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In \(1998,\) NASA launched Deep Space \(1(\mathrm{DS}-1),\) a spacecraft that successfully flew by the asteroid named 1992 \(\mathrm{KD}\) (which orbits the sun millions of miles from the earth). The propulsion system of DS- 1 worked by ejecting high-speed argon ions out the rear of the engine. The engine slowly increased the velocity of DS- 1 by about \(+9.0 \mathrm{m} / \mathrm{s}\) per day. (a) How much time (in days) would it take to increase the velocity of \(\mathrm{DS}-1 \mathrm{by}+2700 \mathrm{m} / \mathrm{s} ? \quad\) (b) What was the acceleration of DS- 1\(\left(\text { in } \mathrm{m} / \mathrm{s}^{2}\right) ?\)

Short Answer

Expert verified
It takes 300 days; acceleration is approximately \(1.0417 \times 10^{-4}\, \mathrm{m/s^2}\).

Step by step solution

01

Define the Problem

We want to find the time it takes for DS-1 to increase its velocity by +2700 m/s and also calculate its acceleration. Given is the velocity increase rate of +9.0 m/s per day.
02

Determine the Time for Velocity Change

To find how many days it takes to reach a velocity change of +2700 m/s: Use the formula \(\text{Time} = \frac{\text{Change in Velocity}}{\text{Velocity Increase per Day}}\). Plug in the values: \(\text{Time} = \frac{2700\, \mathrm{m/s}}{9.0\, \mathrm{m/s/day}} = 300\, \text{days}\).
03

Calculate the Acceleration

Acceleration is the change in velocity over time. Since the velocity change is constant, the acceleration is \(a = \frac{\text{Velocity Increase per Day}}{\text{Seconds in a Day}}\). There are 86400 seconds in a day, so plug the values into the formula: \(a = \frac{9.0\, \mathrm{m/s}}{86400\, \mathrm{s}} \approx 1.04166667 \times 10^{-4}\, \mathrm{m/s^2}\).

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Key Concepts

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

Kinematics
Kinematics is a branch of physics that deals with the motion of objects without considering the forces that cause the motion. It describes how an object moves in terms of its position, velocity, and acceleration over time.
  • Position: Where an object is located in space relative to a reference point.
  • Velocity: The rate at which an object changes its position. It is a vector quantity, meaning it has both magnitude and direction.
  • Acceleration: The rate at which an object changes its velocity. Like velocity, it is also a vector quantity.

In the case of Deep Space 1 (DS-1), understanding kinematics is crucial to determine how long it takes to achieve a certain change in velocity. The velocity increase of DS-1 was constant at +9.0 m/s per day. By dividing the total desired change in velocity by this daily increase, we can find how many days it takes for DS-1 to accelerate by a specific amount.
Acceleration Calculation
Acceleration is defined as the change in velocity over time. In DS-1’s propulsion system, acceleration is the result of the constant ejection of high-speed argon ions.

To calculate the acceleration of DS-1, we need to understand that the spacecraft's velocity increase is constant. This makes the calculation straightforward, as we can use the following formula:
\[ a = \frac{\text{Velocity Increase per Day}}{\text{Seconds in a Day}} \]
For DS-1, the velocity increase per day is +9.0 m/s. Knowing there are 86,400 seconds in a day, we plug these values into the formula:
\[ a = \frac{9.0 \, \mathrm{m/s}}{86400 \, \mathrm{s}} \approx 1.04166667 \times 10^{-4} \, \mathrm{m/s^2}\]
This tells us that each second, the spacecraft's velocity increases by a tiny amount due to the propulsion. Understanding this calculation is key to grasping how propulsion systems incrementally build up speed over time.
Spacecraft Propulsion
Spacecraft propulsion is a critical aspect of effectively moving within and beyond our solar system. For DS-1, propulsion was achieved through the ejection of charged particles, specifically argon ions.

This propulsion method works based on Newton's third law of motion: for every action, there is an equal and opposite reaction. When argon ions are ejected backward from the spacecraft, they push the spacecraft forward. This type of propulsion is known as ion propulsion, characterized by:
  • High Efficiency: Ion propulsion is more efficient than conventional methods because it uses gas much more effectively, providing continuous thrust over long periods.
  • Low Thrust: Although efficient, ion engines produce much less thrust compared to chemical rockets. They achieve their speeds over lengthy durations.
  • Durability: Since ion propulsion systems have fewer moving parts, they are often more reliable for long missions.

Deep Space 1's propulsion system exemplifies how modern spacecraft can improve mission endurance and efficiency, allowing scientists to explore distant parts of the universe with prolonged missions that were once thought impossible.

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Most popular questions from this chapter

Electrons move through a certain electric circuit at an average speed of \(1.1 \times 10^{-2} \mathrm{m} / \mathrm{s}\) . How long (in minutes) does it take an electron to traverse 1.5 \(\mathrm{m}\) of wire in the filament of a light bulb?

While standing on a bridge 15.0 \(\mathrm{m}\) above the ground, you drop a stone from rest. When the stone has fallen \(3.20 \mathrm{m},\) you throw a second stone straight down. What initial velocity must you give the second stone if they are both to reach the ground at the same instant? Take the downward direction to be the negative direction.

In a historical movie, two knights on horseback start from rest 88.0 \(\mathrm{m}\) apart and ride directly toward each other to do battle. Sir George's acceleration has a magnitude of 0.300 \(\mathrm{m} / \mathrm{s}^{2}\) , while Sir Alfred's has a magnitude of 0.200 \(\mathrm{m} / \mathrm{s}^{2}\) . Relative to Sir George's starting point, where do the knights collide?

The Kentucky Derby is held at the Churchill Downs track in Louisville, Kentucky. The track is one and one-quarter miles in length. One of the most famous horses to win this event was Secretariat. In 1973 he set a Derby record that would be hard to beat. His average acceleration during the last four quarter-miles of the race was \(+0.0105 \mathrm{m} / \mathrm{s}^{2}\) . His velocity at the start of the final mile \((x=+1609 \mathrm{m})\) was about \(+16.58 \mathrm{m} / \mathrm{s}\) . The acceleration, although small, was very important to his victory. To assess its effect, determine the difference between the time he would have taken to run the final mile at a constant velocity of \(+16.58 \mathrm{m} / \mathrm{s}\) and the time he actually took. Although the track is oval in shape, assume it is straight for the purpose of this problem.

Before starting this problem, review Multiple-Concept Example \(6 .\) The left ventricle of the heart accelerates blood from rest to a velocity of \(+26 \mathrm{cm} / \mathrm{s}\) . (a) If the displacement of the blood during the acceleration is \(+2.0 \mathrm{cm},\) determine its acceleration (in \(\mathrm{cm} / \mathrm{s}^{2} )\) (b) How much time does blood take to reach its final velocity?

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