what is the angular speed, in rad/s, of a point on the surface of Jupiter at latitude 72 degrees N?
It would be 2 pi divided by the rotation period of Jupiter, in seconds. Although Jupiter is not a rigid body, the differences between rotational speeds at different latitudes or "shear" is not large.
For the Jupiter rotation period, see http://www.windows.ucar.edu/tour/link=/jupiter/statistics.html
To calculate the angular speed of a point on the surface of Jupiter at a given latitude, we first need to know the rotation period of Jupiter. The rotation period of Jupiter is approximately 9.93 hours, or 35460 seconds.
The angular speed is a measure of how quickly an object rotates per unit time. In this case, we want to find the angular speed in radians per second (rad/s).
To calculate the angular speed, we can use the formula:
Angular Speed (ω) = 2π / Time Period (T)
In this case, the Time Period (T) is the rotation period of Jupiter.
Now let's calculate the angular speed.
Angular Speed (ω) = 2π / 35460 seconds
Angular Speed (ω) ≈ 0.000177 rad/s
Therefore, the angular speed of a point on the surface of Jupiter at latitude 72 degrees N is approximately 0.000177 rad/s.