Chapter 9.1, Problem 104E

(a)

To determine

To find: The escape velocity of the satellite.

Answer to Problem 104E

The escape velocity of the satellite is $24748.74\text{mi}/\text{hr}$ .

Explanation of Solution

Given information:

A satellite in a $100$ -mile-high circular orbit around earth has the velocity of approximately $17500$ miles per hour. When this velocity is multiplied by $\sqrt{2}$ , the satellite has the minimum velocity necessary to escape earth’s gravity and follows a parabolic path with the center of earth as focus.

  

Calculation:

As given, the velocity of the satellite is $17500$ miles per hour. When this velocity is multiplied by $\sqrt{2}$ , it becomes the necessary velocity to escape earth’s gravity.

Use the given for the necessary velocity to escape earth’s gravity.

  $\begin{array}{c}v=17500\times \sqrt{2}\\ =17500\times 1.4142\\ =24748.74\text{mi}/\text{hr}\end{array}$

Therefore, the escape velocity of the satellite is $24748.74\text{mi}/\text{hr}$ .

(b)

To determine

To find: The equation of the parabolic path.

Answer to Problem 104E

The equation of the parabolic path is $y^2=-16400\left(x-4100\right)$ .

Explanation of Solution

Given information:

A satellite in a $100$ -mile-high circular orbit around earth has the velocity of approximately $17500$ miles per hour. When this velocity is multiplied by $\sqrt{2}$ , the satellite has the minimum velocity necessary to escape earth’s gravity and follows a parabolic path with the center of earth as focus.

  

Calculation:

As the center of the earth is at origin and it acts as focus of the parabolic path.

The general equation of the parabolic path is given as ${\left(y-k\right)}^2=4p\left(x-h\right)$ .

As the distance of the focus from the parabola is $4100\text{mi}$ . So, the value of $p$ is $-4100$ .

As the vertex of parabola lies on $x$ -axis and is at a distance of $4100\text{mi}$ from origin. So, the vertex of parabola is at $\left(4100,0\right)$ .

Substitute $4100$ for $h$ , $0$ for $k$ and $-4100$ for $p$ in the general equation.

  $\begin{array}{c}{\left(y-k\right)}^2=4p\left(x-h\right)\\ {\left(y-0\right)}^2=4\left(-4100\right)\left(x-4100\right)\\ y^2=-16400\left(x-4100\right)\end{array}$

Therefore, the equation of the parabolic path is $y^2=-16400\left(x-4100\right)$ .