What is the eccentricity of the orbit of a planet whose distance from the sun varies 180 million to 220 million km?

Eccentricity is found by the distance from a focus to the center of an ellipse along the semi-major axis divided by the semi-major axis length.

Remember that the sun is always a focus point to a planet orbiting around the sun.

11)  The semi-major axis length is (180+220)/2 = 200

      220-200= 20 <= distance from focus to the center of the ellipse along the semi-major axis length

        e= 20/200 = 0.1

Here's the answer for the next question too :)

12)  (147.2+152.1)/2= 149.65

       152.1-149.65= 2.45

        e= 2.45/149.65 = 0.0164

Explanation:

yahoo answered these questions 8 years ago O.o

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

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