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An exoplanet is in an elliptical orbit around a distant star. At its closest approach, the exoplanet is 0.540 AU from the star and has a speed of 54.0 km/s. When the exoplanet is at its farthest distance from the star of 41.0 AU, what is its speed (in km/s)? (1 AU is the average distance from the Earth to the Sun and is equal to 1.496 ✕ 1011 m. You may assume that other planets and smaller objects in the star system exert negligible forces on the exoplanet.)

Respuesta :

boffeemadrid

Answer:

0.71121 km/s

Explanation:

[tex]v_1[/tex] = Velocity of planet initially = 54 km/s

[tex]r_1[/tex] = Distance from star = 0.54 AU

[tex]v_2[/tex] = Final velocity of planet

[tex]r_2[/tex] = Final distance from star = 41 AU

As the angular momentum of the system is conserved

[tex]mv_1r_1=mv_2r_2\\\Rightarrow v_1r_1=v_2r_2\\\Rightarrow v_2=\frac{v_1r_1}{r_2}\\\Rightarrow v_2=\frac{54\times 0.54}{41}\\\Rightarrow v_2=0.71121\ km/s[/tex]

When the exoplanet is at its farthest distance from the star the speed is 0.71121 km/s.

okpalawalter8

The speed of the expoplanet  is mathematically given as

v_2=0.71121 km/s

What is its speed?

Question Parameter(s):

At its closest approach, the exoplanet is 0.540 AU from the star and has a speed of 54.0 km/s.

At its closest approach, the exoplanet is 0.540 AU from the star and has a speed of 54.0 km/s.

Generally, the equation for the angular momentum   is mathematically given as

mv_1r_1=mv_2r_2

Thereofore

v_1r_1=v_2r_2

[tex]v_2=\frac{v_1r_1}{r_2}\\\\ v_2=\frac{54*0.54}{41}[/tex]

v_2=0.71121 km/s

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