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.