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The planet Mercury travels in an elliptical orbit with eccentricity 0.203. Its minimum distance from the Sun is 4.5 x 10^7 km. If the perihelion distance from a planet to the Sun is a(1 - e) and the aphelion distance is a(1 + e), find the maximum distance (in km) from Mercury to the Sun.Pick from the following:1. 7.7 x 10^7 km.2. 6.6 x 10^7 km.3. 6.8 x 10^7 km.

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okpalawalter8

Answer:

Option C

Step-by-step explanation:

From the question we are told that:

Eccentricity [tex]e=0.203[/tex]

Minimum distance from the Sun [tex]d_s= 4.5 x 10^7 km[/tex]

Perihelion distance from a planet to the Sun is [tex]r= a(1 - e)[/tex]

Aphelion distance [tex]r'=a(1 + e)[/tex]

Generally the equation for Perihelion distance is mathematically given by

 [tex]4.5 * 10^7= a(1 - 0.203)[/tex]

 [tex]4.5 * 10^7 = 0.797a[/tex]

 [tex]a = 56.46 * 10^6 km[/tex]

Generally the equation for Aperihelion distance is mathematically given by

 [tex]r' = a(1 + e)[/tex]

 [tex]r' = 56.4617 * 10^6 (1 + 0.203)[/tex]

 [tex]r'=6.8 * 10^7 km[/tex]

Option C

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