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Chapter 7: Problem 69

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 must be burned and ejected.

Step by step solution

01

Identify given values

Extract the given values from the problem statement. The initial speed of the spacecraft is given as \(400 \mathrm{~m} / \mathrm{s}\), the desired speed increase is \(2.2 \mathrm{~m} / \mathrm{s}\), and the speed of the exhaust products is \(1000 \mathrm{~m} / \mathrm{s}\) relative to the spacecraft.
02

Understand the rocket equation

The rocket equation is given by \[ \Delta v = v_e \, \ln \left( \frac{m_0}{m_f} \right) \] where \( \Delta v \) is the change in velocity, \( v_e \) is the exhaust velocity, \( m_0 \) is the initial mass, and \( m_f \) is the final mass.
03

Substitute the given values into the rocket equation

Substitute \( \Delta v = 2.2 \mathrm{~m} / \mathrm{s} \), \( v_e = 1000 \mathrm{~m} / \mathrm{s} \) into the equation: \[ 2.2 = 1000 \, \ln \left( \frac{m_0}{m_f} \right) \]
04

Solve for the mass ratio

Rearrange the equation to solve for the mass ratio \( \frac{m_0}{m_f} \). \[ \ln \left( \frac{m_0}{m_f} \right) = \frac{2.2}{1000} \] Thus, \[ \frac{m_0}{m_f} = e^{0.0022} \]
05

Calculate the mass ratio

Calculate the value of the exponential: \[ \frac{m_0}{m_f} = e^{0.0022} \approx 1.0022 \]
06

Determine the fraction of initial mass burned and ejected

The fractional mass burned and ejected is given by \[ 1 - \frac{m_f}{m_0} = 1 - \frac{1}{1.0022} \approx 0.0022 \]

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Key Concepts

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

rocket propulsion
Rocket propulsion refers to the mechanism by which a rocket is thrust forward. It operates on Newton鈥檚 Third Law, which states that for every action, there is an equal and opposite reaction. In this case, the rocket engine expels exhaust gases at high speeds, generating a forward force propelling the spacecraft. This fundamental principle allows rockets to travel even in the vacuum of space, where traditional propulsion methods used in atmospheric flight become ineffective.
mass ratio
The mass ratio is crucial for understanding rocket propulsion. It is defined as the ratio of the initial mass of the spacecraft (\(m_0\)) to its final mass (\(m_f\)) after fuel has been burned and expelled. Mathematically, it is expressed as \(\frac{m_0}{m_f}\). A higher mass ratio implies that more fuel has been burned, increasing the spacecraft鈥檚 velocity. Proper calculation of the mass ratio is vital in designing efficient rockets and planning space missions to ensure they have enough fuel for acceleration and deceleration phases.
velocity change calculation
Velocity change, or $$\text{Delta v}$$, is a key factor in space navigation. It indicates the change in velocity needed to maneuver a spacecraft. To calculate \(\text{Delta v}\), we use the rocket equation: \( \text{Delta v} = v_e \times \text{ln}(\frac{m_0}{m_f}) \) , where \(v_e\) is the exhaust velocity. By substituting given values, you can determine the velocity change achievable for a specific mass ratio and exhaust velocity, critical for mission planning and executing orbital adjustments.
exhaust velocity
Exhaust velocity (\(v_e\)) is the speed at which gas exits the rocket engine, relative to the spacecraft. It directly impacts the rocket鈥檚 thrust and efficiency. The higher the exhaust velocity, the greater the momentum change, allowing for more significant adjustments to the spacecraft鈥檚 speed. For example, in our lunar mission problem, an exhaust velocity of \(1000 \text{ m/s}\) was used to achieve the desired velocity increase.
lunar mission physics
Lunar missions involve precise calculations and application of physics principles to navigate the spacecraft. For instance, when increasing the spacecraft鈥檚 speed by \(2.2 \text{ m/s}\) on our mission, it鈥檚 essential to account for the initial velocity, the exhaust velocity, and the mass ratio. These calculations ensure the correct amount of fuel is burned and ejected to accomplish the required velocity change, ultimately facilitating safe and accurate lunar landings and returns.

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Most popular questions from this chapter

\(\mathrm{A}\) barge with mass \(1.50 \times\) \(10^{5} \mathrm{~kg}\) is proceeding downriver at \(6.2 \mathrm{~m} / \mathrm{s}\) in heavy fog when it collides with a barge heading directly across the river (see Fig. 7-31). The second barge has mass \(2.78\) \(\times 10^{5} \mathrm{~kg}\) and before the collision is moving at \(4.3\) \(\mathrm{m} / \mathrm{s} . \quad\) Immediately after impact, the second barge finds its course deflected by \(18^{\circ}\) in the downriver direction and its speed increased to \(5.1 \mathrm{~m} / \mathrm{s}\). The river current is approximately zero at the time of the accident. What are the speed and direction of motion of the first barge immediately after the collision?

A collision occurs between a \(2.00 \mathrm{~kg}\) particle traveling with velocity \(\vec{v}_{A 1}=(-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(-5.00 \mathrm{~m} / \mathrm{s}) \mathrm{j}\) and a \(4.00 \mathrm{~kg}\) particle traveling with velocity \(\vec{v}_{B 1}=(6.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+\) \((-2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). The collision connects the two particles. What then is their velocity in (a) unit-vector notation and (b) magnitude-angle notation?

A \(4.0 \mathrm{~kg}\) mess kit sliding on a frictionless surface explodes into two \(2.0 \mathrm{~kg}\) parts, one moving at \(3.0 \mathrm{~m} / \mathrm{s}\), due north, and the other at \(5.0 \mathrm{~m} / \mathrm{s}, 30^{\circ}\) north of east. What is the original speed of the mess kit?

The principle of conservation of momentum is useful in some situations and not in others. Describe how you obtain the impulse-momentum theorem from Newton's Second Law and what situations lead to momentum conservation. How would you decide whether conservation of momentum could be useful in a particular problem?

In February 1955, a paratrooper fell \(370 \mathrm{~m}\) from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was \(56 \mathrm{~m} / \mathrm{s}\) (terminal speed), that his mass (including gear) was \(85 \mathrm{~kg}\), and that the magnitude of the force on him from the snow was at the survivable limit of \(1.2 \times 10^{5} \mathrm{~N}\). What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?
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