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Chapter 9: Problem 76

A \(6090 \mathrm{~kg}\) space probe moving nose-first toward Jupiter at \(105 \mathrm{~m} / \mathrm{s}\) relative to the Sun fires its rocket engine, ejecting \(80.0 \mathrm{~kg}\) of exhaust at a speed of \(253 \mathrm{~m} / \mathrm{s}\) relative to the space probe. What is the final velocity of the probe?

Short Answer

Expert verified
The final velocity of the probe is approximately 108.33 m/s.

Step by step solution

01

Understand Conservation of Momentum

In space, the law of conservation of momentum applies. The total momentum before the space probe fires its rocket must equal the total momentum after firing.
02

Define Initial Momentum

The initial momentum (\(p_{ ext{initial}}\)) of the system is the product of the probe's mass and its velocity. It is given by \(p_{ ext{initial}} = m_{ ext{probe, initial}} \cdot v_{ ext{probe, initial}} = 6090 \times 105\).
03

Calculate Initial Momentum

Compute the initial momentum: \[\text{Initial Momentum} = 6090 \times 105 = 639450 \, \text{kg} \cdot \text{m/s}\]
04

Define Final Momentum

The final momentum (\(p_{ ext{final}}\)) involves both the remaining mass of the probe and the mass of the exhaust. It accounts for their velocities, where \(m_{ ext{probe, final}} = 6090 - 80\) and the exhaust velocity is relative to the probe.
05

Relative Velocities

The velocity of the ejected exhaust relative to the probe is given as 253 m/s. Therefore, the velocity of the exhaust relative to the Sun becomes \(v_{ ext{exhaust}} = v_{ ext{probe, final}} - 253\).
06

Set Up Conservation Equation

Set up the equation for conservation of momentum: \[m_{ ext{probe, final}} \cdot v_{ ext{probe, final}} + m_{ ext{exhaust}} \cdot v_{ ext{exhaust}} = p_{ ext{initial}}\]Substitute in the values: \[(6090 - 80) \cdot v_{ ext{probe, final}} + 80 \cdot (v_{ ext{probe, final}} - 253) = 639450\]
07

Simplify the Equation

Simplify and solve for \(v_{ ext{probe, final}}\): \[6010v_{ ext{probe, final}} + 80v_{ ext{probe, final}} - 20240 = 639450\]Combine like terms: \[6090v_{ ext{probe, final}} - 20240 = 639450\]
08

Solve for Final Velocity

Rearrange and solve for \(v_{\text{probe, final}}\):\[6090v_{\text{probe, final}} = 639450 + 20240\]\[v_{\text{probe, final}} = \frac{659690}{6090}\]
09

Calculate Final Velocity

Perform the final calculation: \[v_{\text{probe, final}} = 108.33 \, \text{m/s}\]

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Key Concepts

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

Space Probe
A space probe is a type of spacecraft that is designed to travel a great distance from Earth to collect scientific information. Usually, these probes are unmanned and equipped with a variety of instruments to observe celestial bodies, such as planets and moons.
Some key features of space probes include:
  • They lack life-support systems since they are not designed to carry humans.
  • They are powered by solar panels or onboard generators.
  • They communicate with Earth using radio signals.
In this exercise, the space probe is traveling to Jupiter. Its journey requires precise calculations to ensure it reaches its destination, using principles like conservation of momentum to navigate through space.
Exhaust Velocity
Exhaust velocity is a crucial concept in rocketry, referring to the speed at which exhaust gases are expelled from a rocket engine. This speed is a key factor in determining how effectively a rocket can propel itself forward.
In the context of our exercise:
  • The space probe ejects exhaust gases at a velocity of 253 m/s relative to itself, which helps it to change its speed and direction.
  • Exhaust velocity is pivotal in calculating the change in velocity of the space probe using the principle of conservation of momentum.
Higher exhaust velocities tend to increase a rocket's efficiency, allowing it to achieve greater speeds or conserve fuel for extended travel.
Rocket Propulsion
Rocket propulsion is the method by which a rocket propels itself through space. This propulsion is based on Newton's Third Law of Motion, which states that for every action, there is an equal and opposite reaction.
The key components involved in rocket propulsion include:
  • Thrust: The force generated by the expulsion of gas to propel the rocket forward.
  • Propellant: The material expelled from the rocket, which can be either liquid or solid.
  • Exhaust Velocity: The speed at which the propellant is ejected, influencing the rocket's thrust and ultimately its velocity.
In the exercise, when the space probe fires its rocket, it ejects exhaust, creating momentum that propels it forward. This principle allows the probe to maneuver and accelerate as needed.

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Most popular questions from this chapter

A \(2100 \mathrm{~kg}\) truck traveling north at \(41 \mathrm{~km} / \mathrm{h}\) turns east and accelerates to \(51 \mathrm{~km} / \mathrm{h}\). (a) What is the change in the truck's kinetic energy? What are the (b) magnitude and (c) direction of the change in its momentum?

block 1 of mass \(m_{1}\) slides along an \(x\) axis on a frictionless floor at speed \(4.00 \mathrm{~m} / \mathrm{s}\). Then it undergoes a one-dimensional elastic collision with stationary block 2 of mass \(m_{2}=2.00 \mathrm{~m}_{1}\). Next, block 2 undergoes a one-dimensional elastic collision with stationary block 3 of mass \(m_{3}=2.00 m_{2}\). (a) What then is the speed of block 3 ? Are (b) the speed, (c) the kinetic energy, and (d) the momentum of block 3 greater than, less than, or the same as the initial values for block \(1 ?\)

block 1 (mass \(2,0 \mathrm{~kg}\) ) is moving rightward at \(10 \mathrm{~m} / \mathrm{s}\) and block 2 (mass \(5.0 \mathrm{~kg}\) ) is moving rightward at \(3.0 \mathrm{~m} / \mathrm{s}\) The surface is frictionless, and a spring with a spring constant of \(1120 \mathrm{~N} / \mathrm{m}\) is fixed to block 2 . When the blocks collide, the compression of the spring is maximum at the instant the blocks have the same velocity. Find the maximum compression.

After a completely inelastic collision, two objects of the same mass and same initial speed move away together at half their initial speed. Find the angle between the initial velocities of the objects.

A \(0.70 \mathrm{~kg}\) ball moving horizontally at \(5.0 \mathrm{~m} / \mathrm{s}\) strikes a vertical wall and rebounds with speed \(2.0 \mathrm{~m} / \mathrm{s}\). What is the magnitude of the change in its linear momentum?
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