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Chapter 17: Problem 93

Spacecraft Reentry. A spacecraft made of aluminum circles the earth at a speed of 7700 \(\mathrm{m} / \mathrm{s} .\) (a) Find the ratio of its kinetic energy to the energy required to raise its temperature from \(0^{\circ} \mathrm{C}\) to \(600^{\circ} \mathrm{C}\) . (The melting point of aluminum is \(660^{\circ} \mathrm{C}\) . Assume a constant specific heat of 910 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K} .\) (b) Discuss the bearing of your answer on the problem of the reentry of a manned space vehicle into the earth's atmosphere.

Short Answer

Expert verified
The kinetic energy is 54 times greater than the needed thermal energy to reach 600\( ^{\circ} \)C, emphasizing the importance of thermal protection during reentry.

Step by step solution

01

Calculate the Kinetic Energy

The kinetic energy (KE) of the spacecraft can be calculated using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the spacecraft and \( v = 7700 \, \text{m/s} \) is its velocity. Since the mass is not specified, we will assume the mass to be an arbitrary \( m \) and calculate the KE in terms of \( m \): \[ KE = \frac{1}{2}m(7700)^2 = 2.9645 \times 10^{7} m \text{ J} \]
02

Calculate the Thermal Energy Required

The energy required to raise the temperature of the spacecraft from \( 0^{\circ} \text{C} \) to \( 600^{\circ} \text{C} \) can be found by \( Q = mc\Delta T \), where \( c = 910 \, \text{J} / \text{kg} \cdot \text{K} \) is the specific heat capacity and \( \Delta T = 600 \, \text{K} \). Substituting these values gives:\[ Q = m \times 910 \times 600 = 546,000 m \text{ J} \]
03

Determine the Ratio of Kinetic Energy to Thermal Energy

The ratio of the kinetic energy to the thermal energy is found by dividing the kinetic energy by the thermal energy:\[ \text{Ratio} = \frac{KE}{Q} = \frac{2.9645 \times 10^{7} m}{546,000 m} = \frac{2.9645 \times 10^{7}}{546,000} \approx 54.26 \]
04

Discuss the Reentry Implication

The ratio of about 54:1 means the kinetic energy is 54 times greater than the energy required to raise the temperature to near the melting point. This indicates that during reentry, a significant portion of the kinetic energy must be absorbed and dissipated to prevent the spacecraft's temperature from reaching or exceeding the melting point of aluminum, highlighting the critical role of thermal protection systems.

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Key Concepts

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

Understanding Kinetic Energy
Kinetic energy is a fundamental concept in physics associated with the motion of an object. It is given by the equation:
  • \( KE = \frac{1}{2}mv^2 \)
where \( m \) is the mass of the object and \( v \) is its velocity. Kinetic energy increases with both the mass and the square of the velocity of the object. This means that even small increases in velocity can result in large increases in kinetic energy.
For a spacecraft traveling at extremely high speeds, such as 7700 meters per second, its kinetic energy is substantial. The energy is pivotal during reentry, as it needs to be managed to avoid damage to the spacecraft. Understanding kinetic energy helps design protective systems that can handle the energy changes experienced during reentry.
The Role of Specific Heat Capacity
Specific heat capacity is the amount of energy required to raise the temperature of one kilogram of a substance by one degree Celsius (or one Kelvin). In our scenario, aluminum has a specific heat capacity of 910 J/kg路K. The formula for calculating the energy required to change the temperature is:
  • \( Q = mc\Delta T \)
This means, to raise the temperature from 0掳C to 600掳C, it requires a calculated amount of energy based on the mass of aluminum and its specific heat capacity. Specific heat capacity is a crucial factor when considering how much temperature change a spacecraft can endure without structural integrity failure.
Importance of Thermal Protection Systems
Thermal protection systems (TPS) are essential for safeguarding spacecraft during reentry into Earth's atmosphere. Given the enormous kinetic energy that must be dissipated, a TPS is designed to absorb and withstand high levels of heat generated as kinetic energy transitions to thermal energy.
These systems use materials with high melting points and adequate specific heat capacities to minimize the heat reaching critical parts of the spacecraft.
  • For instance, ablative materials that burn off and take excess heat with them or insulating materials that reflect heat.
A well-designed TPS ensures the spacecraft doesn't reach the melting point of materials like aluminum, thus preserving its structural integrity.
Understanding Temperature Change in Reentry
The temperature change a spacecraft undergoes during reentry is drastic because of the conversion of kinetic energy into thermal energy.
  • This conversion is due to aerodynamic heating as the spacecraft interacts with the Earth's atmosphere.
Using the calculated energy needed to raise the spacecraft to near its melting point, we can appreciate the challenge of managing significant heat increases. Understanding this helps engineers create cooling strategies and select materials that will handle the extreme temperature changes effectively without compromising the spacecraft.
The study of these temperature changes is key in ensuring a safe and successful return to Earth, balancing the kinetic energy transformation into heat and the ability of the spacecraft to dissipate it without overheating.

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Most popular questions from this chapter

On a cool \(\left(4.0^{\circ} \mathrm{C}\right)\) Saturday morning, a pilot fills the fuel tanks of her Pitts \(\mathrm{S}-2 \mathrm{C}(\) a two-seat aerobatic airplane) to their full capacity of 106.0 \(\mathrm{L} .\) Before flying on Sunday morning, when the temperature is again \(4.0^{\circ} \mathrm{C},\) she checks the fuel level and finds only 103.4 \(\mathrm{L}\) of gasoline in the tanks. She realizes that it was hot on Saturday afternoon, and that thermal expansion of the gasoline caused the missing fuel to empty out of the tank's vent. (a) What was the maximum temperature (in \(^{\circ} \mathrm{C} )\) reached by the fuel and the tank on Saturday afternoon? The coefficient of volume expansion of gasoline is \(9.5 \times 10^{-4} \mathrm{K}^{-1}\) , and the tank is made of aluminum. (b) In order to have the maximum amount of fuel available for flight, when should the pilot have filled the fuel tanks?

Why Do the Seasons Lag? In the northern hemisphere, June 21 (the summer solstice) is both the longest day of the year and the day on which the sun's rays strike the earth most vertically, hence delivering the greatest amount of heat to the surface. Yet the hottest summer weather usually occurs about a month or so later. Let us see why this is the case. Because of the large specific heat of water, the oceans are slower to warm up than the land (and also slower to cool off in winter). In addition to perusing pertinent information in the tables included in this book, it is useful to know that approximately two-thirds of the earth's surface is ocean composed of salt water having a specific heat of 3890 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and that the oceans, on the average, are 4000 m deep. Typically, an average of 1050 \(\mathrm{W} / \mathrm{m}^{2}\) of solar energy falls on the earth's surface, and the oceans absorb essentially all of the light that strikes them. However, most of that light is absorbed in the upper 100 \(\mathrm{m}\) of the surface. Depths below that do not change temperature seasonally. Assume that the sunlight falls on the surface for only 12 hours per day and that the ocean retains all the heat it absorbs. What will be the rise in temperature of the upper 100 \(\mathrm{m}\) of the oceans during the month following the summer solstice? Does this seem to be large enough to be perceptible?

One end of an insulated metal rod is maintained at \(100.0^{\circ} \mathrm{C},\) and the other end is maintained at \(0.00^{\circ} \mathrm{C}\) by an ice-water mixture. The rod is 60.0 \(\mathrm{cm}\) long and has a cross-sectional area of 1.25 \(\mathrm{cm}^{2} .\) The heat conducted by the rod melts 8.50 \(\mathrm{g}\) of ice in 10.0 \(\mathrm{min} .\) Find the thermal conductivity \(k\) of the metal.

In a container of negligible mass, 0.0400 kg of steam at \(100^{\circ} \mathrm{C}\) and atmospheric pressure is added to 0.200 \(\mathrm{kg}\) of water at \(50.0^{\circ} \mathrm{C} .\) (a) If no heat is lost to the surroundings, what is the final temperature of the system? (b) At the final temperature, how many kilograms are there of steam and how many of liquid water?

You pour 108 \(\mathrm{cm}^{3}\) of ethanol, at a temperature of \(-10.0^{\circ} \mathrm{C},\) into a graduated cylinder initially at \(20.0^{\circ} \mathrm{C},\) filling it to the very top. The cylinder is made of glass with a specific heat of 840 \(\mathrm{J} / \mathrm{kg} \cdot \mathrm{K}\) and a coefficient of volume expansion of \(1.2 \times 10^{-5} \mathrm{K}^{-1} ;\) its mass is 0.110 \(\mathrm{kg} .\) The mass of the ethanol is 0.0873 \(\mathrm{kg} .\) (a) What will be the final temperature of the ethanol, once thermal equilibrium is reached? (b) How much ethanol will overflow the cylinder before thermal equilibrium is reached?
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