## Related questions with answers

Jupiter is about $320$ times as massive as the Earth. Thus, it has been claimed that a person would be crushed by the force of gravity on a planet the size of Jupiter because people cannot survive more than a few $g$ 's. Calculate the number of $g$ 's a person would experience at Jupiter's equator, using the following data for Jupiter: mass = $1.9 \times 10^{27} \mathrm{~kg}$, equatorial radius $=7.1 \times 10^4 \mathrm{~km}$, rotation period $=9 \mathrm{~hr~} 55 \mathrm{~min}$. Take the centripetal acceleration into account.

Solution

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$\begin{align*} m_J&=1.9\cdot 10^{27}\text{ kg},\\ r_J&=7.1\cdot 10^4\text{ km},\\ T&=9\text{ hr}\ 55\text{ min}= 35\ 700 \text{ s}. \end{align*}$

Let us first find the free fall acceleration on Jupiter's equator taking into accound the rotation of Jupiter. Consider an object of mass $m$ on the surface of the Jupiter. Its rotational speed can be found from its period:

$v=\frac{2\pi r_J}{T}.$

The weight of the body will be equal to the normal force the surface of the Jupiter exerts on the object $F_N$. The free fall acceleration is then weight divided by mass $m$ which, in turn, is $F_N/m$.

Since the object is rotating $F_N$ is less than the gravitational force since their difference has to provide the centripetal force:

$F_g-F_N=\frac{mv^2}{r_J}=m\frac{4\pi^2 r_J}{T^2}.$

This yields

$F_N=F_g-m\frac{4\pi^2 r_J}{T^2}=G\frac{mm_J}{r_J^2}-m\frac{4\pi^2 r_J}{T^2},$

where we have applied the universal law of gravity for $F_g$. As we have said earlier, the free fall acceleration will be

$g_J=\frac{F_N}{m}=G\frac{m_J}{r_J^2}-\frac{4\pi^2 r_J}{T^2}.$

Substituting given values we find

$\boxed{g_J=23\text{ m/s}^2=2.3g.}$

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