/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 72 The space probe Deep Space \(I\)... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The space probe Deep Space \(I\) was launched on October \(24,1998 .\) Its mass was \(474 \mathrm{~kg}\). The goal of the mission was to test a new kind of engine called an ion propulsion drive. This engine generated only a weak thrust, but it could do so over long periods of time with the consumption of only small amounts of fuel. The mission was spectacularly successful. At a thrust of \(56 \mathrm{mN}\) how many days were required for the probe to attain a velocity of \(805 \mathrm{~m} / \mathrm{s}(1800 \mathrm{mi} / \mathrm{h})\), assuming that the probe started from rest and that the mass remained nearly constant?

Short Answer

Expert verified
It takes approximately 79 days for the space probe to reach 805 m/s.

Step by step solution

01

Understand Given Data

We need to find the time in days for the space probe to reach a velocity of \(805 \text{ m/s}\) starting from rest. We are given:- Initial velocity \(v_0 = 0 \text{ m/s}\)- Final velocity \(v_f = 805 \text{ m/s}\)- Thrust \(F = 56 \text{ mN} = 0.056 \text{ N}\)- Mass of the probe \(m = 474 \text{ kg}\)
02

Calculate Acceleration

The thrust of the engine can be used to find the acceleration using Newton's second law, \(F = ma\).\[ a = \frac{F}{m} = \frac{0.056}{474} \text{ m/s}^2 \approx 0.000118 \text{ m/s}^2 \]
03

Use Kinematic Equation to Find Time

Since the motion starts from rest, we use the kinematic equation \(v_f = v_0 + at\) to find time:\[ 805 = 0 + 0.000118t \]Solving for \(t\):\[ t = \frac{805}{0.000118} \approx 6,822,034 \text{ seconds} \]
04

Convert Time from Seconds to Days

We convert seconds to days since time is often better understood in larger units.\[ \text{Number of days} = \frac{6,822,034}{60 \times 60 \times 24} \approx 78.96 \approx 79 \text{ days} \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Newton's Second Law
Newton's second law is a cornerstone of physics, providing a relationship between the motion of an object and the forces acting on it. It is formulated as: \( F = ma \) where \( F \) is the force applied to an object, \( m \) is the mass of the object, and \( a \) is the acceleration produced. This law explains how the velocity of an object changes when it is subjected to external forces. In our exercise, we use this formula to find the acceleration by rearranging it to: \( a = \frac{F}{m} \). For the space probe Deep Space I, we know the thrust (F) is small, but constant, and the mass (m) is nearly unchanging, allowing a straightforward calculation of acceleration.
Kinematic Equations
Kinematic equations describe the motion of objects without considering the forces that cause this motion. They provide a link between different motion parameters: displacement, velocity, acceleration, and time. One of the key equations useful in our exercise is: \( v_f = v_0 + at \) where \( v_f \) is the final velocity, \( v_0 \) is the initial velocity, \( a \) is the acceleration, and \( t \) is time. Since our space probe starts from rest, the initial velocity \( v_0 \) is zero. We rearrange this equation to solve for time \( t \), given that we know the final velocity and the acceleration from previous calculations. This forms the basis of calculating how long it takes the probe to reach its target speed.
Space Probe Mission
A space probe mission like that of Deep Space I involves sending a spacecraft beyond Earth's atmosphere to gather data. Deep Space I's mission was distinctive for employing an ion propulsion drive, a technology promising efficient, long-duration thrust. This engine operates by ionizing a propellant and expelling ions at high speed, generating thrust. While the thrust is weak compared to chemical rockets, it can be sustained for extended periods, precisely why it was a highlight of the mission. This mission served a dual purpose: to test ion propulsion technology and to contribute to our understanding of the outer solar system. The design required careful calculations, especially regarding acceleration and velocity over time, ensuring the mission could achieve its objectives with minimal fuel usage.
Acceleration Calculation
Acceleration is the rate of change of velocity with respect to time. Calculating acceleration was essential in solving our problem, as it determined how quickly the velocity target could be reached. Utilizing the relationship from Newton's second law, the formula becomes: \( a = \frac{F}{m} \) where \( F \) is the constant thrust in Newtons and \( m \) is the mass of the space probe. For Deep Space I, the calculated acceleration was roughly \( 0.000118 \) m/s², demonstrating how gentle the propulsion was. Given the constraints of the mass and the constant force, we can see how this low acceleration means the probe requires a significant amount of time to reach its full speed. This property exemplifies the efficiency in long-term space missions, showcasing one benefit of ion propulsion.
Velocity Calculation
Velocity calculation involves determining the speed and direction of the probe over time. Since the thrust was constant and the initial velocity was zero, we used the kinematic equation to relate velocity to time and acceleration: \( v_f = v_0 + at \). In plugging in known values: - Initial velocity \( v_0 = 0 \) m/s- Final velocity \( v_f = 805 \) m/s- Acceleration \( a = 0.000118 \) m/s² we solve for time \( t \), leading to a result of approximately 6,822,034 seconds, which translates to about 79 days when converted from seconds to days. This conversion is an important step, providing a more intuitive understanding of the time required and exemplifying a real-world application of physics in astronautics.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A \(95.0-\mathrm{kg}\) person stands on a scale in an elevator. What is the apparent weight when the elevator is (a) accelerating upward with an acceleration of \(1.80 \mathrm{~m} / \mathrm{s}^{2},\) (b) moving upward at a constant speed, and (c) accelerating downward with an acceleration of \(1.30 \mathrm{~m} / \mathrm{s}^{2}\) ?

In a supermarket parking lot, an employee is pushing ten empty shopping carts, lined up in a straight line. The acceleration of the carts is \(0.050 \mathrm{~m} / \mathrm{s}^{2}\). The ground is level, and each cart has a mass of \(26 \mathrm{~kg}\). (a) What is the net force acting on any one of the carts? (b) Assuming friction is negligible, what is the force exerted by the fifth cart on the sixth cart?

A \(1380-\mathrm{kg}\) car is moving due east with an initial speed of \(27.0 \mathrm{~m} / \mathrm{s}\). After \(8.00 \mathrm{~s}\) the car has slowed down to \(17.0 \mathrm{~m} / \mathrm{s}\). Find the magnitude and direction of the net force that produces the deceleration.

A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of the book are perpendicular to the book and are horizontal. The book weighs \(31 \mathrm{~N}\). The coefficient of static friction between his hands and the book is \(0.40 .\) To keep the book from falling, what is the magnitude of the minimum pressing force that each hand must exert?

The earth exerts a gravitational force on a falling raindrop, pulling it down. Does the raindrop exert a gravitational force on the earth, pulling it up? If so, is this force greater than, less than, or equal to the force that the earth exerts on the raindrop? Provide a reason for your answer. Problem A raindrop has a mass of \(5.2 \times 10^{-7} \mathrm{~kg}\), and is falling near the surface of the earth. Calculate the magnitude of the gravitational force exerted (a) on the raindrop by the earth and (b) on the earth by the raindrop.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.