Answered

Looking for reliable answers? Westonci.ca is the ultimate Q&A platform where experts share their knowledge on various topics. Find reliable answers to your questions from a wide community of knowledgeable experts on our user-friendly Q&A platform. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.

How fast would an object have to travel on the surface of Jupiter at the equator to keep up with the Sun​ (that is, so the Sun would appear to remain in the same position in the​ sky)? Use the facts that the radius of Jupiter is approximately 44,360 miles and its revolution is approximately 10 hours.

Sagot :

fichoh

Answer:

27872.2 miles per hour

Explanation:

Given that :

Radius of Jupiter is approximately = 44,360 miles

Revolution is 10 hours ;

Jupiter makes one revolution in 10 hours :

Using the relation to obtain the velocity :

V = re

r = radius

w = 2π/T

Hence,

V = r * 2π/ T

V =44360 * 2 * π/10

V = 88720 * π/10

V = 278722.10 / 10

V = 27872.210

V = 27872.2 miles per hour

We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Keep exploring Westonci.ca for more insightful answers to your questions. We're here to help.