/*! 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 105 The gravitational slingshot effe... [FREE SOLUTION] | 91Ó°ÊÓ

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The gravitational slingshot effect. Figure 55 shows the planet Saturn moving in the negative \(x\) direction at its orbital speed (with respect to the Sun) of 9.6 \(\mathrm{km} / \mathrm{s}\) . The mass of Saturn is \(5.69 \times 10^{26} \mathrm{kg} .\) A spacecraft with mass 825 \(\mathrm{kg}\) approaches Saturn. When far from Saturn, it moves in the \(+x\) direction at 10.4 \(\mathrm{km} / \mathrm{s} .\) The gravitational attraction of Saturn (a conservative force) acting on the spacecraft causes it to swing around the planet (orbit shown as dashed line ) and head off in the opposite direction. Estimate the final speed of the spacecraft after it is far enough away to be considered free of Saturn's gravitational pull.

Short Answer

Expert verified
The spacecraft's final speed is 10.4 km/s.

Step by step solution

01

Identify Initial Conditions

Initially, the spacecraft is moving towards Saturn with a speed of 10.4 km/s in the positive x-direction, while Saturn moves in the negative x-direction at 9.6 km/s. The goal is to find the spacecraft's speed after the encounter.
02

Apply Conservation of Momentum

Since gravitational force is conservative and there are no external forces, the total momentum along the x-direction is conserved. Initially, total momentum is: \( p_{initial} = m_{spacecraft} \cdot v_{spacecraft,initial} + m_{Saturn} \cdot v_{Saturn,initial} \) which is: \( 825 \times 10.4 + 5.69 \times 10^{26} \times (-9.6) \).
03

Apply Conservation of Energy

The kinetic energy of the system is conserved throughout the interaction. The initial kinetic energy is: \( KE_{initial} = \frac{1}{2} m_{spacecraft} v_{spacecraft,initial}^2 + \frac{1}{2} m_{Saturn} v_{Saturn,initial}^2 \).
04

Analyze Post-Interaction Motion

Post interaction, the spacecraft will have reversed direction. To find the spacecraft's final velocity \( v_{spacecraft,final} \), recognize that its speed relative to Saturn swaps with its initial approach speed by conservation laws. Initial relative velocity of spacecraft to Saturn was: \( 10.4 + 9.6 = 20 \) km/s. Post turn, this remains the same, and the spacecraft heads away at relative speed of -20 km/s to Saturn.
05

Calculate Final Speed

Now, compute the spacecraft's final speed with respect to the Sun, using its relative speed to Saturn: \( v_{spacecraft,final} = v_{Saturn,initial} - v_{spacecraft,Saturn,relative,final} = -9.6 + 20 = 10.4 \) km/s.

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Key Concepts

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

Conservation of Momentum
When studying the gravitational slingshot effect, understanding the conservation of momentum is crucial. Momentum is the product of an object's mass and velocity, and it's conserved in an isolated system. This means that the total momentum before and after the spacecraft and Saturn interact is the same since no external forces act on their system.
For the given problem, the initial momentum is calculated by \[p_{initial} = m_{spacecraft} \cdot v_{spacecraft,initial} + m_{Saturn} \cdot v_{Saturn,initial}\]Here, the spacecraft and Saturn exert forces only on each other, maintaining the momentum constant in the closed system. Using this principle helps predict the changes in velocities after an event like the slingshot maneuver. It's a powerful tool in physics that makes space travel efficient.
  • Momentum is calculated as the product of mass and velocity.
  • In the absence of external forces, total momentum remains constant.
  • In this scenario, despite the spacecraft's direction reversal, overall momentum is preserved.
Conservation of Energy
Energy conservation also plays a vital role in understanding how a spacecraft harnesses gravitational fields to alter its velocity. Energy, particularly kinetic energy in this context, must remain constant in a closed system without external forces. The spacecraft's kinetic energy transformation through the gravitational slingshot involves \[KE_{initial} = \frac{1}{2} m_{spacecraft} v_{spacecraft,initial}^2 + \frac{1}{2} m_{Saturn} v_{Saturn,initial}^2\]The interaction doesn’t change the total kinetic energy, but simply redistributes it between the spacecraft and Saturn.
Gravitational slingshots leverage the gravitational energy exchanges to modify spacecraft velocities without using additional onboard resources. This characteristic is pivotal in space explorations allowing spacecraft to reach distant destinations using planets' gravitational fields as cosmic catapults.
  • Kinetic energy helps measure the system's movement energy state.
  • Energy conservation means the kinetic energy sum before and after remains identical.
  • The slingshot effect allows free velocity gains by tapping into a planet’s gravitational field.
Orbital Mechanics
Grasping orbital mechanics is essential to comprehend how gravitational assists or slingshots work. It's the field that studies the motion of spacecraft in the gravitational fields of planets, leveraging the planets' significant masses to alter spacecraft trajectories and speeds. The key principle here is that a spacecraft's trajectory around a planet is influenced by the planet's gravity in a predictable way. This offers an economical way to propel a spacecraft to further distances.
A spacecraft approaching a planet gains speed as it falls into the planet's gravitational well and then slingshots off by updating its velocity and direction. The relative velocity of the spacecraft to the planet dictates the new path post-encounter. In the exercise's solution, the spacecraft approaches and leaves Saturn, essentially "borrowing" some of Saturn’s orbital velocity around the Sun to achieve a new speed at minimal fuel expense.
  • Orbital mechanics dictate the path and speed of spacecraft within a gravitational system.
  • Slingshot maneuvers depend heavily on relative velocities and trajectories.
  • These maneuvers illustrate the practical applications of celestial mechanics.

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Most popular questions from this chapter

A \(4800-\mathrm{kg}\) open railroad car coasts along with a constant speed of \(8.60 \mathrm{~m} / \mathrm{s}\) on a level track. Snow begins to fall vertically and fills the car at a rate of \(3.80 \mathrm{~kg} / \mathrm{min}\). Ignoring friction with the tracks, what is the speed of the car after \(60.0 \mathrm{~min} ?\) (See Section 9-2.)

(II) A rocket of mass \(m\) traveling with speed \(v_{0}\) along the \(x\) axis suddenly shoots out fuel equal to one-third its mass, perpendicular to the \(x\) axis (along the \(y\) axis) with speed \(2 v_{0}\). Express the final velocity of the rocket in \(\hat{\mathbf{i}}, \hat{\mathbf{j}}, \hat{\mathbf{k}}\) notation.

(I) A \(0.145-\mathrm{kg}\) baseball pitched at \(35.0 \mathrm{~m} / \mathrm{s}\) is hit on a horizontal line drive straight back at the pitcher at \(56.0 \mathrm{~m} / \mathrm{s}\). If the contact time between bat and ball is \(5.00 \times 10^{-3} \mathrm{~s},\) calculate the force (assumed to be constant) between the ball and bat.

(III) A particle of mass \(m_{\mathrm{A}}\) traveling with speed \(v_{\mathrm{A}}\) collides elastically head-on with a stationary particle of smaller mass \(m_{\mathrm{B}} \cdot(a)\) Show that the speed of \(m_{\mathrm{B}}\) after the collision is $$ v_{\mathrm{B}}^{\prime}=\frac{2 v_{\mathrm{A}}}{1+m_{\mathrm{B}} / m_{\mathrm{A}}} $$ (b) Consider now a third particle of mass \(m_{\mathrm{C}}\) at rest between \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}}\) so that \(m_{\mathrm{A}}\) first collides head on with \(m_{\mathrm{C}}\) and then \(m_{\mathrm{C}}\) collides head on with \(m_{\mathrm{B}}\). Both collisions are elastic. Show that in this case, $$ v_{\mathrm{B}}^{\prime}=4 v_{\mathrm{A}} \frac{m_{\mathrm{C}} m_{\mathrm{A}}}{\left(m_{\mathrm{C}}+m_{\mathrm{A}}\right)\left(m_{\mathrm{B}}+m_{\mathrm{C}}\right)} $$ (c) From the result of part (b), show that for $$ \operatorname{maximum} v_{\mathrm{B}}^{\prime}, m_{\mathrm{C}}=\sqrt{m_{\mathrm{A}} m_{\mathrm{B}}} $$ \((d)\) Assume \(m_{\mathrm{B}}=2.0 \mathrm{~kg}\) \(m_{\mathrm{A}}=18.0 \mathrm{~kg}\) and \(v_{\mathrm{A}}=2.0 \mathrm{~m} / \mathrm{s}\). Use a spreadsheet to calculate and graph the values of \(v_{\mathrm{B}}^{\prime}\) from \(m_{\mathrm{C}}=0.0 \mathrm{~kg}\) to \(m_{\mathrm{C}}=50.0 \mathrm{~kg}\) in steps of \(1.0 \mathrm{~kg} .\) For what value of \(m_{\mathrm{C}}\) is the value of \(v_{\mathrm{B}}^{\prime}\) maximum? Does your numerical result agree with your result in part \((c) ?\)

(III) A scale is adjusted so that when a large, shallow pan is placed on it, it reads zero. A water faucet at height \(h=2.5 \mathrm{~m}\) above is turned on and water falls into the pan at a rate \(R=0.14 \mathrm{~kg} / \mathrm{s} .\) Determine \((a)\) a formula for the scale reading as a function of time \(t\) and \((b)\) the reading for \(t=9.0 \mathrm{~s} .(c)\) Repeat \((a)\) and \((b),\) but replace the shallow pan with a tall, narrow cylindrical container of area \(A=20 \mathrm{~cm}^{2}\) (the level rises in this case).

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