The planet jupiter moves in an elliptical orbit with the sun at one focus. given that jupiter's closest approach to the sun is approximately 740.52 million kilometers and that the eccentricity of jupiter's orbit is approximately 0.048, estimate this planet's maximum distance from the sun. express your answer as a decimal rounded to two decimal places.

814.62 million kilometers

Explanation:

eccentricity of elliptical orbit =0.048

Let semi major axis of elliptical path be a.

From center of the ellipse focus is at a distance = a*eccentricity

Now, closest approach from focus is = a(1-e)

After plugging values = a(1-0.048)=740.52

                                         a=777.31 million kilometers

Maximum distance from sun (apihalian) =a+a*e =a(1+e)

                                                 = 777.31(1+0.048)

                                                 =814.62 million kilometers

the answer would be c. motion and direction

i hope this : )