A Ferris wheel with a diameter of 35 m starts from rest and achieves its maximum operational tangential speed of 2.3 m/s in a ti
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This question can be solved from the Kepler's law of planetary motion.
As per this law the square of time period of a planet is proportional to the cube of semi major axis.
Mathematically it can be written as
⇒
Here K is the proportionality constant.
If and
are the orbital periods of the planets and
and
are the distance of the planets from the sun, then Kepler's law can be written as-
⇒
Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.
Let the distance from sun and orbital period of Saturn is denoted as and
respectively.
Let the distance from sun and orbital period of earth is denoted as and
respectively.
we are given that
we know that 1 AU and
1 year.
1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.
Hence distance of Saturn from sun is calculated as -
From Kepler's law as mentioned above-
=
⇒
=9.5386 AU [ans]