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What multiple of the energy needed to escape from Earth gives the energy needed to escape from (a) a moon of mass \(1.472 \times 10^{23} \mathrm{~kg}\) and radius \(3.480 \times 10^{6} \mathrm{~m}\) and (b) Jupiter?

Short Answer

Expert verified
The moon requires approximately 0.17 times Earth's energy; Jupiter requires about 11.2 times Earth's energy.

Step by step solution

01

Understanding the Gravitational Potential Energy Formula

The formula for gravitational potential energy for an object at the surface of a celestial body is \( E = \frac{GMm}{r} \), where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, \( m \) is the mass of the object, and \( r \) is the radius of the celestial body.
02

Defining Energy to Escape from Earth

The escape energy required from Earth can be calculated using the formula \( E_{Earth} = \frac{GM_{Earth}m}{R_{Earth}} \) where \( M_{Earth} \) is the mass of Earth and \( R_{Earth} \) its radius. For Earth, \( M_{Earth} = 5.972 \times 10^{24} \mathrm{~kg} \) and \( R_{Earth} = 6.371 \times 10^{6} \mathrm{~m} \).
03

Calculating Escape Energy for the Moon

Using the formula \( E_{Moon} = \frac{GM_{Moon}m}{R_{Moon}} \) where \( M_{Moon} = 1.472 \times 10^{23} \mathrm{~kg} \) and \( R_{Moon} = 3.480 \times 10^{6} \mathrm{~m} \). The energy will be \( E_{Moon} = \frac{G \times 1.472 \times 10^{23}}{3.480 \times 10^{6}} \). Use the value of \( G = 6.674 \times 10^{-11} \frac{\mathrm{m}^3}{\mathrm{kg}\cdot \mathrm{s}^2} \).
04

Calculating Escape Energy for Jupiter

The energy required to escape from Jupiter is given by \( E_{Jupiter} = \frac{GM_{Jupiter}m}{R_{Jupiter}} \) where \( M_{Jupiter} = 1.898 \times 10^{27} \mathrm{~kg} \) and \( R_{Jupiter} = 6.991 \times 10^{7} \mathrm{~m} \). Calculate \( E_{Jupiter} = \frac{6.674 \times 10^{-11} \times 1.898 \times 10^{27}}{6.991 \times 10^{7}} \).
05

Calculating the Ratio of Energies

For both the moon and Jupiter, calculate the ratio \( \text{Ratio} = \frac{E_{Body}}{E_{Earth}} \). For the moon and for Jupiter, the energy for Earth \( E_{Earth} = \frac{6.674 \times 10^{-11} \times 5.972 \times 10^{24}}{6.371 \times 10^{6}} \). Evaluate each ratio to determine the multiple of Earth's escape energy needed.

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Key Concepts

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

Celestial Mechanics
Celestial mechanics is the branch of physics that deals with the motions and gravitational interactions of celestial bodies, like planets, moons, and stars. It helps us understand how these bodies move in space, influenced primarily by gravity. The gravitational force between any two masses in the universe is described by Newton's law of universal gravitation. This law states that every point mass attracts every other point mass in the universe with a force that is proportional to the product of their masses and inversely proportional to the square of the distance between their centers.
  • This is crucial in calculating the energy required for motions such as an object escaping a planet's gravitational pull.
  • The gravitational potential energy on the surface of a celestial body represents the work done against gravity to bring an object from infinity to that point.
Understanding celestial mechanics is fundamental when analyzing how spacecraft travel orbits and when considering efforts to reach other planets or celestial bodies.
Escape Velocity
Escape velocity is the minimum speed a body must achieve to break free from the gravitational attraction of a massive object, without any additional propulsion. This means once that speed is attained, no further force need be applied. The formula for escape velocity from a celestial body is derived using the concept of gravitational potential energy and kinetic energy. The equation is given by \( v_e = \sqrt{\frac{2GM}{r}} \) where \( G \) is the gravitational constant, \( M \) is the mass of the object to be escaped from, and \( r \) is the radius from the center of the object to the point of escape.
  • The larger the mass and the smaller the radius of the celestial body, the greater the escape velocity required.
  • Planets and celestial bodies with larger masses generally require higher escape velocities.
Escape velocity is a fundamental concept when considering launching spacecraft or objects into space.
Gravitational Constant
The gravitational constant, denoted as \( G \), is a key parameter in the universal law of gravitation. It is what makes the calculations of gravitational force possible in real-world terms. The gravitational constant has a standard value of \( 6.674 \times 10^{-11} \frac{\mathrm{m}^3}{\mathrm{kg} \cdot \mathrm{s}^2} \). This constant ensures that we can relate the masses and the distances in the universe to the forces experienced by those masses.
  • Without \( G \), it would be impossible to calculate accurately the forces or potential energies involved in celestial mechanics.
  • It provides a bridge between the macro world of celestial bodies and the more familiar terrestrial world with its standard units.
Understanding \( G \) is crucial for solving any physics problem involving gravitational forces.
Physics Problem Solving
Physics problem solving is an essential skill that involves identifying the underlying concepts needed to understand and work with real-world phenomena. When dealing with problems related to gravitational forces, celestial mechanics, and energy needs for space travel, the following steps are typically involved:
  • Clearly understand the problem and identify known and unknown variables.
  • Apply appropriate formulas, such as those for gravitational potential energy or escape velocity.
  • Calculate step-by-step, ensuring each calculation is correct before proceeding to the next.
  • Analyze the results to ensure they make sense within the context of the problem.
Developing strong problem-solving skills in physics can aid in tackling various challenges from determining the energetics of space missions to understanding everyday gravitational interactions.

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