Answered

Westonci.ca is the premier destination for reliable answers to your questions, provided by a community of experts. Discover reliable solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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 = 2Ï€/T

Hence,

V = r * 2Ï€/ T

V =44360 * 2 * π/10

V = 88720 * π/10

V = 278722.10 / 10

V = 27872.210

V = 27872.2 miles per hour

We appreciate your visit. Hopefully, the answers you found were beneficial. Don't hesitate to come back for more information. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for using Westonci.ca. Come back for more in-depth answers to all your queries.