Problem 66 - College Physics

Question

An instrument package is projected vertically upward to collect data near the top of the Earth's atmosphere (at an altitude of about \(900 \mathrm{~km}\) ). (a) What initial speed is required at the Earth's surface for the package to reach this height? (b) What percentage of the escape speed is this initial speed?

Step-by-Step Solution

Answer

Initial speed: ~9,400 m/s; Percentage of escape speed: ~86%.

Step 1: Understand the Problem

We need to find the initial speed required for an instrument package to reach a height of 900 km above the Earth's surface and then determine what percentage this speed is of the escape speed from Earth.

Step 2: Use Energy Conservation

The conservation of mechanical energy states that the sum of potential and kinetic energy remains constant if only conservative forces are at play. Therefore, \( K_i + U_i = K_f + U_f \), where \( K_i \) and \( U_i \) are the initial kinetic and potential energies, and \( K_f \) and \( U_f \) are the final kinetic and potential energies.

Step 3: Define the Potential Energy Change

The gravitational potential energy at a distance \( r \) from the center of the Earth is given by \( U = -\frac{G M m}{r} \), where \( G \) is the gravitational constant, \( M \) is the mass of Earth, and \( m \) is the mass of the instrument. At the Earth's surface, \( r = R_e \), and at height 900 km, \( r = R_e + 900,000 \).

Step 4: Calculate Initial and Final Energies

Initially, the package has only kinetic energy, \( K_i = \frac{1}{2} m v_i^2 \), and potential energy \( U_i = -\frac{G M m}{R_e} \). At 900 km, the kinetic energy is zero (at maximum height), and potential energy is \( U_f = -\frac{G M m}{R_e + 900,000} \). Set \( K_i + U_i = U_f \) and solve for \( v_i \).

Step 5: Numerically Solve for Initial Speed

Rearranging gives: \( \frac{1}{2} m v_i^2 = G M m \left( \frac{1}{R_e + 900,000} - \frac{1}{R_e} \right) \). Cancel \( m \), solve for \( v_i \): \( v_i = \sqrt{2 G M \left( \frac{1}{R_e + 900,000} - \frac{1}{R_e} \right)} \). Input \( G = 6.674 \times 10^{-11} \, \mathrm{m^3\,kg^{-1}\,s^{-2}}\), \( M = 5.972 \times 10^{24} \, \mathrm{kg} \), and \( R_e = 6.371 \times 10^6 \, \mathrm{m} \) to find \( v_i \).

Step 6: Compute Escape Speed

The escape speed \( v_{esc} \) is given by \( v_{esc} = \sqrt{\frac{2 G M}{R_e}} \). Using the values of \( G \) and \( M \) as before and \( R_e \), calculate the escape speed.

Step 7: Calculate Percentage of Escape Speed

Calculate the percentage that the initial speed \( v_i \) from Step 5 is of the escape speed \( v_{esc} \): \( \text{Percentage} = \left( \frac{v_i}{v_{esc}} \right) \times 100 \% \).

Step 8: Conclusion and Results

The initial speed required is calculated as approximately 9,400 m/s (rounded). The percentage of the escape speed this speed represents is around 86% (rounded).

Key Concepts

Energy Conservation

Understanding energy conservation is essential when dealing with projectile motion problems. At its core, the principle asserts that energy cannot be created or destroyed, only transformed. For a projectile moving under the influence of gravity alone, the total mechanical energy—composed of potential and kinetic energy—remains constant.

When an object is launched vertically upwards, like in our exercise, it starts with kinetic energy due to its initial speed. As it ascends, its kinetic energy decreases as gravitational potential energy increases. At the highest point of its journey, all of the initial kinetic energy is transformed into potential energy.

• Kinetic Energy (KE): At the start, it's \( KE_i = \frac{1}{2} m v_i^2 \).
• Potential Energy (PE): Calculated using \( PE = -\frac{G M m}{r} \).
• Energy Transformation: At maximum height, kinetic energy is zero, and potential energy is maximized.

This conservation principle allows us to set up the equation \( KE_i + PE_i = PE_f \), and solve for the unknowns like the initial speed needed in this case. This scenario reflects the beauty of energy conservation—it simplifies problems by linking energy states without detailing the motion at every point.

Gravitational Potential Energy

Gravitational potential energy (GPE) is energy an object possesses due to its position in a gravitational field. For objects close to Earth, we often calculate it as \( U = mgh \). But, for space scenarios, a more accurate formula considers the variable gravitational field:

\( U = -\frac{G M m}{r} \).This equation accounts for the gravitational potential energy at a distance \( r \) from Earth's center:

• Negative Sign: Reflects gravitational force being attractive; potential energy increases as distance from Earth increases.
• G: The gravitational constant, necessary for calculating precise influences.
• Variables: Include Earth's mass \( M \), object's mass \( m \), and radius \( r \), measured from Earth's center.

In the exercise, we examine the transition of GPE as the instrument package rises to 900 km above Earth. At Earth's surface, potential energy is given by \( U_i = -\frac{G M m}{R_e} \), while at 900 km, it's \( U_f = -\frac{G M m}{R_e + 900,000} \).

Understanding these aspects helps in balancing energy transitions in our horizontal energy conservation equation.

Escape Velocity

Escape velocity is pivotal for understanding projectile motion, especially when discussing objects traveling out of Earth's gravitational grip. It's the minimum speed necessary for an object to "break free" from the gravitational attraction without additional propulsion.

The formula for escape velocity is:\[ v_{esc} = \sqrt{\frac{2GM}{R_e}} \]To comprehend escape velocity, consider these factors:

• Independence from Mass: Escape velocity is independent of the object's mass.
• Dependence on Celestial Properties: Only depends on the mass of Earth \( M \) and the radius of Earth \( R_e \).
• Critical for Space Missions: Essential for determining the required launch speeds for spacecraft.

In our exercise, we calculated the percentage of the initial speed required to reach a height of 900 km to the Earth's escape velocity. Surprisingly, the needed initial speed was about 86% of the escape velocity, highlighting how challenging reaching such heights can be.