Problem 61 - University Physics with Modern Physics

Question

A 70 -kg astronaut floating in space in a 110 -kg MMU (manned maneuvering unit) experiences an acceleration of 0.029 \(\mathrm{m} / \mathrm{s}^{2}\) when he fires one of the MMU's thrusters. (a) If the speed of the escaping \(\mathrm{N}_{2}\) gas relative to the astronaut is \(490 \mathrm{m} / \mathrm{s},\) how much gas is used by the thruster in 5.0 \(\mathrm{s} ?\) (b) What is the thrust of the thruster?

Step-by-Step Solution

Answer

0.0532 kg of gas is used, and thrust is 5.22 N.

Step 1: Understanding the Problem

We are given an astronaut with an MMU that has a combined mass of 180 kg (70 kg for the astronaut and 110 kg for the MMU). We need to calculate the amount of gas used and the thrust when the thruster is fired for 5.0 seconds, with the given acceleration and known speed of gas escape.

Step 2: Calculate the Thrust (b)

We use Newton's Second Law to find the thrust (\( F \)) of the thruster. The formula is \( F = m \cdot a \), where \( m \) is the total mass (180 kg) and \( a \) is the acceleration (0.029 m/s²).\[ F = 180 \cdot 0.029 = 5.22 \, \text{N} \]

Step 3: Calculate Mass of Gas Used (a)

The thrust produced by the expelled gas can also be expressed as \( F = \Delta p/\Delta t \), where \( \Delta p \) is the change in momentum. Since \( \Delta p = v_g \cdot \Delta m \) (where \( v_g = 490 \text{ m/s} \)), and \( F = v_g \cdot (\Delta m/\Delta t) \), rearranging gives \( \Delta m = \frac{F \cdot \Delta t}{v_g} \). Plugging the values: \( \Delta m = \frac{5.22 \cdot 5.0}{490} \) gives \( \Delta m \approx 0.0532 \text{ kg} \).

Step 4: Presenting the Results

After performing the calculations, in 5 seconds, approximately 0.0532 kg of gas is used by the thruster, and the thrust of the thruster is 5.22 N.

Key Concepts

Momentum

Momentum is a fundamental concept in physics, defined as the product of an object's mass and its velocity. Mathematically, this is expressed as \( p = m \times v \). In our space scenario, momentum helps us understand how the escaping gas changes the movement of the astronaut and MMU. When the gas is expelled, it carries momentum with it. According to the law of conservation of momentum, the total momentum of the system (astronaut, MMU, and expelled gas) remains constant before and after the gas is released.

• The expelled gas gains momentum in one direction,
• while the astronaut and MMU gain equal and opposite momentum.

This principle allows the astronaut to move, by shifting momentum from the MMU to the gas. The change of momentum \( \Delta p \) in the thruster can be calculated using the formula \( \Delta p = v_g \times \Delta m \). Here, \( v_g \) is the velocity of the gas and \( \Delta m \) is the mass of the gas used. Understanding this helps explain why even a small amount of expelled gas can result in significant movement of the astronaut.

Thrust

Thrust is the force exerted by the thruster to propel the astronaut and MMU. It is the result of the high-speed ejection of mass from the thruster. According to Newton's Second Law, the force can be calculated using \( F = m \cdot a \), where \( m \) is the mass and \( a \) is the acceleration caused by the thrust.
In the exercise, calculating thrust is crucial to understanding how the MMU moves. When the gas is expelled at high speeds (490 m/s in the problem), it creates an equal and opposite reaction force that moves the astronaut and MMU in the opposite direction to the expelled gas.

• This creates a thrust of 5.22 Newtons.
• Newton's Third Law (action-reaction) is at work here, ensuring that every action (ejection of gas) has an equal and opposite reaction (movement of the astronaut).

Thrust helps define how much force the thruster can exert at any time during the propulsion. Understandably, knowing how thrust works allows engineers to design efficient propulsion systems in spacecraft.

Acceleration

Acceleration describes how quickly the velocity of an object changes. It's a critical concept when considering propulsion and movement in the vacuum of space. For the astronaut and MMU, acceleration is necessary to change their speed or direction. It's calculated by using Newton's Second Law as \( a = \frac{F}{m} \).
The acceleration of 0.029 m/s² in the given problem tells us how fast the astronaut and MMU will speed up once the thruster is fired.

• This acceleration occurs due to the force (or thrust) exerted by expelling the gas.
• Even with a small acceleration, sustained thrust can lead to significant speed over time, which is essential for maneuvering in space.

Understanding acceleration in this context helps to predict how quickly or slowly the astronaut and MMU change their velocity. Additionally, it clarifies how various factors like mass and thrust impact movement in dynamic environments like outer space.