods.
Kepler enunciated in 1619 this third law in a laborious attempt to determine what he viewed as the "music of the spheres" according to precise laws, and express it in terms of musical notation. So it was known as the harmonic law.
Using Newton's Law of gravitation (published 1687), this relation can be found in the case of a circular orbit by setting the centripetal force equal to the gravitational force:
{\displaystyle mr\omega ^{2}=G{\frac {mM}{r^{2}}}} {\displaystyle mr\omega ^{2}=G{\frac {mM}{r^{2}}}}
Then, expressing the angular velocity in terms of the orbital period and then rearranging, we find Kepler's Third Law:
{\displaystyle mr\left({\frac {2\pi }{T}}\right)^{2}=G{\frac {mM}{r^{2}}}\rightarrow T^{2}=\left({\frac {4\pi ^{2}}{GM}}\right)r^{3}\rightarrow T^{2}\propto r^{3}} {\displaystyle mr\left({\frac {2\pi }{T}}\right)^{2}=G{\frac {mM}{r^{2}}}\rightarrow T^{2}=\left({\frac {4\pi ^{2}}{GM}}\right)r^{3}\rightarrow T^{2}\propto r^{3}}
A more detailed derivation can be done with general elliptical orbits, instead of circles, as well as orbiting the center of mass, instead of just the large mass. This results in replacing a circular radius, {\displaystyle r} r, with the elliptical semi-major axis, {\displaystyle a} a, as well as replacing the large mass {\displaystyle M} M with {\displaystyle M+m} {\displaystyle M+m}. However, with planet masses being so much smaller than the Sun, this correction is often ignored. The full corresponding formula is:
{\displaystyle {\frac {a^{3}}{T^{2}}}={\frac {G(M+m)}{4\pi ^{2}}}\approx {\frac {GM}{4\pi ^{2}}}\approx 7.496\cdot 10^{-6}\left({\frac {{\text{AU}}^{3}}{{\text{days}}^{2}}}\right){\text{ is constant}}} {\displaystyle {\frac {a^{3}}{T^{2}}}={\frac {G(M+m)}{4\pi ^{2}}}\approx {\frac {GM}{4\pi ^{2}}}\approx 7.496\cdot 10^{-6}\left({\frac {{\text{AU}}^{3}}{{\text{days}}^{2}}}\right){\text{ is constant}}}
where {\displaystyle M} M is the mass of the Sun, {\displaystyle m} m is the mass of the planet, and {\displaystyle G} G is the gravitational constant, {\displaystyle T} T is the orbital period and {\displaystyle a} a is the elliptical semi-major axis.
Explanation:
Answer is refer in Attachment
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