An astronaut is performing a space walk outside the International Space Station. The total mass of the astronaut with her space suit and all her gear is \(115 \mathrm{~kg} .\) A small leak develops in her propulsion system and \(7.00 \mathrm{~g}\) of gas are ejected each second into space with a speed of \(800 \mathrm{~m} / \mathrm{s}\). She notices the leak 6.00 s after it starts. How much will the gas leak have caused her to move from her original location in space by that time?

Short Answer

Expert verified
The astronaut will have a displacement of approximately 0.878 meters from her original location in space within 6 seconds of the gas leak starting.

Step by step solution

01

Calculate the force exerted by the gas leak

The first step is to calculate the force exerted by the gas leak. According to Newton's third law, the force exerted on the astronaut by the gas leak is equal and opposite to the force exerted by the astronaut on the gas. The force exerted by the gas leak can be found using the following equation: \(F = \frac{dp}{dt}\) where \(F\) is the force exerted, \(dp\) is the change in momentum, and \(dt\) is the change in time. In this case, we know the amount of gas ejected per second and the speed of the gas: - Mass of gas ejected per second: \(7.00 \mathrm{~g/s} = 0.00700 \mathrm{~kg/s}\) - Speed of gas: \(800 \mathrm{~m/s}\) Since the speed of the gas is constant, the force exerted by the gas leak can be calculated by using the equation: \(F = \frac{dm}{dt} \times v\) Substituting the given values: \(F = 0.00700 \mathrm{~kg/s} \times 800 \mathrm{~m/s} = 5.60 \mathrm{~N}\)
02

Calculate the acceleration of the astronaut

Now that we have the force exerted by the gas leak, we can calculate the acceleration of the astronaut: Newton's second law states: \(F = ma\) Here \(F\) is the force exerted on the astronaut, \(m\) is her total mass, and \(a\) is the acceleration experienced. - Mass of the astronaut (with gear): \(115 \mathrm{~kg}\) We want to find the acceleration \(a\). Rearranging the equation: \(a = \frac{F}{m} = \frac{5.60 \mathrm{~N}}{115 \mathrm{~kg}} = 0.0487 \mathrm{~m/s^2}\)
03

Calculate the displacement of the astronaut

Now that we know the acceleration of the astronaut, we can determine her displacement during the 6 seconds she noticed the gas leak: \( x =\frac{1}{2}at^2 \) Here \(x\) is the displacement, \(a\) is the acceleration, and \(t\) is the time. Substituting the values: \(x =\frac{1}{2} \times 0.0487 \mathrm{~m/s^2} \times (6.00 \mathrm{s})^2 = 0.878 \mathrm{~m}\) So, the gas leak will have caused the astronaut to move approximately \(0.878\) meters from her original location in space within 6 seconds of the leak starting.

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Key Concepts

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

Newton's Third Law
When we discuss movements in space, Newton's third law is a fundamental concept that comes into play. This law states that for every action, there is an equal and opposite reaction. In the context of our space walk physics problem, this means that as the gas is expelled forcefully from the astronaut's propulsion system, the astronaut herself will experience a force in the opposite direction.

This interaction allows the astronaut to maneuver in the void of space. Since there's no air or surface to push off of in space, astronauts rely on this principle to start, stop, or change direction. The law applies to all interactions; from an astronaut using a jet pack to a rocket being launched into space, all act under the premise of action-reaction pairs.
Momentum Conservation
Momentum conservation is a principle stating that the total momentum of an isolated system remains constant if no external forces act upon it. This is a direct consequence of Newton's third law, and it has profound implications for objects in space.

In our problem, momentum conservation is key to determining the movement of the astronaut after the gas leak. The mass of gas ejected and its speed contribute to the system's total momentum. As gas escapes, it carries momentum away from the astronaut, and conserving momentum means the astronaut gains an equal amount of momentum in the opposite direction. Therefore, despite being in a frictionless environment, the astronaut can move and indeed has moved due to the leaking gas.
Astronaut Acceleration
Acceleration is the rate at which an object changes its velocity. In the astronaut's case, her acceleration is due to the force applied by the escaping gas. According to Newton's second law of motion, which states that force equals mass times acceleration (\( F = ma \)), we're able to determine the astronaut's acceleration by dividing the applied force by the astronaut's mass.

Acceleration here is not just about speed but also direction. With no gravity to anchor her and no resistance to impede her movement, the astronaut's acceleration is entirely dependent on the force exerted by the gas and her own mass. This simple yet powerful concept underlies most of what we understand about motion in space, particularly on how astronauts are able to control their spatial orientation and movements during space walks.
Kinematics Equations
Kinematics equations describe the motion of objects without reference to its causes. These equations are invaluable tools for solving problems related to displacement, velocity, and acceleration. In the given solution, we used a kinematic equation specific for acceleration at a constant rate to find the astronaut's displacement.

The equation \( x = \frac{1}{2}at^2 \) is derived from the principles of kinematics and is used when the initial velocity (\( u \) is zero and acceleration (\( a \) is constant. The displacement (\( x \) is the distance moved in a particular direction, 't' is the time, and 'a' is the acceleration. By knowing the time duration and the acceleration of the astronaut, we could calculate how far she moved as a result of the gas leak, illustrating how kinematic equations serve as the link between the motion we observe and the numerical values that describe it.

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Most popular questions from this chapter

A toy car of mass \(2.0 \mathrm{~kg}\) is stationary, and a child rolls a toy truck of mass 3.5 kg straight toward it with a speed of \(4.0 \mathrm{~m} / \mathrm{s}\) a) What is the velocity of the center of mass of the system consisting of the two toys? b) What are the velocities of the truck and the car with respect to the center of mass of the system consisting of the two toys?

A catapult on a level field tosses a 3 -kg stone a horizontal distance of \(100 \mathrm{~m}\). A second 3 -kg stone tossed in an identical fashion breaks apart in the air into 2 pieces, one with a mass of \(1 \mathrm{~kg}\) and one with a mass of \(2 \mathrm{~kg} .\) Both of the pieces hit the ground at the same time. If the 1 -kg piece lands a distance of \(180 \mathrm{~m}\) away from the catapult, how far away from the catapult does the 2 -kg piece land? Ignore air resistance. a) \(20 \mathrm{~m}\) c) \(100 \mathrm{~m}\) e) \(180 \mathrm{~m}\) b) \(60 \mathrm{~m}\) d) \(120 \mathrm{~m}\)

A student with a mass of \(40.0 \mathrm{~kg}\) can throw a \(5.00-\mathrm{kg}\) ball with a relative speed of \(10.0 \mathrm{~m} / \mathrm{s}\). The student is standing at rest on a cart of mass \(10.0 \mathrm{~kg}\) that can move without friction. If the student throws the ball horizontally, what will the velocity of the ball with respect to the ground be?

A \(1000 .-\mathrm{kg}\) cannon shoots a \(30.0-\mathrm{kg}\) shell at an angle of \(25.0^{\circ}\) above the horizontal and a speed of \(500 . \mathrm{m} / \mathrm{s}\). What is the recoil velocity of the cannon?

You are piloting a spacecraft whose total mass is \(1000 \mathrm{~kg}\) and attempting to dock with a space station in deep space. Assume for simplicity that the station is stationary, that your spacecraft is moving at \(1.0 \mathrm{~m} / \mathrm{s}\) toward the station, and that both are perfectly aligned for docking. Your spacecraft has a small retro-rocket at its front end to slow its approach, which can burn fuel at a rate of \(1.0 \mathrm{~kg} / \mathrm{s}\) and with an exhaust velocity of \(100 \mathrm{~m} / \mathrm{s}\) relative to the rocket. Assume that your spacecraft has only \(20 \mathrm{~kg}\) of fuel left and sufficient distance for docking. a) What is the initial thrust exerted on your spacecraft by the retro-rocket? What is the thrust's direction? b) For safety in docking, NASA allows a maximum docking speed of \(0.02 \mathrm{~m} / \mathrm{s}\). Assuming you fire the retro-rocket from time \(t=0\) in one sustained burst, how much fuel (in kilograms) has to be burned to slow your spacecraft to this speed relative to the space station? c) How long should you sustain the firing of the retrorocket? d) If the space station's mass is \(500,000 \mathrm{~kg}\) (close to the value for the ISS), what is the final velocity of the station after the docking of your spacecraft, which arrives with a speed of \(0.02 \mathrm{~m} / \mathrm{s}\) ?

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