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NEET PHYSICSGRAVITATIONEasy

Question

Kepler's third law states that the square of the period of revolution (T) of a planet around the sun, is proportional to the third power of the average distance r between the sun and planet i.e. T2=Kr3T^2 = Kr^3, here K is constant. If the masses of the sun and planet are M and m respectively, then as per Newton's law of gravitation, the force of attraction between them is F=GMm/r2F = GMm/r^2, here G is gravitational constant. The relation between G and K is described as:

A

GK = 4\pi ^2

B

GMK = 4\pi ^2

C

K = G

D

K = 1/G

Step-by-Step Solution

The gravitational force provides the necessary centripetal force for the planet's circular motion around the Sun: Fg=FcGMmr2=mv2rF_g = F_c \Rightarrow \frac{GMm}{r^2} = \frac{mv^2}{r}. Solving for orbital velocity vv, we get v=GMrv = \sqrt{\frac{GM}{r}}. The time period of revolution TT is given by T=2πrvT = \frac{2\pi r}{v}. Substituting vv: T=2πrGM/r=2πr3GMT = \frac{2\pi r}{\sqrt{GM/r}} = 2\pi \sqrt{\frac{r^3}{GM}}. Squaring both sides: T2=(4π2GM)r3T^2 = \left(\frac{4\pi^2}{GM}\right) r^3. Comparing this with the given Kepler's law equation T2=Kr3T^2 = Kr^3, we can see that the constant KK is: K=4π2GMK = \frac{4\pi^2}{GM}. Rearranging the terms, we get: GMK=4π2GMK = 4\pi^2.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from GRAVITATION. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSGRAVITATIONkeplersstatessquareperiodrevolution

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