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Zero, a hypothetical planet, has a mass of \(5.0 \times 10^{23} \mathrm{~kg},\) a radius of \(3.0 \times 10^{6} \mathrm{~m}\) and no atmosphere. A \(10 \mathrm{~kg}\) space probe is to be launched vertically from its surface. (a) If the probe is launched with an initial energy of \(5.0 \times 10^{7} \mathrm{~J},\) what will be its kinetic energy when it is \(4.0 \times 10^{6} \mathrm{~m}\) from the center of Zero? (b) If the probe is to achieve a maximum distance of \(8.0 \times 10^{6} \mathrm{~m}\) from the center of Zero, with what initial kinetic energy must it be launched from the surface of Zero?

Short Answer

Expert verified
a) The kinetic energy is approximately 8.35 million J. b) Initial kinetic energy is approximately 1.22 million J.

Step by step solution

01

Understanding Gravitational Potential Energy

The gravitational potential energy (U) at a distance (r) from the center of a planet with mass M is given by the formula \( U = -\frac{G M m}{r} \), where G is the gravitational constant and m is the mass of the object.
02

Calculating the Change in Gravitational Potential Energy (a)

For part (a), we calculate the gravitational potential energy at the surface \( (r = 3.0 \times 10^6 \text{ m}) \) and at \( 4.0 \times 10^6 \text{ m} \) from the center of Zero. The initial potential energy: \( U_i = -\frac{G \times 5.0 \times 10^{23} \times 10}{3.0 \times 10^6} \) and the potential energy at \( 4.0 \times 10^6 \text{ m} \): \( U_f = -\frac{G \times 5.0 \times 10^{23} \times 10}{4.0 \times 10^6} \). The change in potential energy is \( \Delta U = U_f - U_i \).
03

Applying Conservation of Energy (a)

According to conservation of energy, the total mechanical energy (sum of kinetic and potential energy) remains constant. Initially, the total energy is \( K_i + U_i \) and at \( 4.0 \times 10^6 \text{ m} \), it is \( K_f + U_f \). Therefore, \( K_f = K_i + U_i - U_f \). Given \( K_i = 5.0 \times 10^7 \text{ J} \), find \( K_f \).
04

Calculating Initial Kinetic Energy for Maximum Distance (b)

For part (b), at maximum distance \( r = 8.0 \times 10^6 \text{ m} \), the probe has zero kinetic energy. The total mechanical energy at the surface is \( K_i + U_i \) and at maximum distance is \( U_{max} \). Set \( K_i + U_i = U_{max} \) to find the required \( K_i \). \( U_{max} = -\frac{G \times 5.0 \times 10^{23} \times 10}{8.0 \times 10^6} \).

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Key Concepts

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

Conservation of Energy
The principle of conservation of energy is fundamental in physics. It states that in an isolated system, energy cannot be created or destroyed; it can only change forms. In the context of a space probe launched from a planet, this concept simplifies calculations. The total mechanical energy of the probe, which is the sum of its kinetic energy and gravitational potential energy, remains constant throughout its journey if we ignore other influences.
The equation expressing this conservation for the probe is:
  • Initial Energy = Final Energy
  • \[ K_i + U_i = K_f + U_f \]
Here, \( K_i \) and \( K_f \) are the initial and final kinetic energies, while \( U_i \) and \( U_f \) are the initial and final gravitational potential energies of the probe.
This equation allows us to solve for unknowns, like the final kinetic energy of the probe, by knowing the others. Thus, we can deduce how energy is conserved as the probe travels further from the planet.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is calculated using the equation:
  • \[ K = \frac{1}{2} m v^2 \]
where \( m \) is the mass of the object, and \( v \) is its velocity.
In the case of our space probe, its kinetic energy changes as it moves further from the planet Zero. Initially, it has a kinetic energy \( K_i \) given to it at launch. As it moves upwards, some of this kinetic energy converts into potential energy due to the gravitational pull of Zero, slowing the probe down.
As the probe reaches higher altitudes, its velocity decreases, reducing its kinetic energy. According to the conservation of energy formula, as the kinetic energy decreases, the potential energy increases to keep the total energy constant. Understanding this balance is crucial to predicting the motion and energy of moving objects.
Gravitational Constant
The gravitational constant, denoted as \( G \), is a key value in calculating the gravitational forces between two masses. Its numerical value is approximately \( 6.674 \times 10^{-11} \, \text{m}^3\, \text{kg}^{-1}\, \text{s}^{-2} \).
In the context of our problem, \( G \) is crucial for finding the gravitational potential energy of the probe at different distances from the center of the planet Zero. The potential energy \( U \) is given by:
  • \[ U = -\frac{G M m}{r} \]
where \( M \) is the mass of the planet, \( m \) is the mass of the probe, and \( r \) is the distance from the center of the planet.
Without the gravitational constant, we couldn't compute how the gravitational attraction affects the potential energy and motion of a space probe. It gives us insight into the universal law of gravitation and its effects on planetary and object interactions.
Mechanical Energy
Mechanical energy is the sum total of kinetic and potential energy in a system. When dealing with the space probe, mechanical energy helps us understand how energy is redistributed as the probe moves.
Mechanical energy can be expressed as:
  • \[ E = K + U \]
where \( E \) is the total mechanical energy, \( K \) is the kinetic energy, and \( U \) is the potential energy.
In the absence of external forces, such as air resistance (since Zero has no atmosphere), the mechanical energy of the probe is preserved. This means, for part (a) of our problem, as the probe moves from the surface to a height of \( 4.0 \times 10^6 \, \text{m} \), its mechanical energy remains unchanged.
By understanding and calculating mechanical energy, we can predict how much kinetic energy remains as the potential energy increases. This insight helps in planning space missions by providing the needed initial energies to achieve desired altitudes or speeds.

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Most popular questions from this chapter

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