Problem 86 - Precalculus Mathematics for Calculus

Question

Suppose a rocket is fired upward from the surface of the earth with an initial velocity \(v\) (measured in meters per second). Then the maximum height \(h\) (in meters) reached by the rocket is given by the function $$ h(v)=\frac{R v^{2}}{2 g R-v^{2}} $$ where \(R=6.4 \times 10^{6} \mathrm{m}\) is the radius of the earth and \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) is the acceleration due to gravity. Use a graphing device to draw a graph of the function \(h .\) (Note that \(h\) and \(v\) must both be positive, so the viewing rectangle need not contain negative values.) What does the vertical asymptote represent physically?

Step-by-Step Solution

Answer

The vertical asymptote represents the critical velocity at which the rocket can escape Earth's gravity.

Step 1: Understand the function

The function given for maximum height is \(h(v)=\frac{R v^{2}}{2 g R-v^{2}}\), where \(R\) is the Earth's radius and \(g\) is the gravitational acceleration. The variables are height \(h\) and initial velocity \(v\).

Step 2: Determine domain considerations

To ensure the denominator is non-zero, \(2gR - v^2 > 0\). Therefore, the maximum initial velocity \(v\) must be such that \(v^2 < 2gR\). This condition avoids division by zero.

Step 3: Calculate the Denominator Limit

Substitute the values of \(R\) and \(g\) into \(2gR\):\[ 2 \times 9.8 \times 6.4\times 10^6 = 1.2544 \times 10^8 \]. This result gives the limit \(v^2 < 1.2544 \times 10^8\), a critical value for plotting.

Step 4: Plot the function using a graphing device

In your graphing device, plot \(h(v)=\frac{R v^{2}}{2 g R-v^{2}}\) with the range \(v \in (0,\sqrt{1.2544 \times 10^8})\). Ensure \(h(v)\) is plotted only for positive values.

Step 5: Identify the vertical asymptote

The vertical asymptote occurs at the velocity where \(v^2 = 2gR\). Physically, this means that if the initial velocity of the rocket equals or exceeds this value, the rocket can escape Earth's gravitational pull and the height tends to infinity.

Key Concepts

Maximum Height Calculation

The concept of calculating the maximum height of a rocket revolves around understanding its initial velocity and how this interacts with Earth's gravitational field. The function given for height is \(h(v) = \frac{Rv^2}{2gR - v^2}\), where \(R\) is the Earth's radius \(6.4 \times 10^6\, \mathrm{m}\) and \(g\) represents the acceleration due to gravity, \(9.8\, \mathrm{m/s}^2\).
To compute the maximum height:

• The initial velocity \(v\) must be plugged into the function.
• The function accounts for how quickly the rocket loses upward momentum under Earth's gravity.

The denominator \(2gR - v^2\) plays a crucial role. It ensures the rocket doesn't reach an infinite height under normal conditions. As the initial velocity increases, the rocket can potentially reach the maximum theoretical height calculated by the function before flattening out at the critical point of escape.

Gravitational Force

Gravitational force is a vital factor in the trajectory and maximum height reachable by a rocket. It represents the pull that Earth exerts on the rocket, pulling it back towards the ground. In the equation for maximum height, gravitational acceleration \(g = 9.8\, \mathrm{m/s}^2\) is key.
Understanding gravitational force involves appreciating:

• It's constant for our calculations, as the rocket remains relatively close to Earth's surface on the way up.
• A larger initial velocity is needed to overcome stronger gravitational force.

If the rocket's velocity squared \(v^2\) approaches \(2gR\), the denominator trends towards zero, and theoretically, the rocket could continue outward without returning. This scenario is when the rocket achieves escape velocity.

Graphing Functions

Graphing functions provides a visual understanding of how a rocket's height changes with different initial velocities. A graphing device visualizes the function \(h(v) = \frac{Rv^2}{2gR - v^2}\) by plotting initial velocity \(v\) against height \(h\).
When graphing the function:

• Set the domain of \(v\) from zero to \(\sqrt{1.2544 \times 10^8}\) to avoid the denominator becoming zero.
• Only consider positive values as negative values don't physically apply.

A vertical asymptote appears at \(v^2 = 2gR\). Physically, this asymptote indicates the escape velocity threshold where the rocket escapes Earth's gravity. The function visually demonstrates how as \(v\) approaches this value, the rocket height approaches infinity, highlighting the critical nature of velocity in rocket flight.