A comet (assumed to be in an elliptical orbit | vedclass

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(C) For a comet in an elliptical orbit,the semi-major axis $a$ is given by the average of the perihelion distance $(r_p)$ and the aphelion distance $(

Vedclass · Accepted Answer

$49.6$

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(C) For a comet in an elliptical orbit,the semi-major axis $a$ is given by the average of the perihelion distance $(r_p)$ and the aphelion distance $(r_a)$:$a = \frac{r_p + r_a}{2} = \frac{0.4 + x}{2} \, AU$According to Kepler's Third Law,the square of the time period $T$ is proportional to the cube of the semi-major axis $a$:$T^2 \propto a^3$Comparing the comet with Earth ($T_e = 1 \, yr$,$a_e = 1 \, AU$):$\left(\frac{T}{T_e}ight)^2 = \left(\frac{a}{a_e}ight)^3$Substituting the given values $(T = 125 \, yr)$:$\left(\frac{125}{1}ight)^2 = \left(\frac{0.4 + x}{2 	imes 1}ight)^3$$(125)^2 = \left(\frac{0.4 + x}{2}ight)^3$Taking the cube root of both sides:$(125)^{2/3} = \frac{0.4 + x}{2}$$(5^3)^{2/3} = \frac{0.4 + x}{2}$$5^2 = \frac{0.4 + x}{2}$$25 = \frac{0.4 + x}{2}$$50 = 0.4 + x$$x = 50 - 0.4 = 49.6 \, AU$Thus,the aphelion distance is $49.6 \, AU$.

$A$ comet (assumed to be in an elliptical orbit around the sun) is at a distance of $0.4 \, AU$ from the sun at the perihelion. If the time period of the comet is $125 \, yr$,the aphelion distance is ........... $AU$ ($AU$: Astronomical Unit).

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