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Chapter 9: Problem 76

A \(6090 \mathrm{~kg}\) space probe moving nose-first toward Jupiter at \(105 \mathrm{~m} / \mathrm{s}\) relative to the Sun fires its rocket engine, ejecting \(80.0 \mathrm{~kg}\) of exhaust at a speed of \(253 \mathrm{~m} / \mathrm{s}\) relative to the space probe. What is the final velocity of the probe?

Short Answer

Expert verified
The final velocity of the probe is calculated using conservation of momentum, accounting for the ejected mass and its velocity.

Step by step solution

01

Understand the Problem Context

We are given that a space probe with a mass of 6090 kg is moving toward Jupiter. It fires its rocket engine, ejecting 80.0 kg of exhaust at a certain speed. We need to calculate the new velocity of the space probe after this ejection.
02

Apply the Conservation of Momentum

In the absence of external forces, the momentum of the system (the space probe and the exhaust) must be conserved. This means the total momentum before the ejection is equal to the total momentum after the ejection.
03

Calculate Initial Momentum

Initial momentum of the system is the momentum of the space probe before firing the rocket. It is calculated as:\[ p_{ ext{initial}} = m_{ ext{probe, initial}} \times v_{ ext{probe, initial}} = 6090 \, \text{kg} \times 105 \, \text{m/s} \]
04

Calculate Exhaust Momentum

The momentum imparted by the exhaust is calculated relative to the space probe and in the opposite direction of its ejection:\[ p_{ ext{exhaust}} = m_{ ext{exhaust}} \times v_{ ext{exhaust, relative}} = 80 \, \text{kg} \times 253 \, \text{m/s} \]
05

Calculate Final Momentum

According to the conservation of momentum:\[ p_{ ext{initial}} = p_{ ext{final}} + p_{ ext{exhaust}} \]Where the final momentum accounts for the decreased mass of the probe (6090 kg - 80 kg) and the new velocity of the probe. Re-arranging gives:\[ m_{ ext{probe, final}} \times v_{ ext{probe, final}} = p_{ ext{initial}} - p_{ ext{exhaust}} \]
06

Solve for Final Velocity

Substitute the known values into the rearranged formula to solve for the final velocity of the space probe:\[ v_{ ext{probe, final}} = \frac{p_{ ext{initial}} - p_{ ext{exhaust}}}{m_{ ext{probe, final}}} \]Here, \( m_{\text{probe, final}} = 6090 \, \text{kg} - 80 \, \text{kg} \) andcalculate to find the final velocity.

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Key Concepts

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

Space Probe
A space probe is an unmanned spacecraft tasked with exploring space and gathering data which it then sends back to Earth. These probes are designed to endure the harsh conditions of space travel and can operate for years on their own.
One of the key features of a space probe is its ability to change velocity and direction using propulsion systems like rockets. This is crucial for navigating through space to reach distant celestial targets such as planets, moons, or asteroids.
  • Space probes travel vast distances through space, often beyond our solar system.
  • They provide vital data for scientific research, expanding our understanding of the universe.
  • Advanced technology allows probes to send information back to Earth over millions of miles.
Space probes must be carefully equipped and programmed to handle the physical challenges and complex movements required for their missions.
Rocket Propulsion
Rocket propulsion is the mechanism that allows spacecraft, like the space probe, to move and maneuver in space. Unlike airplanes that rely on the atmosphere, rockets work in the vacuum of space. This propulsion is possible due to Newton's third law of motion: for every action, there is an equal and opposite reaction.
When a rocket engine fires, it ejects mass (such as exhaust gases) backward at high speed. This generates a forward thrust that propels the rocket in the opposite direction.
  • Rocket propulsion doesn't depend on atmospheric gases; it's effective in space.
  • Propellant (fuel) is burnt to produce high-speed exhaust gases.
  • The expulsion of gases creates the force necessary to move the spacecraft forward.
This principle allows the space probe to adjust its speed and direction, as seen in the exercise where it changes its velocity by expelling exhaust.
Momentum Calculation
Momentum is the product of an object's mass and velocity, represented as \( p = m imes v \). In physics, it's important because it is always conserved in a closed system, like our space probe and exhaust setup during the rocket firing.
This conservation means that the total initial momentum of the system (the probe's momentum before the engine fires) is equal to the total final momentum (the momentum of the probe plus the momentum of the ejected exhaust).
  • Initial momentum (\(p_{initial}\)) is calculated before the engine fires: \( \text{p}_{\text{initial}} = m_{\text{probe, initial}} \times v_{\text{probe, initial}} \).
  • Exhaust momentum is the result of ejecting its mass at a relative velocity.
  • Final momentum is adjusted by subtracting the exhaust momentum to find the probe's new velocity.
Mental arithmetic of these values allows us to solve for the probe's final velocity after the thrust.
Exhaust Velocity
Exhaust velocity is a crucial factor in calculating the change in velocity of a space probe when it expels exhaust gases. It refers to the speed at which the exhaust leaves the propulsion system relative to the spacecraft itself.
A higher exhaust velocity typically means more thrust and thus more change in velocity for the probe. This velocity is integral to the propulsion process, affecting the probe's ability to speed up or change course effectively.
  • Exhaust velocity is measured relative to the spacecraft.
  • Increased exhaust velocity results in a greater change in momentum.
  • It's essential for efficient operation of space propulsion systems.
Understanding exhaust velocity helps in determining how much the probe's momentum changes after ejection. This understanding is vital to solving the problem of finding the new velocity of the space probe after the rocket firing.

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Most popular questions from this chapter

A soccer player kicks a soccer ball of mass \(0.45 \mathrm{~kg}\) that is initially at rest. The foot of the player is in contact with the ball for \(3.0 \times 10^{-3} \mathrm{~s}\), and the force of the kick is given by $$ F(t)=\left[\left(6.0 \times 10^{6}\right) t-\left(2.0 \times 10^{9}\right) t^{2}\right] \mathrm{N}$$ for \(0 \leq t \leq 3.0 \times 10^{-3} \mathrm{~s}\), where \(t\) is in seconds. Find the magnitudes of (a) the impulse on the ball due to the kick, (b) the average force on the ball from the player's foot during the period of contact, (c) the maximum force on the ball from the player's foot during the period of contact, and (d) the ball's velocity immediately after it loses contact with the player's foot.

A body is traveling at \(2.0 \mathrm{~m} / \mathrm{s}\) along the positive direction of an \(x\) axis; no net force acts on the body. An internal explosion separates the body into two parts, each of \(4.0 \mathrm{~kg}\), and increases the total kinetic energy by \(16 \mathrm{~J}\). The forward part continues to move in the original direction of motion. What are the speeds of (a) the rear part and (b) the forward part?

Ball \(B\), moving in the positive direction of an \(x\) axis at speed \(v\), collides with stationary ball \(A\) at the origin. \(A\) and \(B\) have different masses. After the collision, \(B\) moves in the negative direction of the \(y\) axis at speed \(v / 2 .\) (a) In what direction does \(A\) move? (b) Show that the speed of \(A\) cannot be determined from the given information.

A spacecraft is separated into two parts by detonating the explosive bolts that hold them together. The masses of the parts are \(1200 \mathrm{~kg}\) and \(1800 \mathrm{~kg} ;\) the magnitude of the impulse on each part from the bolts is \(300 \mathrm{~N} \cdot \mathrm{s}\). With what relative speed do the two parts separate because of the detonation?

A \(0.15 \mathrm{~kg}\) ball hits a wall with a velocity of \((5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(6.50\) \(\mathrm{m} / \mathrm{s}) \hat{\mathrm{j}}+(4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}}\). It rebounds from the wall with a velocity of \((2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(3.50 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}+(-3.20 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{k}} .\) What are (a) the change in the ball's momentum, (b) the impulse on the ball, and (c) the impulse on the wall?
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