Question

Jupiter, the largest planet in the solar system, has an equatorial radius of about 7.1 x 104 km (more than 10 times that of Earth). Its period of rotation, however, is only 9 h 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.

Answer 1

The average speed of an equatorial point is given by the ratio between the distance covered during one complete rotation (therefore, the circumference of Jupiter at the equator) and the time to complete one rotation:
$v= \frac{S}{t}$
where S is the circumference and t the time.

The radius of Jupiter is
$r=7.1 \cdot 10^4 km = 7.1 \cdot 10^7 m$
so the circumference at the equator is:
$p=2 \pi r = 2 \pi (7.1 \cdot 10^7 m)=4.46 \cdot 10^8 m$

While the time to complete one rotation is (in seconds)
$t=9h 50 min=590 min = 35400 s$

Therefore the average speed of a point at the equator is
$v= \frac{4.46 \cdot 10^8 m}{35400 s}=1.26 \cdot 10^4 m/s$