Question

The function \(h(v)=\frac{6378 v^{2}}{125-v^{2}}\) gives the maximum height, \(h,\) in kilometres, as a function of the initial velocity, \(v,\) in kilometres per second, for an object launched upward from Earth's surface, if the object gets no additional propulsion and air resistance is ignored. a) Graph the function. What parts of the graph are applicable to this situation? b) Explain what the graph indicates about how the maximum height is affected by the initial velocity. c) The term escape velocity refers to the initial speed required to break free of a gravitational field. Describe the nature of the graph for its non- permissible value, and explain why it represents the escape velocity for the object.

Short answer

Graph the function avoiding the vertical asymptote at \[\sqrt{125} \approx 11.18\]. Increased initial velocity leads to higher maximum heights. The escape velocity is about 11.18 km/s, where the function becomes undefined, indicating an infinite height.

Step-by-step solution

Understand the Function

The function given is \[h(v)=\frac{6378 v^{2}}{125 - v^{2}},\] where \[h\] is the maximum height in kilometers and \[v\] is the initial velocity in kilometers per second. This equation models the height an object will reach given an initial velocity.

Graph the Function

To graph the function, recognize the behavior of rational functions. Specifically, the denominator becomes zero when \[v = \sqrt{125},\] creating a vertical asymptote. Plot points for values of \[v\] that do not make the denominator zero.

Identify Relevant Parts of the Graph

On the graph, only positive values for the initial velocity \[v\] are physically meaningful since negative values don't apply in this context. Also, consider velocities less than \[\sqrt{125} \approx 11.18\] km/s for the object to remain within the modeled constraints.

Analyze Maximum Height Dependency

From the graph, observe that as \[v\] approaches \[\sqrt{125},\] the maximum height \[h\] increases rapidly. This signifies that increasing the initial velocity leads to a higher maximum height, but there’s a cutoff point.

Non-Permissible Value and Escape Velocity

The non-permissible value for the function is \[v = \sqrt{125} \approx 11.18\] km/s. At this velocity, the function is undefined, indicating an infinite height, which corresponds to the escape velocity -- the lowest velocity needed to escape Earth's gravitational field.

Key concepts

Rational Functions

In mathematics, a rational function is a function that can be expressed as the quotient of two polynomials. Specifically, it is written as \[f(x) = \frac{P(x)}{Q(x)}\]where both \(P(x)\) and \(Q(x)\) are polynomials, and \(Q(x) eq 0\).
The function \(h(v) = \frac{6378v^2}{125 - v^2}\) is an example of a rational function since it is the ratio of the polynomial \(6378v^2\) and \(125 - v^2\).
Rational functions are important because they can model real-life situations, such as the height an object reaches when launched upwards.
Understanding the behavior of both the numerator and the denominator is critical in analyzing such functions, particularly in identifying key features like the domain and asymptotes.

Vertical Asymptote

A vertical asymptote is a line that a graph approaches but never touches or crosses. In other words, it represents a value of \(x\) that makes the function undefined.
For the function \(h(v) = \frac{6378v^2}{125 - v^2}\), the denominator becomes zero when \(v = \sqrt{125} \approx 11.18\) km/s. When the denominator is zero, the function's value tends to infinity, indicating a vertical asymptote at this point.
The vertical asymptote is significant because it marks the boundary where the function changes behavior drastically.
In real-life terms, this means that the initial velocity \(v\) of 11.18 km/s is crucial since the function is undefined at this point, representing the escape velocity.

Initial Velocity

Initial velocity is the speed and direction at which an object starts its motion. For the function \(h(v)\), the initial velocity \(v\) is expressed in kilometers per second (km/s).
The initial velocity plays a crucial role in determining how high the object will go. The higher the initial velocity, the greater the height the object can achieve.
However, there is a limit. As described in the exercise, if \(v\) approaches \(\sqrt{125}\), the height increases infinitely, indicating that around this velocity, the object is on the verge of escaping Earth's gravity.
Thus, the dependency on initial velocity is not linear, especially as it nears the threshold represented by the vertical asymptote.

Escape Velocity

Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a celestial body without further propulsion. For Earth, this value is approximately 11.18 km/s.
In the context of the function \(h(v) = \frac{6378v^2}{125 - v^2}\), the escape velocity is represented by the vertical asymptote at \(v = \sqrt{125}\).
When the initial velocity reaches this critical value, the function becomes undefined, metaphorically indicating that an object launched at this speed can reach an infinitely high position. This is why this specific velocity is termed escape velocity – it allows the object to overcome Earth's gravitational pull completely.