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Chapter 10: Problem 86

Voyager 2 Spacecraft Voyager 2 (of mass \(m\) and speed \(v\) relative to the Sun) approaches the planet Jupiter (of mass \(M\) and speed \(V_{J}\) relative to the Sun) as shown in Fig. \(10-60 .\) The spacecraft rounds the planet and departs in the opposite direction. What is its speed, relative to the Sun, after this slingshot encounter, which can be analyzed as a collision? Assume \(v=12 \mathrm{~km} / \mathrm{s}\) and \(V_{J}=13 \mathrm{~km} / \mathrm{s}\) (the orbital speed of Jupiter). The mass of Jupiter is very much greater than the mass of the spacecraft \((M \gg m)\).

Short Answer

Expert verified
The final speed of Voyager 2 relative to the Sun after the encounter is 25 km/s.

Step by step solution

01

Understand the scenario

The problem involves the Voyager 2 spacecraft performing a gravitational slingshot maneuver around Jupiter. This can be analyzed as an elastic collision due to the significant mass difference between Jupiter and Voyager 2.
02

Identify known values

It is given that the initial speed of Voyager 2 relative to the Sun is \(v = 12 \mathrm{~km/s}\) and the orbital speed of Jupiter relative to the Sun is \(V_J = 13 \mathrm{~km/s}\). Additionally, the mass of Jupiter \(M\) is much greater than the mass of Voyager 2 \(m\), so \(M \gg m\).
03

Apply the principle of elastic collision

In an elastic collision where one object (Jupiter) is significantly more massive than the other (Voyager 2), the less massive object effectively 'bounces' off the more massive one. Thus, the speed of Voyager 2 after encounter relative to Jupiter will be the same as before the encounter, but the direction will be reversed.
04

Calculate speed after the encounter relative to Jupiter

Before the encounter, Voyager 2 has speed \(v\) relative to the Sun. After the encounter, it will have speed \(v\) relative to Jupiter, but in the opposite direction.
05

Determine the final speed relative to the Sun

After the encounter, to find the speed of Voyager 2 relative to the Sun, sum the velocities: \[ \text{Final speed} = |V_J + (-v)| = V_J + v \]Since the direction is opposite, we use the absolute value. Therefore, \[v' = v + V_J \]
06

Substitute the known values

Given \(v = 12 \mathrm{~km/s}\) and \(V_J = 13 \mathrm{~km/s}\), \[v' = 12 \mathrm{~km/s} + 13 \mathrm{~km/s} = 25 \mathrm{~km/s} \]

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Key Concepts

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

Elastic Collision
An elastic collision is an event where two bodies collide and bounce off each other without losing kinetic energy. In the context of the Voyager 2 spacecraft and Jupiter, since the mass of Jupiter is enormously greater than Voyager 2, it can be considered a perfectly elastic collision. Here, Voyager 2 'bounces' off Jupiter. The change is primarily in its direction due to the gravitational interaction, but its speed relative to Jupiter remains the same. The massive difference in masses means Jupiter's velocity remains mostly unchanged.
Relative Velocity
Relative velocity is the velocity of an object as observed from a particular reference frame. In this exercise, we consider the velocities of both Voyager 2 and Jupiter relative to the Sun. Initially, Voyager 2 travels at 12 km/s and Jupiter at 13 km/s. After performing the maneuver (elastic collision), Voyager 2's velocity relative to Jupiter becomes opposite in direction but retains the same magnitude. To find the spacecraft's final speed relative to the Sun, we essentially add the speed of Voyager 2 to that of Jupiter, given by the formula:
Orbital Mechanics
Orbital mechanics examines the motions of spacecraft and celestial bodies under the influence of gravitational forces. The gravitational slingshot maneuver utilized by Voyager 2 is a practical application of these principles. By approaching a massive planet like Jupiter and using its gravity, Voyager 2 gains additional energy, allowing it to travel faster relative to the Sun. This maneuver relies on the conservation of momentum and energy principles, effectively giving the spacecraft a speed boost without fuel consumption. The interaction changes Voyager 2's trajectory and speed utilizing Jupiter's immense gravitational field.
Spacecraft Dynamics
Spacecraft dynamics involves understanding the motion and control of spacecraft. In the case of Voyager 2, the gravitational slingshot maneuver around Jupiter is meticulously planned to ensure that Voyager gains enough speed to continue its journey. This maneuver must account for various factors such as the planet's gravity, the spacecraft's speed, and the desired outbound trajectory. The dynamics of managing this include calculating the correct approach angle and timing the interaction perfectly to achieve the desired end speed. The objective is to maximize the gain in kinetic energy imparted to the spacecraft by Jupiter's motion relative to the Sun.

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Most popular questions from this chapter

Nonconforming Spring A certain spring is found \(n o t\) to conform to Hooke's law. The force (in newtons) it exerts when stretched a distance \(x\) (in meters) is found to have magnitude \((52.8 \mathrm{~N} / \mathrm{m}) x+\left(38.4 \mathrm{~N} / \mathrm{m}^{2}\right) x^{2}\) in the direction opposing the stretch. (a) Compute the work required to stretch the spring from \(x_{1}=0.500\) \(\mathrm{m}\) to \(x_{2}=1.00 \mathrm{~m} .\) (b) With one end of the spring fixed, a particle of mass \(2.17 \mathrm{~kg}\) is attached to the other end of the spring when it is extended by an amount \(x_{2}=1.00 \mathrm{~m}\). If the particle is then released from rest, what is its speed at the instant the spring has returned to the configuration in which the extension is \(x_{1}=0.500 \mathrm{~m} ?(\mathrm{c})\) Is the force exerted by the spring conservative or nonconservative? Explain.

. Playground Slide A girl whose weight is \(267 \mathrm{~N}\) slides down a \(6.1 \mathrm{~m}\) playground slide that makes an angle of \(20^{\circ}\) with the horizontal. The coefficient of kinetic friction between slide and child is \(0.10\). (a) How much energy is transferred to thermal energy? (b) If the girl starts at the top with a speed of \(0.457 \mathrm{~m} / \mathrm{s}\), what is her speed at the bottom?

Worker Pushes Block A worker pushed a \(27 \mathrm{~kg}\) block \(9.2 \mathrm{~m}\) along a level floor at constant speed with a force directed \(32^{\circ}\) below the horizontal. If the coefficient of kinetic friction between block and floor was \(0.20\), what were (a) the work done by the worker's force and (b) the increase in thermal energy of the block-floor system?

Spring at the Top of an Incline a spring with spring constant \(k=170 \mathrm{~N} / \mathrm{m}\) is at the top of a \(37.0^{\circ}\) frictionless incline. The lower end of the incline is \(1.00 \mathrm{~m}\) from the end of the spring, which is at its relaxed length. A \(2.00 \mathrm{~kg}\) canister is pushed against the spring until the spring is compressed \(0.200 \mathrm{~m}\) and released from rest. (a) What is the speed of the canister at the instant the spring returns to its relaxed length (which is when the canister loses contact with the spring)? (b) What is the speed of the canister when it reaches the lower end of the incline?

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