Jupiter turns once in 9.93 hours, its equator's radius is 7.15 × 10⁷ m, and its pull there gives…

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Try it on paper first. A hint, the key idea and the worked solution are here when you need them.

Jupiter turns once in $9.93$ hours, its equator's radius is $7.15 \times 10^{7}\ \text{m}$ , and its pull there gives $24.8\ \text{m/s}^2$ .

How much of that is needed for the circular motion at the equator?
What would a scale read under an $80\ \text{kg}$ probe resting there, compared with $80\ \text{kg} \times 24.8\ \text{m/s}^2$ ?

Need a reminder? The key ideaShow

At a planet's surface $g = \dfrac{GM}{R^2}$ . At a height $h$ : $g\left(\dfrac{R}{R + h}\right)^2$ . At a depth $d$ in a uniform planet: $g\left(1 - \dfrac{d}{R}\right)$ . A spinning planet's equator needs $\omega^2R$ of it for the circular motion.

The worked solution

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Work it out on paper first. It is fine to get stuck or make a mistake: that is how we learn. When you are ready, go through the solution one step at a time, and check each step against your own.

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