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Chapter 9: Problem 127

During a lunar mission, it is necessary to increase the speed of a spacecraft by \(2.2 \mathrm{~m} / \mathrm{s}\) when it is moving at \(400 \mathrm{~m} / \mathrm{s}\) relative to the Moon. The speed of the exhaust products from the rocket engine is \(1000 \mathrm{~m} / \mathrm{s}\) relative to the spacecraft. What fraction of the initial mass of the spacecraft must be burned and ejected to accomplish the speed increase?

Short Answer

Expert verified
About 0.22% of the initial mass needs to be burned and ejected.

Step by step solution

01

Understand the Problem

We need to determine what fraction of the spacecraft's initial mass must be ejected as exhaust to achieve a velocity increase. This can be calculated using the rocket equation, which involves the change in velocity, exhaust speed, and mass.
02

Introduce the Rocket Equation

The rocket equation is given by:\[v_f = v_i + v_e \ln\left(\frac{m_i}{m_f}\right)\]where \(v_f\) is the final velocity, \(v_i\) is the initial velocity, \(v_e\) is the exhaust velocity, \(m_i\) is the initial mass, and \(m_f\) is the final mass of the spacecraft.
03

Substitute Known Values

We know the change in velocity \(\Delta v = v_f - v_i = 2.2 \text{ m/s}\), \(v_e = 1000 \text{ m/s}\), and \(v_f = v_i + 2.2 \text{ m/s}\). Substitute these values into the rocket equation to get:\[2.2 = 1000 \ln\left(\frac{m_i}{m_f}\right)\]
04

Solve for the Mass Ratio

Rearrange the equation to solve for the mass ratio \(\frac{m_i}{m_f}\):\[\ln\left(\frac{m_i}{m_f}\right) = \frac{2.2}{1000}\]Calculate this to find:\[\frac{m_i}{m_f} = e^{\frac{2.2}{1000}}\]
05

Calculate the Exponent

Compute the value of \(e^{\frac{2.2}{1000}}\) using a calculator:\[\frac{m_i}{m_f} \approx e^{0.0022} \approx 1.0022\]
06

Determine the Fraction of Mass Ejected

The fraction of the initial mass that must be burned and ejected is given by:\[\text{Fraction} = 1 - \frac{m_f}{m_i}\]Substitute \(\frac{m_f}{m_i} = \frac{1}{1.0022}\):\[\text{Fraction} \approx 1 - \frac{1}{1.0022} \approx 0.0022\]
07

Conclusion: Final Result

Only a tiny fraction of the spacecraft's initial mass, approximately 0.22%, must be burned and ejected to increase its speed by 2.2 m/s.

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Key Concepts

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

Rocket Propulsion
Rocket propulsion is the mechanism that allows spacecraft to accelerate in space. This propulsion method works on the basic principle of Newton's Third Law of Motion, which states that for every action there is an equal and opposite reaction. When a rocket engine expels exhaust gases at high speed, this action propels the spacecraft in the opposite direction.
  • The gases expelled are the result of combustion inside the rocket engine.
  • The speed of the exhaust gases directly influences how fast or how much the rocket accelerates.
  • In our problem, the exhaust velocity is 1000 m/s.
The expulsion of mass creates thrust, which is the force that changes the craft's velocity. This is crucial in space, where resistance is minimal, and efficient propulsion is necessary for any adjustments in trajectory or speed.
Lunar Mission
Lunar missions refer to space explorations that are aimed at studying the Moon either by orbiting it or landing on its surface. These missions often involve precise calculations and maneuvers due to the gravitational influences of both Earth and the Moon. Drastic changes cannot be accommodated easily; thus efficient propulsion systems become indispensable.
  • Velocity adjustments like the one in our problem ensure that the spacecraft can enter, orbit, or leave the Moon's vicinity effectively.
  • The Moon has a weaker gravitational pull than Earth, which influences fuel and mass calculations needed for propulsion.
  • Such missions highlight the need for precise calculations to ensure a safe and successful journey.
Careful planning and execution are imperative in lunar missions, making the study of dynamics such as velocity change pivotal.
Mass Ejection
Mass ejection is a critical concept in rocket propulsion. It refers to the process of expelling a portion of the spacecraft's total mass to generate the required thrust. This expulsion must be carefully calculated to achieve the desired velocity increase without depleting the spacecraft's resources.
  • The rocket equation helps determine how much mass needs to be ejected.
  • In the given problem, only a small fraction, about 0.22%, of the initial mass is required to achieve the velocity change.
  • Mass ejection must be balanced to ensure that the spacecraft has enough remaining mass to complete its mission.
Understanding mass ejection is crucial for planning missions, as it directly impacts the spacecraft's ability to make precise maneuvers in space.
Velocity Change
The change in velocity, or \( \Delta v \), is a key parameter in space missions and acts as a measure of how much speed adjustment the spacecraft needs to reach its target. Any mission in space, including lunar ones, may require changes in speed to enter or exit orbits or to make corrections along the path.
  • The rocket equation: \( v_f = v_i + v_e \ln\left(\frac{m_i}{m_f}\right) \) is used to calculate the necessary mass ejection for a given \( \Delta v \).
  • In our example, \( \Delta v = 2.2 \,\text{m/s} \), necessitated by the mission parameters.
  • This minor adjustment makes a significant difference in trajectory and timing.
Efficient planning of velocity changes ensures the mission's objectives are achieved with optimal use of resources, and knowing how to calculate these changes is a fundamental skill in astronautics.

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Most popular questions from this chapter

A rocket sled with a mass of \(2900 \mathrm{~kg}\) moves at \(250 \mathrm{~m} / \mathrm{s}\) on a set of rails. At a certain point, a scoop on the sled dips into a trough of water located between the tracks and scoops water into an empty tank on the sled. By applying the principle of conservation of linear momentum, determine the speed of the sled after \(920 \mathrm{~kg}\) of water has been scooped up. Ignore any retarding force on the scoop.

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 force in the negative direction of an \(x\) axis is applied for \(27 \mathrm{~ms}\) to a \(0.40 \mathrm{~kg}\) ball initially moving at \(14 \mathrm{~m} / \mathrm{s}\) in the positive direction of the axis. The force varies in magnitude, and the impulse has magnitude \(32.4 \mathrm{~N} \cdot \mathrm{s}\). What are the ball's (a) speed and (b) direction of travel just after the force is applied? What are (c) the average magnitude of the force and (d) the direction of the impulse on the ball?

A body of mass \(2.0 \mathrm{~kg}\) makes an elastic collision with another body at rest and continues to move in the original direction but with one-fourth of its original speed. (a) What is the mass of the other body? (b) What is the speed of the two-body center of mass if the initial speed of the \(2.0 \mathrm{~kg}\) body was \(4.0 \mathrm{~m} / \mathrm{s}\) ?

A \(5.20 \mathrm{~g}\) bullet moving at \(672 \mathrm{~m} / \mathrm{s}\) strikes a \(700 \mathrm{~g}\) wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to \(428 \mathrm{~m} / \mathrm{s}\). (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?
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