A planet of mass m moves around the sun of ma | vedclass

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(C) According to Kepler's Third Law (Law of Periods),the square of the time period $T$ of revolution of a planet is proportional to the cube of the se

Vedclass · Accepted Answer

$(r_1+r_2)^{3/2}$

Vedclass · Answer

(C) According to Kepler's Third Law (Law of Periods),the square of the time period $T$ of revolution of a planet is proportional to the cube of the semi-major axis $a$ of the elliptical orbit.$T^2 \propto a^3$For an elliptical orbit with the sun at one focus,the maximum distance (aphelion) is $r_1 = a(1+e)$ and the minimum distance (perihelion) is $r_2 = a(1-e)$,where $e$ is the eccentricity.Adding these two distances:$r_1 + r_2 = a(1+e) + a(1-e) = 2a$Therefore,the semi-major axis is:$a = \frac{r_1 + r_2}{2}$Substituting this into Kepler's Third Law:$T^2 \propto \left(\frac{r_1 + r_2}{2}\right)^3$Since $2^3$ is a constant,we have:$T^2 \propto (r_1 + r_2)^3$Taking the square root of both sides:$T \propto (r_1 + r_2)^{3/2}$Thus,the correct option is $C$.

$A$ planet of mass $m$ moves around the sun of mass $M$ in an elliptical orbit. The maximum and minimum distances of the planet from the sun are $r_1$ and $r_2$ respectively. The time period of the planet is proportional to ..........

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