A satellite with a mass of \(274 \mathrm{~kg}\) approaches a large planet at a speed \(v_{i, 1}=13.5 \mathrm{~km} / \mathrm{s}\). The planet is moving at a speed \(v_{i, 2}=10.5 \mathrm{~km} / \mathrm{s}\) in the opposite direction. The satellite partially orbits the planet and then moves away from the planet in a direction opposite to its original direction (see the figure). If this interaction is assumed to approximate an elastic collision in one dimension, what is the speed of the satellite after the collision? This so-called slingshot effect is often used to accelerate space probes for journeys to distance parts of the solar system (see Chapter 12).

Short Answer

Expert verified
After an elastic collision with a much larger planet, the final speed of the satellite is approximately 22,350 m/s.

Step by step solution

01

Convert speeds to the same units

To solve the problem, first convert the given speeds from km/s to m/s: $$ v_{i, 1} = 13.5 \frac{\mathrm{km}}{\mathrm{s}} \times \frac{1000 \mathrm{m}}{1 \mathrm{km}} = 13500 \frac{\mathrm{m}}{\mathrm{s}} $$ $$ v_{i, 2} = 10.5 \frac{\mathrm{km}}{\mathrm{s}} \times \frac{1000 \mathrm{m}}{1 \mathrm{km}} = 10500 \frac{\mathrm{m}}{\mathrm{s}} $$ Now the speeds are in the same units.
02

Apply linear momentum conservation

The first principle to apply is the conservation of linear momentum. The linear momentum before the collision is equal to the linear momentum after the collision. First, consider the satellite's linear momentum: $$ m_1v_{i, 1} + m_2v_{i, 2} = m_1v_{f, 1} + m_2v_{f, 2} $$ In this problem, we are only interested in the final speed of the satellite (i.e. \(v_{f, 1}\)). We also know that the planet is much larger and heavier than the satellite, so we can assume its speed remains almost unchanged after the collision. Thus, \(v_{f, 2} \approx v_{i, 2}\). Therefore, we can rewrite the above equation as: $$ v_{f, 1} = \frac{m_1v_{i, 1} + m_2(v_{i, 2} - v_{f, 2})}{m_1} $$
03

Apply kinetic energy conservation

The second principle to apply is the conservation of kinetic energy. The kinetic energy before the collision (initial) is equal to the kinetic energy after the collision (final): $$ \frac{1}{2} m_1v_{i, 1}^2 + \frac{1}{2} m_2v_{i, 2}^2 = \frac{1}{2} m_1v_{f, 1}^2 + \frac{1}{2} m_2v_{f, 2}^2 $$ As we assumed in Step 2 that the planet's speed remains unchanged, we can rewrite the kinetic energy equation as: $$ \frac{1}{2} m_1(v_{i, 1}^2 - v_{f, 1}^2) = \frac{1}{2} m_2(v_{f, 2}^2 - v_{i, 2}^2) $$
04

Solve for the final velocity of the satellite

Now, we have 2 equations with 2 variables (\(v_{f, 1}\) and \(v_{f, 2}\)) and we need to solve for \(v_{f, 1}\). To simplify the expression, notice that the linear momentum equation (Step 2) can be rewritten as: $$ v_{f, 2} = v_{i, 2} + \frac{m_1(v_{i, 1} - v_{f, 1})}{m_2} $$ Plug this expression into the kinetic energy equation (Step 3): $$ \frac{1}{2} m_1(v_{i, 1}^2 - v_{f, 1}^2) = \frac{1}{2} m_2\left(v_{i, 2} + \frac{m_1(v_{i, 1} - v_{f, 1})}{m_2}\right)^2 - \frac{1}{2} m_2v_{i, 2}^2 $$ After simplifying and solving the equation for \(v_{f, 1}\), we find: $$ v_{f, 1} = \frac{(2m_1+m_2)v_{i, 1} - m_2v_{i, 2}}{2m_1} $$ Now plug in the known values: $$ v_{f, 1} = \frac{(2(274 \,\mathrm{kg})+(274 \,\mathrm{kg}))(13500 \,\mathrm{m/s}) - (274 \,\mathrm{kg})(10500 \,\mathrm{m/s})}{2(274 \,\mathrm{kg})} $$ Finally, compute the speed of the satellite after the collision: $$ v_{f, 1} \approx 22350 \,\mathrm{m/s} $$ Hence, the speed of the satellite after the collision is approximately \(22350 \,\mathrm{m/s}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Conservation of Linear Momentum
The concept of conservation of linear momentum is essential in understanding physical interactions where no external forces act on the system. According to this principle, the total linear momentum of a system remains constant if it is not influenced by external forces.

When a satellite slingshot maneuver occurs, it is an excellent example of this conservation law in action. The satellite and the planet form an isolated system where, for a brief moment, their mutual gravitational forces are the only significant interactions. During this time, the total linear momentum before and after the encounter must be the same.

Mathematically, this is expressed as: \[ m_1v_{i, 1} + m_2v_{i, 2} = m_1v_{f, 1} + m_2v_{f, 2} \] where \(m_1\) and \(m_2\) are the masses, and \(v_{i, 1}\), \(v_{i, 2}\), \(v_{f, 1}\), and \(v_{f, 2}\) are the initial and final velocities of the satellite and the planet, respectively.

In our exercise, by recognizing that the planet’s mass is much greater than the satellite's mass, we simplify our calculations by assuming the planet's velocity does not significantly change, leading us to find the final velocity of the satellite post-interaction.
Conservation of Kinetic Energy
The conservation of kinetic energy is another fundamental principle, which states that in an isolated system with no non-conservative forces (like friction or air resistance), the total kinetic energy remains constant during an elastic collision.

An elastic collision is one in which both momentum and kinetic energy are conserved. This assumption is crucial in the satellite slingshot effect since it allows the satellite to gain speed without any external energy input, simply by interacting with a moving planet.

The formula that expresses this conservation is: \[ \frac{1}{2} m_1v_{i, 1}^2 + \frac{1}{2} m_2v_{i, 2}^2 = \frac{1}{2} m_1v_{f, 1}^2 + \frac{1}{2} m_2v_{f, 2}^2 \] This is fundamental in our problem for determining the satellite's final speed, as we equate the initial total kinetic energy with the final total kinetic energy to obtain the necessary equations for solving.
Elastic Collision
An elastic collision is particularly significant when talking about satellite maneuvers, like the slingshot effect. In an elastic collision, two bodies bounce off each other without losing kinetic energy to other forms, like heat or sound.

During a satellite's slingshot maneuver around a planet, it can be modeled as an elastic collision because the gravitational interaction does not generate significant heat or other energy losses; instead, the satellite's path is just redirected, and its speed is changed.

Understanding elastic collisions enables scientists and engineers to predict the outcomes of such maneuvers accurately, ensuring the satellite achieves the required velocity to reach its destination. The assumption of an elastic collision simplifies our equations and allows us to solve for the satellite’s final velocity post-collision.
Final Velocity Calculation
The final velocity calculation of the satellite after the slingshot effect combines the conservation principles discussed above. By setting up equations for the conservation of linear momentum and kinetic energy and assuming an elastic collision, we can solve for the satellite's final velocity.

In our exercise, we derived a simplified expression for the final speed of the satellite: \[ v_{f, 1} = \frac{(2m_1+m_2)v_{i, 1} - m_2v_{i, 2}}{2m_1} \] After inserting the values, we can calculate the final velocity that the satellite will have after the slingshot maneuver.

This step is crucial for space missions, as it determines how the satellite will benefit from the slingshot effect to reach farther destinations with less fuel consumption. The calculations are complex but essential for mission success.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following statements about car collisions are true and which are false? a) The essential safety benefit of crumple zones (parts of the front of a car designed to receive maximum deformation during a head-on collision) results from absorbing kinetic energy, converting it into deformation, and lengthening the effective collision time, thus reducing the average force experienced by the driver. b) If car 1 has mass \(m\) and speed \(v\), and car 2 has mass \(0.5 m\) and speed \(1.5 v\), then both cars have the same momentum. c) If two identical cars with identical speeds collide head on, the magnitude of the impulse received by each car and each driver is the same as if one car at the same speed had collided head on with a concrete wall. d) Car 1 has mass \(m,\) and car 2 has mass \(2 m .\) In a head-on collision of these cars while moving at identical speeds in opposite directions, car 1 experiences a bigger acceleration than car 2 . e) Car 1 has mass \(m\), and car 2 has mass \(2 m\). In a headon collision of these cars while moving at identical speeds in opposite directions, car 1 receives an impulse of bigger magnitude than that received by car 2.

Using momentum and force principles, explain why an air bag reduces injury in an automobile collision.

NASA has taken an increased interest in near Earth asteroids. These objects, popularized in recent blockbuster movies, can pass very close to Earth on a cosmic scale, sometimes as close as 1 million miles. Most are small-less than \(500 \mathrm{~m}\) across \(-\) and while an impact with one of the smaller ones could be dangerous, experts believe that it may not be catastrophic to the human race. One possible defense system against near Earth asteroids involves hitting an incoming asteroid with a rocket to divert its course. Assume a relatively small asteroid with a mass of \(2.10 \cdot 10^{10} \mathrm{~kg}\) is traveling toward the Earth at a modest speed of \(12.0 \mathrm{~km} / \mathrm{s}\). a) How fast would a large rocket with a mass of \(8.00 \cdot 10^{4} \mathrm{~kg}\) have to be moving when it hit the asteroid head on in order to stop the asteroid? b) An alternative approach would be to divert the asteroid from its path by a small amount to cause it to miss Earth. How fast would the rocket of part (a) have to be traveling at impact to divert the asteroid's path by \(1.00^{\circ}\) ? In this case, assume that the rocket hits the asteroid while traveling along a line perpendicular to the asteroid's path.

To solve problems involving projectiles traveling through the air by applying the law of conservation of momentum requires evaluating the momentum of the system immediately before and immediately after the collision or explosion. Why?

A 60.0 -kg astronaut inside a 7.00 -m-long space capsule of mass \(500 . \mathrm{kg}\) is floating weightlessly on one end of the capsule. He kicks off the wall at a velocity of \(3.50 \mathrm{~m} / \mathrm{s}\) toward the other end of the capsule. How long does it take the astronaut to reach the far wall?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free