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Any object, be it a space satellite or a molecule, must attain an initial upward velocity of at least \(11.2 \mathrm{~km} / \mathrm{s}\) in order to escape the gravitational attraction of the earth. What would be the kinetic energy in joules of a satellite weighing \(2354 \mathrm{lb}\) that has the speed equal to this escape velocity of \(11.2 \mathrm{~km} / \mathrm{s}\) ?

Short Answer

Expert verified
The kinetic energy is approximately \(6.69 \times 10^{10}\) Joules.

Step by step solution

01

Convert Mass from Pounds to Kilograms

First, convert the mass of the satellite from pounds to kilograms. We know that 1 pound is approximately equal to 0.453592 kilograms. Thus, the satellite's mass in kilograms is obtained by multiplying 2354 pounds by 0.453592.\[mass (kg) = 2354 \times 0.453592 = 1067.41 \text{ kg}\]
02

Convert Velocity from km/s to m/s

The escape velocity is given as 11.2 km/s. To convert this velocity to meters per second, multiply by 1000 (since 1 km = 1000 m).\[velocity (m/s) = 11.2 \times 1000 = 11200 \text{ m/s}\]
03

Use the Kinetic Energy Formula

The formula for kinetic energy \( KE \) is given by: \[KE = \frac{1}{2} m v^2\]where \( m \) is the mass in kilograms and \( v \) is the velocity in meters per second. Substitute the converted mass and velocity into the equation to find the kinetic energy.
04

Calculate the Kinetic Energy

Substitute the values from the previous steps into the kinetic energy formula:\[KE = \frac{1}{2} \times 1067.41 \times (11200)^2\]\[KE = \frac{1}{2} \times 1067.41 \times 125440000\]\[KE = 6.6927 \times 10^{10} \text{ Joules}\]

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Key Concepts

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

Escape Velocity
Escape velocity is the minimum speed that an object must have in order to break free from the gravitational pull of a celestial body without any additional propulsion. For Earth, this speed is approximately 11.2 kilometers per second (km/s). This is the speed required to ensure that an object can enter space and not fall back due to Earth's gravity. Achieving escape velocity means that the object's kinetic energy is sufficient to surpass the gravitational energy that holds it back. Hence, it doesn't require further energy input from rockets after reaching that velocity.
When designing space missions, engineers must consider escape velocity to plan how much fuel and initial thrust are needed. Failing to reach this speed means that the spacecraft will eventually be pulled back to Earth by gravity.
Gravitational Attraction
Gravitational attraction is the force that pulls two masses toward each other. It is governed by Isaac Newton's Law of Universal Gravitation, which states that every mass attracts every other mass in the universe. The force of attraction is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
For anything launched from Earth, gravitational attraction plays a crucial role in its trajectory. It is a constant force trying to pull objects back toward the center of the Earth. In our exercise context, overcoming this gravitational attraction with sufficient kinetic energy will allow an object, like a satellite, to escape into space.
Mass Conversion
Mass conversion is a crucial step in calculating physical properties such as kinetic energy. In our exercise, the mass of the satellite was initially given in pounds, a unit commonly used in the United States. However, scientific calculations often require mass in kilograms, which is part of the metric system.
To convert from pounds to kilograms, it is essential to use the conversion factor: 1 pound is approximately 0.453592 kilograms. This conversion ensures that calculations are consistent with the International System of Units (SI) used in science and engineering. Such consistency is vital for accurate computations and comparisons of data across different systems.
Velocity Conversion
Velocity conversion is necessary when dealing with different measurement units. In the exercise, the escape velocity is provided in kilometers per second (km/s), but the kinetic energy formula requires velocity in meters per second (m/s).
The conversion is straightforward since there are 1000 meters in a kilometer. Therefore, to convert kilometers per second to meters per second, you multiply the velocity by 1000. This ensures that the kinetic energy can be accurately calculated in joules, as the units for kinetic energy require mass in kilograms and velocity in meters per second, following the standard scientific conventions.

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Most popular questions from this chapter

Define the heat capacity of a substance. Define the specific heat of a substance.

A piece of iron was heated to \(95.4^{\circ} \mathrm{C}\) and dropped into a constant-pressure calorimeter containing \(284 \mathrm{~g}\) of water at \(32.2^{\circ} \mathrm{C}\). The final temperature of the water and iron was \(51.9^{\circ} \mathrm{C}\). Assuming that the calorimeter itself absorbs a negligible amount of heat, what was the mass (in grams) of the piece of iron? The specific heat of iron is \(0.449 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\), and the specific heat of water is \(4.18 \mathrm{~J} /\left(\mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)\).

Hydrogen cyanide is a highly poisonous, volatile liquid. It can be prepared by the reaction $$ \mathrm{CH}_{4}(g)+\mathrm{NH}_{3}(g) \longrightarrow \mathrm{HCN}(g)+3 \mathrm{H}_{2}(g) $$ What is the heat of reaction at constant pressure? Use the following thermochemical equations: $$ \begin{gathered} \mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{NH}_{3}(g) ; \Delta H=-91.8 \mathrm{~kJ} \\ \mathrm{C}(\text { graphite })+2 \mathrm{H}_{2}(g) \longrightarrow \mathrm{CH}_{4}(g) ; \Delta H=-74.9 \mathrm{~kJ} \\ \mathrm{H}_{2}(g)+2 \mathrm{C}(\text { graphite })+\mathrm{N}_{2}(g) \longrightarrow 2 \mathrm{HCN}(g) \\ \Delta H=270.3 \mathrm{~kJ} \end{gathered} $$

You discover that you cannot carry out a particular reaction for which you would like the enthalpy change. Does this mean that you will be unable to obtain this enthalpy change? Explain.

When \(15.3 \mathrm{~g}\) of sodium nitrate, \(\mathrm{NaNO}_{3}\), was dissolved in water in a calorimeter, the temperature fell from \(25.00^{\circ} \mathrm{C}\) to \(21.56^{\circ} \mathrm{C}\). If the heat capacity of the solution and the calorimeter is \(1071 \mathrm{~J} /{ }^{\circ} \mathrm{C}\), what is the enthalpy change when \(1 \mathrm{~mol}\) of sodium nitrate dissolves in water? The solution process is $$ \mathrm{NaNO}_{3}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) ; \Delta H=? $$

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