Problem 102 - Thomas Calculus in SI Units

Question

The hammer and the feather When Apollo 15 astronaut David Scott dropped a hammer and a feather on the moon to demonstrate that in a vacuum all bodies fall with the same (constant) acceleration, he dropped them from about \(1.2 \mathrm{m}\) above the ground. The television footage of the event shows the hammer and the feather falling more slowly than on Earth, where, in a vacuum, they would have taken only half a second to fall the \(1.2 \mathrm{m}\). How long did it take the hammer and feather to fall \(1.2 \mathrm{m}\) on the moon? To find out, solve the following initial value problem for \(s\) as a function of \(t .\) Then find the value of \(t\) that makes \(s\) equal to \(0 .\). Differential equation: \(\frac{d^{2} s}{d t^{2}}=-1.6 \mathrm{m} / \mathrm{s}^{2}\) Initial conditions: \(\frac{d s}{d t}=0\) and \(s=1.2\) when \(t=0\)

Step-by-Step Solution

Answer

It took about 1.22 seconds for the hammer and feather to fall 1.2 meters on the moon.

Step 1: Identify the Problem Type

We have a second-order differential equation with initial conditions, commonly known as an initial value problem. This equation models the motion of an object under constant acceleration.

Step 2: Write the Differential Equation

The differential equation given is \( \frac{d^2 s}{dt^2} = -1.6 \ m/s^2 \). This represents the acceleration \( a \) of the hammer or feather on the moon.

Step 3: Integrate to Find Velocity Function

First, integrate the acceleration equation to find the velocity \( \frac{ds}{dt} \). Integrating \( -1.6 \) with respect to \( t \) gives: \[ \frac{ds}{dt} = -1.6t + C_1 \] where \( C_1 \) is a constant of integration.

Step 4: Apply Initial Condition for Velocity

We know that \( \frac{ds}{dt} = 0 \) when \( t = 0 \). Substitute these values into the velocity function: \[ 0 = -1.6(0) + C_1 \], solving gives \( C_1 = 0 \). So, \( \frac{ds}{dt} = -1.6t \).

Step 5: Integrate to Find Position Function

Integrate the velocity function to find the position function \( s(t) \). Integrating \( -1.6t \) with respect to \( t \): \[ s(t) = -0.8t^2 + C_2 \] where \( C_2 \) is another constant of integration.

Step 6: Apply Initial Condition for Position

We also have \( s = 1.2 \) when \( t = 0 \). Substitute these into the position function: \[ 1.2 = -0.8(0)^2 + C_2 \], solving gives \( C_2 = 1.2 \). Thus, \( s(t) = -0.8t^2 + 1.2 \).

Step 7: Solve for Time When Position is Zero

To find when the hammer and feather hit the ground (\( s = 0 \)), set the position function to zero: \[ 0 = -0.8t^2 + 1.2 \]. Solve for \( t \): \[ 0.8t^2 = 1.2 \] so \[ t^2 = \frac{1.2}{0.8} = 1.5 \]. Thus, \( t = \sqrt{1.5} \approx 1.22 \ s \).

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate some function with its derivatives. In the context of physics, they often describe how a particular quantity changes over time. These equations are essential for modeling various physical scenarios, such as motion under gravity. For our problem, we have a second-order differential equation: \( \frac{d^2 s}{dt^2} = -1.6 \ m/s^2 \). This equation tells us that the acceleration of the hammer and feather is constant on the moon. The solution to this equation will give us expressions for velocity and position as functions of time.

Constant Acceleration

Constant acceleration means the rate of change of velocity is steady over time. On the moon, the gravitational acceleration is \(-1.6 \ m/s^2\), which is much less than Earth's \( -9.8 \ m/s^2\). This slower acceleration is why objects take longer to fall the same distance on the moon compared to Earth.
In the problem, due to this constant acceleration, the velocity changes linearly over time. This makes it easier to predict the motion path of the falling objects, using simple integration techniques.

Integration

Integration is the mathematical process of finding the whole from the sum of its parts. In physics, it is used to determine quantities like displacement (position) and velocity from a given acceleration.
- **Velocity Function:** By integrating the acceleration function, we derive the velocity function. In our situation: \[ \frac{ds}{dt} = \int -1.6 \, dt = -1.6t + C_1 \] - **Position Function:** Similarly, we integrate the velocity function to calculate the position: \[ s(t) = \int -1.6t \, dt = -0.8t^2 + C_2 \] Applying initial conditions helps determine the constants \(C_1\) and \(C_2\), giving us precise equations for this specific scenario.

Physics Applications

Physics relies heavily on mathematical concepts to explain real-world phenomena. By understanding and applying principles like differential equations and integration, we can model the behavior of objects under certain conditions. This involves calculating how they move, what paths they take, and how long it takes for them to reach a particular point, like reaching the ground.
In the Apollo 15 experiment, the differential equation and its solutions allowed us to predict the path and duration of the hammer and feather's fall on the moon, showcasing that mathematical tools can precisely model physical events. This demonstration emphasized how gravity affects objects regardless of their mass, as per Galileo's principle, specifically when no air resistance is present, like on the moon.