Related questions with answers

A satellite orbiting above the equator of the earth is geosynchronous if the period is T = 24 hours (in this case, the satellite stays over a fixed point on the equator). Use Kepler's Third Law to show that in a circular geosynchronous orbit, the distance from the center of the earth is

R \approx 42,246 \mathrm { km }.

Then compute the altitude h of the orbit above the earth's surface. The earth has mass

M \approx 5.974 \times 10 ^ { 24 } \mathrm { kg }

and radius

R \approx 6371 \mathrm { km }.

The planet 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 Jupiter since people can't survive more than a few $g$ 's. Calculate the number of $g$ 's a person would experience if she could stand on the equator of Jupiter. Use the following astronomical data for Jupiter: mass $= 1.9\times10^{27}$ kg, equatorial radius $= 7.1\times10^4$ km, rotation period $9$ hr $55$ min. Take the centripetal acceleration into account.

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= $1.9\times 10^{27}$ kg, equatorial radius= $7.1\times 10^4$ km, rotation period= $9$ hr $55$ min. Take the centripetal acceleration into account.

Question

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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Set

\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 $F_N/m$ .