
According to the Kepler’s Law of Planetary motion, if \[T\] represents the time period and \[r\] is the orbital radius, then for two planets these are related as:(a) \[\dfrac{{{{T_1}^{\dfrac{3}{2}}}}}{{{T_2}}} = \dfrac{{{r_1}}}{{{r_2}}}\](b) \[\dfrac{{{T^{}}_1}}{{{T_2}}} = (\dfrac{{{r_1}}}{{{r_2}}})\](c) \[{(\dfrac{{{T^{}}_1}}{{{T_2}}})^2} = {(\dfrac{{{r_1}}}{{{r_2}}})^3}\](d) \[\dfrac{{{T^2}_1}}{{{T_2}}} = {\dfrac{{{r_1}}}{{{r_2}}}^3}\]
Answer
506.4k+ views
Hint: Kepler’s third law of planetary motion, Law of periods.
Complete Step-By-Step-Answer:
Note: Semi major axis of an elliptical path is the arithmetic mean of the minimum and the maximum distance covered by the planet. The semi-minor axis is the geometric mean of minimum and the maximum distance covered by the planet; these two must not be confused.
It states that, 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.
Kepler stated three laws that govern planetary motion. These laws describe the motion of planets around the sun. The first law named the Law of orbits, states that every planet moves in elliptical orbits with the sun at one of its foci. The second law, named the law of areas, states that the line joining the sun and any planet sweeps equal areas in equal intervals of time.
The third law, the law of periods states that, , 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 is mathematically expressed as:
\[{T^2}\] is proportional to \[{a^3}\]
Where ‘$a$’ is the semi-major axis.
Removing the proportionality we get:
\[{T^2} = \dfrac{{2\pi {a^3}}}{{GM}}\]
Where,
$T$= Time period of the satellite
$A$= semi-major axis
$G$= gravitational constant
$M$= mass of the earth (in this case)
Here, \[\dfrac{{2\pi }}{{GM}}\] is the constant of proportionality.
So: \[T = \dfrac{{2\pi {a^{\dfrac{3}{2}}}}}{{GM}}\]
Now, we know the semi-major axis of an elliptical path is the arithmetic mean of the minimum and the maximum distance covered by the planet. Thus if we proceed keeping the ratios constant, we arrive at :
\[{(\dfrac{{{T^{}}_1}}{{{T_2}}})^2} = {(\dfrac{{{r_1}}}{{{r_2}}})^3}\]
Thus, option (c) is correct.
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