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Chapter 13: Problem 50

Submarines on Europa. Some scientists are eager to send a remote-controlled submarine to Jupiter's moon Europa to search for life in its oceans below an icy crust. Europa's mass has been measured to be \(4.8 \times 10^{22}\) kg, its diameter is \(3138 \mathrm{km},\) and it has no appreciable atmosphere. Assume that the layer of ice at the surface is not thick enough to exert substantial force on the water. If the windows of the submarine you are designing are 25.0 \(\mathrm{cm}\) square and can stand a maximum inward force of 9750 N per window, what is the greatest depth to which this submarine can safely dive?

Short Answer

Expert verified
The submarine can dive safely to a depth of about 118.18 km in Europa's ocean.

Step by step solution

01

Calculate the radius of Europa

First, we find the radius of Europa using its diameter. The formula for radius is \( r = \frac{d}{2} \), where \( d \) is the diameter. Diameter of Europa, \( d = 3138 \text{ km} = 3138 \times 10^3 \text{ m} \). Thus, the radius \( r = \frac{3138 \times 10^3}{2} = 1569 \times 10^3 \text{ m} \).
02

Calculate the gravitational acceleration on Europa

The gravitational acceleration \( g \) on the surface of a celestial body is given by the formula: \[ g = \frac{G \, M}{r^2} \] where \( G = 6.674 \times 10^{-11} \text{ m}^3\text{/kg/s}^2 \) is the gravitational constant, \( M = 4.8 \times 10^{22} \text{ kg} \) is the mass of Europa, and \( r = 1569 \times 10^3 \text{ m} \) is the radius of Europa. Substituting these values, we get: \[ g = \frac{6.674 \times 10^{-11} \times 4.8 \times 10^{22}}{(1569 \times 10^3)^2} \approx 1.32 \text{ m/s}^2 \].
03

Determine the maximum pressure the window can withstand

The pressure \( P \) exerted by a force on an area is given by the formula: \[ P = \frac{F}{A} \] where \( F = 9750 \text{ N} \) is the maximum force the window can withstand, and \( A = 0.25 \times 0.25 \text{ m}^2 = 0.0625 \text{ m}^2 \) is the area of the window (since it is 25 cm square).So, the maximum pressure is: \[ P = \frac{9750}{0.0625} = 156000 \text{ Pa} \] (Pascals).
04

Calculate the depth to which pressure equals maximum pressure

The pressure under a fluid is given by the equation: \[ P = \rho g h \] where \( \rho \) is the density of the fluid, \( g \) is the gravitational acceleration, and \( h \) is the depth. Rearrange to solve for \( h \): \[ h = \frac{P}{\rho g} \]Assuming the density of Europa's ocean is similar to water, \( \rho = 1000 \text{ kg/m}^3 \). Using \( P = 156000 \text{ Pa} \) and \( g = 1.32 \text{ m/s}^2 \), we find: \[ h = \frac{156000}{1000 \times 1.32} \approx 118181.82 \text{ meters} \].
05

Final Calculation and Result

Rewriting the depth \( h \) in kilometers (since it's a very deep depth), we have:\[ h \approx 118.18 \text{ km} \]This is the maximum depth the submarine can safely dive without breaking its windows.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Europa's gravitational acceleration
When exploring celestial bodies, it's crucial to understand the local gravity, which affects everything from vehicle design to surface operations. Europa, one of Jupiter's moons, has a distinct gravitational feature due to its mass and size. The gravitational acceleration on Europa can be calculated using Newton's law of universal gravitation.By knowing Europa's mass, which is approximately \(4.8 \times 10^{22}\) kg, and its radius (calculated from its diameter of \(3138\) km), we find its gravitational acceleration using the equation: \[ g = \frac{G \cdot M}{r^2} \]Here, \(G\) is the gravitational constant \(6.674 \times 10^{-11}\) \(\text{m}^3/\text{kg/s}^2\). This equation helps us determine that Europa's surface gravity is about \(1.32 \text{ m/s}^2\), significantly lower than Earth's gravity. This relatively weak gravitational pull impacts how objects move and interact on Europa's icy surface.
Pressure Calculation
In designing submarines for extraterrestrial environments like those of Europa, understanding pressure is essential. Pressure is the force exerted per unit area, and it increases with depth in a fluid, owing to the weight of the fluid above pushing down.For a submarine's window, the maximum pressure it can withstand without breaking can be calculated by dividing the maximum force by the window's area. Using the formula:\[ P = \frac{F}{A} \]For our scenario, where \(F = 9750\) N and the window area \(A = 0.0625\) \(\text{m}^2\), the maximum pressure is \(156000\) Pascals. This number tells us how deep the submarine can safely go without the windows being compromised by water pressure.Understanding this helps engineers develop safer and more efficient designs, crucial for exploring the vast and treacherous environments beneath extraterrestrial ice sheets.
Fluid Mechanics in Space
Fluid mechanics is a cornerstone of engineering practices, especially crucial in unusual extraterrestrial conditions, such as Europa's under-ice oceans. In these situations, the principles are similar to those on Earth but must account for unique variables like lower gravitational forces.Using the formula for pressure due to a fluid's depth:\[ P = \rho g h \]We can determine how deep a vehicle can submerge. Here, \(\rho\) is the fluid density (assumed to be similar to water on Earth at \(1000 \text{ kg/m}^3\)), \(g\) is the gravitational acceleration we calculated earlier, and \(h\) is the sought depth.By rearranging the formula to solve for depth (\(h\)), it becomes:\[ h = \frac{P}{\rho g} \]This tells us that for a submarine on Europa, supported by \(1.32 \text{ m/s}^2\) gravity, reaching depths of approximately \(118.18\) kilometers is feasible without structural failure, considering the limitations on pressure. This comprehensive understanding is vital for ensuring the reliability of scientific operations in such extreme environments.

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Most popular questions from this chapter

A thin, uniform rod has length \(L\) and mass \(M . \mathrm{A}\) small uniform sphere of mass \(m\) is placed a distance \(x\) from one end of the rod, along the axis of the rod (Fig. E13.32). (a) Calculate the gravitational potential energy of the rod-sphere system. Take the potential energy to be zero when the rod and sphere are infinitely far apart. Show that your answer reduces to the expected result when \(x\) is much larger than \(L .\) (Hint: Use the power series expansion for \(\ln (1+x)\) given in Appendix B.) Use \(F_{x}=-d U / d x\) to find the magnitude and direction of the gravitational force exerted on the sphere by the rod (see Section 7.4\()\) . Show that your answer reduces to the expected result when \(x\) is much larger than \(L .\)

Falling Hammer. A hammer with mass \(m\) is dropped from rest from a height \(h\) above the earth's surface. This height is not necessarily small compared with the radius \(R_{\mathrm{E}}\) of the earth. If you ignore air resistance, derive an expression for the speed \(v\) of the hammer when it reaches the surface of the earth. Your expression should involve \(h, R_{\mathrm{E}},\) and \(m_{\mathrm{E}},\) the mass of the earth.

(a) Suppose you are at the earth's equator and observe a satellite passing directly overhead and moving from west to east in the sky. Exactly 12.0 hours later, you again observe this satellite to be directly overhead. How far above the earth's surface is the satellite's orbit? (b) You observe another satellite directly overhead and traveling east to west. This satellite is again overhead in 12.0 hours. How far is this satellite's orbit above the surface of the earth?

An object in the shape of a thin ring has radius \(a\) and mass \(M .\) A uniform sphere with mass \(m\) and radius \(R\) is placed with its center at a distance \(x\) to the right of the center of the ring, along a line through the center of the ring, and perpendicular to its plane (see Fig. E13.33). What is the gravitational force that the sphere exerts on the ring-shaped object? Show that your result reduces to the expected result when \(x\) is much larger than \(a\).

The planet Uranus has a radius of \(25,560 \mathrm{km}\) and a surface acceleration due to gravity of 11.1 \(\mathrm{m} / \mathrm{s}^{2}\) at its poles. Its moon Miranda (discovered by Kuiper in 1948 is in a circular orbit about Uranus at an altitude of \(104,000 \mathrm{km}\) above the planet's surface. Miranda has a mass of \(6.6 \times 10^{19} \mathrm{kg}\) and a radius of 235 \(\mathrm{km}\). (a) Calculate the mass of Uranus from the given data. (b) Calculate the magnitude of Miranda's acceleration due to its orbital motion about Uranus. (c) Calculate the acceleration due to Miranda's gravity at the surface of Miranda. (d) Do the answers to parts (b) and (c) mean that an object released 1 \(\mathrm{m}\) above Miranda's surface on the side toward Uranus will fall \(u p\) relative to Miranda? Explain.
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