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What is the Newton version of Kepler’s third law?

Answer
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Hint: We first write the gravitational force and then write Kepler's third law which is a relationship between time period and the semi-axis. Now, write the centrifugal force and then convert the centrifugal in the terms of T and R. Then, compare gravitational force and centrifugal force. We then get Newton's version of Kepler’s third law.

Complete answer:
Newton’s Law,
$F_{g}=G\dfrac{M_{s}M_{p}}{R^{2}}$
Where $ M_{s}$ and $ M_{s}$ are the mass of the sun and mass of the planet respectively.
G is a constant value and R is the distance between Sun and Planet.
Kepler's Law is
$\dfrac{T^{2}}{R^{3}} = K$
K is a constant.
T is the period of translation in orbit and R is the distance between Sun and Planet.
As we know the centrifugal force,
$F_{c} = M_{p} \times a = M_{p} \left(\dfrac{2 \pi}{T}\right)^{2}R$
where a is acceleration in orbit.
Combine both forces,
We get,
$\dfrac{T^{2}}{R^{3}} = \dfrac{4 \pi^{2}}{M_{s}}$
Hence, it is the Newton version of Kepler’s third law.

Additional Information:
Kepler's third law says that the Sun cubed's average length is directly proportional to the orbital period squared. Newton saw that his gravitation force law could describe Kepler's laws. Kepler observed this law acted for the planets because they all revolve around the same star.

Note:
Newton's version of Kepler's Law permits us to measure distances to far objects. It can be applied to find the masses of many far objects. It describes why objects rotate faster when they contract in size. It informs us that faraway planets revolve around the Sun more slowly. Small the field of the planet around the Sun, smaller the time taken to complete one orbit.