Problem 43 - University Physics with Modern Physics

Question

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).

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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: $$ v_{f, 1} = \frac{m_1v_{i, 1} + m_2(v_{i, 2} - v_{f, 2})}{m_1} $$

Step 3: 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: $$ \frac{1}{2} m_1(v_{i, 1}^2 - v_{f, 1}^2) = \frac{1}{2} m_2(v_{f, 2}^2 - v_{i, 2}^2) $$

Step 4: 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 \(22350 \,\mathrm{m/s}\).

Key Concepts

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.