Problem 54 - Calculus with Concepts in Calculus

Question

Suppose a rocket is launched vertically upward and its velocity \(v(r)\) in miles per second at a distance of \(r\) miles from the center of the earth is given by the formula $$ v(r)=\sqrt{\frac{192,000}{r}+v_{0}^{2}-48} \text { for } r \geq 4000 $$ where \(v_{0}\) is constant and represents the velocity of the rocket at burnout. a. If \(v_{0}=8\) (miles per second), what is the limit of the velocity as the distance grows without bound? b. Find the number \(v_{0}>0\) for which \(\lim _{r \rightarrow \infty} v(r)=0\).

Step-by-Step Solution

Answer

a) 4 miles/s. b) \( v_0 = 4\sqrt{3} \).

Step 1: Identify the Limit Calculation

We need to determine \( \lim_{r \to \infty} v(r) \) for the given function \( v(r) = \sqrt{\frac{192,000}{r} + v_0^2 - 48} \). This will find the velocity as the distance \( r \) from the center of the Earth grows infinitely large.

Step 2: Simplify the Limit Expression

Compute the limit as \( r \to \infty \). As \( r \) gets very large, the term \( \frac{192,000}{r} \) approaches zero. Thus, the expression inside the square root becomes \( v_0^2 - 48 \). Therefore, \( \lim_{r \to \infty} v(r) = \lim_{r \to \infty} \sqrt{v_0^2 - 48} = \sqrt{v_0^2 - 48} \).

Step 3: Part A: Substitute Known Values

For part a, substitute \( v_0 = 8 \) into the limit expression. Calculate: \( \sqrt{8^2 - 48} = \sqrt{64 - 48} = \sqrt{16} = 4 \). So when \( v_0 = 8 \), \( \lim_{r \to \infty} v(r) = 4 \).

Step 4: Part B: Set the Limit to Zero

For part b, find \( v_0 \) such that \( \lim_{r \to \infty} v(r) = 0 \). Therefore, set \( \sqrt{v_0^2 - 48} = 0 \) giving \( v_0^2 - 48 = 0 \). Solving this gives \( v_0^2 = 48 \) which means \( v_0 = \sqrt{48} \). Thus, \( v_0 = 4\sqrt{3} \) or approximately \( v_0 \approx 6.93 \) miles per second.

Key Concepts

rocket velocity

Velocity is a critical concept in understanding rocket dynamics. Rocket velocity refers to the speed at which a rocket moves. It is influenced by several factors such as the initial thrust, gravity, and resistance from the atmosphere. In the context of the given exercise, we are examining how this velocity behaves when considering the distance from the Earth's center. Initially, the velocity is determined by the rocket's engines and decreases due to gravitational pull and other resistances. As the rocket moves upwards, gaining distance from the Earth, its velocity changes, showcasing an interesting dynamic that we explore using mathematical functions.

velocity at burnout

When a rocket runs out of fuel, it reaches a point known as 'burnout.' At burnout, the velocity of the rocket is the maximum velocity it achieves solely by its engines before the fuel depletes completely. This is denoted by the constant \( v_0 \) in the velocity formula provided in the problem. Understanding this velocity is crucial for predicting the rocket's subsequent motion. Once the fuel is exhausted, the rocket relies on inertia to continue its movement, which is commonly analyzed in situations involving \'limit calculation\' where the aftereffects of reaching this burnout velocity are considered. The correct prediction of this velocity helps ensure the success and safety of space missions.

limit calculation

The concept of limit calculations helps us understand the behavior of functions as the input grows indefinitely large or approaches a specific value. In the exercise, limit calculations are used to predict how the rocket's velocity \( v(r) \) behaves as it travels farther away from the Earth without end. Using limit concepts, we know that as \( r \rightarrow \infty \), the term \( \frac{192,000}{r} \) approaches zero, simplifying the equation inside the square root to \( v_0^2 - 48 \). This effectively allows us to determine the velocity of the rocket in a state far removed from Earth's influence, proving crucial for calculating escape velocities and understanding long-term dynamics in space travel.

distance from Earth center

Distance from the Earth's center plays a key role in determining a rocket's velocity. As a rocket gains altitude (i.e., increases its distance \( r \) from the center of Earth), several variables in the velocity formula start to change. For instance, gravitational pull decreases with increasing distance, which affects the velocity. The formula \( v(r)=\sqrt{\frac{192,000}{r}+v_{0}^{2}-48} \) explicitly incorporates the distance from the Earth's center into the calculation of velocity. This equation allows us to analyze how velocity changes relatively to increasing distance and helps in evaluating key scenarios like burnout velocity and escape velocity, vital for routing space journeys correctly.