Question

Jupiter, the largest planet in the solar system, has an equatorial radius of about 7.1 x 10^4km (more than 10 times that of Earth). Its period of rotation, however, is only 9h, 50 min. That means that every point on Jupiter's equator "goes around the planet" in that interval of time. Calculate the average speed (in m/s) of an equatorial point during one period of Jupiter's rotation. Is the average velocity different from the average speed in this case?

Answer 1

Answer:

The average speed is $v = 1260 \ km/s$

The average speed is different from the average velocity in this question

Explanation:

From the question we are told that

   The equatorial  radius of Jupiter is  $R_j = 7.1*10^{4} \ km$

   The period of oscillation of Jupiter is $T_J = 9 \ hours , 50 \ min = 35400 \ seconds$

Generally the average speed is mathematically represented as

      $v = \frac{2 \pi * R_j }{T_J}$

=>   $v = \frac{2 *3.142 * 7.1*10^{4} }{35400}$

=>   $v = 1260 \ km/s$

Generally in average speed the direction is not considered while in average velocity the direction is considered for the  case of this question the movement equitorial point has no direction in that it start from one point and after its periodic motion it still remains at that point