Question

Show that orbital time period of planet is proportional to the 3/2power of radius of its orbit

Answer 1

Kepeler's third law of planetary motion

Explanation:

Kepler's third law has been used to find the relationship between the distance of planets from the Sun and their orbital periods.

which can be calculated by putting centripetal force equal to the gravitational force:  

${\displaystyle mr\omega ^{2}=G{\frac {mM}{r^{2}}}}$                (1)

Substitute the value of angular velocity (ω) in terms of the orbital period

ω=$\frac{2\pi }{T}$ in equation (1)

${\displaystyle mr\left({\frac {2\pi }{T}}\right)^{2}=G{\frac {mM}{r^{2}}}\T$

${\displaystyle mr\left({\f{2\pi }{}}\right)^{2}=G{\frac {mM}{r^{2}}}T^{2}$

${\displaystyle \left{\f{4\\\pi ^{2} }{}}\right=G{\frac {M}{r^{3}}}T^{2}$

${\displaystyle \left{\f{\frac{4\pi ^{2} }{GM} }{}}\right={\frac {T^{2} }{r^{3}}}$

${\displaystyle \left{\f{r ^{3} }{}}\right={\frac {4\pi ^{2} T^{2} }{GM}}$

${\displaystyle \left{\f{r ^{3} }{}}\right\alpha {\T^{} }{}} T^{2}$

$T \alpha r^{\frac{3}{2}}$

Hence T is directly proportional to 3/2 power of the radius.