A planet revolves around the Sun in an ellipt | vedclass

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(A) According to the law of conservation of angular momentum,the angular momentum of the planet remains constant at all points in its orbit.$L = mvr \

Vedclass · Accepted Answer

$12$

Vedclass · Answer

(A) According to the law of conservation of angular momentum,the angular momentum of the planet remains constant at all points in its orbit.$L = mvr \sin 	heta = 	ext{constant}$.At the perihelion (closest) and aphelion (farthest) points,the velocity vector is perpendicular to the radius vector,so $	heta = 90^\circ$.Therefore,$m v_{\max} r_{\min} = m v_{\min} r_{\max}$.Given:$r_{\min} = 1.6 	imes 10^{12} \ m$$v_{\max} = 60 \ m/s$$r_{\max} = 8 	imes 10^{12} \ m$Substituting the values:$60 	imes (1.6 	imes 10^{12}) = v_{\min} 	imes (8 	imes 10^{12})$$v_{\min} = \frac{60 	imes 1.6 	imes 10^{12}}{8 	imes 10^{12}}$$v_{\min} = 60 	imes 0.2 = 12 \ m/s$.

$A$ planet revolves around the Sun in an elliptical orbit. When it is closest to the Sun,its distance from the Sun is $1.6 \times 10^{12} \ m$ and its velocity is $60 \ m/s$. What is its velocity in $m/s$ when it is farthest from the Sun,given its distance is $8 \times 10^{12} \ m$?

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