Chapter 11.1, Problem 11.29P

The acceleration due to gravity at an altitude y above the surface of the earth be expressed as
   $a=\frac{-32.2}{{\left[1+\left(y/20.9\times {10}^6\right)\right]}^2}$ where a and y are expressed in ft/s² and feet, respectively. Using this expression, compute the height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is (a) 1800 ft/s, (b) 3000ft/s, (c) 36,700 ft/s.

  

To determine

To calculate:

The height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is 1800 ft/s.

Answer to Problem 11.29P

If the initial velocity is $1800 \text{ft}/\text{s}$, the maximum height reached is $50,431.96 \text{ft}$.

Explanation of Solution

Given information:

The acceleration due to gravity at an altitude $y$ above the surface of the earth can be expressed as $a=\frac{-32.2}{{\left[1+\left(\frac{y}{20.9}\times {10}^6\right)\right]}^2}$ where $a$ and $y$ are expressed in $\text{ft}/s^2$ and feet, respectively.

Concept used:

Substitute $a=v\frac{dv}{dy}$ to the equation $a=\frac{-32.2}{{\left[1+\left(\frac{y}{20.9}\times {10}^6\right)\right]}^2}$ and integrate to find a relation between displacement and velocity.

Calculation:

$a=\frac{-32.2}{{\left[1+\left(\frac{y}{20.9}\times {10}^6\right)\right]}^2}$

$v\frac{dv}{dy}=\frac{-32.2}{{\left[1+\left(\frac{y}{20.9}\times {10}^6\right)\right]}^2}$

$\int v dv=\int \frac{-32.2}{{\left[1+\left(\frac{y}{20.9}\times {10}^6\right)\right]}^2} dy$

$\frac{v^2}{2}+k=672.98\times {10}^6{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^{-1}$; Here $k$ is a constant.

Let $v=v_0$ when $y=0$;

$\frac{v_0^2}{2}+k=672.98\times {10}^6{\left[1+0\right]}^{-1}$

$k=-\frac{v_0^2}{2}+672.98\times {10}^6$

Therefore, the equation can be written as;

$\frac{v^2}{2}-\frac{v_0^2}{2}+672.98\times {10}^6=672.98\times {10}^6{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^{-1}$; where $v_0$ is the initial velocity.

Calculate the maximum height the projectile will reach; when this happens $v=0$.

$\frac{v^2}{2}-\frac{v_0^2}{2}+672.98\times {10}^6=672.98\times {10}^6{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^{-1}$

$-\frac{v_0^2}{2}+672.98\times {10}^6=672.98\times {10}^6{\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^{-1}$

${\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^{-1}=\left[1-\left(\frac{v_0^2}{1345.96\times {10}^6}\right)\right]\leftarrow \left(1\right)$

When $v_0=1800 \text{ft}/{\text{s}}^2$; using equation $\left(1\right)$;

${\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^{-1}=\left[1-\left(\frac{{1800}^2}{1345.96\times {10}^6}\right)\right]$

$y=50,431.96$ ft

Conclusion:

If the initial velocity is $1800 \text{ft}/\text{s}$, the maximum height reached is $50,431.96 \text{ft}$.

If the initial velocity is $3000 \text{ft}/\text{s}$, the maximum height reached is $140,692.32 \text{ft}$.

To determine

To calculate:

The height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is 3000 ft/s.

Explanation of Solution

Calculation:

When $v_0=3000 \text{ft}/{\text{s}}^2$; using equation $\left(1\right)$ in sub part (i);

${\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^{-1}=\left[1-\left(\frac{{3000}^2}{1345.96\times {10}^6}\right)\right]$

$y=140,692.32$ ft

Conclusion:

If the initial velocity is $3000 \text{ft}/\text{s}$, the maximum height reached is $140,692.32 \text{ft}$.

To determine

To calculate:

The height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is 36700 ft/s.

Answer to Problem 11.29P

When projected at $36,700 \text{ft}/\text{s}$, the projectile will escape the gravitational pull of the earth.

Explanation of Solution

Calculation:

Consider equation $\left(1\right)$ in sub part (i);

${\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]}^{-1}=\left[1-\left(\frac{v_0^2}{1345.96\times {10}^6}\right)\right]\leftarrow \left(1\right)$

Rearrange the equation;

$\left[1+\left(\frac{y}{20.9\times {10}^6}\right)\right]=\frac{1}{\left[1-\left(\frac{v_0^2}{1345.96\times {10}^6}\right)\right]}$

When the right hand side denominator is zero, $y$ will reach an infinite value. Find $v_0$ such that the left hand side numerator is zero.

$1-\left(\frac{v_0^2}{1345.96\times {10}^6}\right)=0$

$v_0=\sqrt{1345.96\times {10}^6}\approx 36687.33 \text{ft}/\text{s}$

Therefore, when the projectile has an initial velocity greater than or equal to $36687.33 \text{ft}/\text{s}$, the projectile will escape the gravitational pull of the earth.

Hence, when projected at $36,700 \text{ft}/\text{s}$, the projectile will escape the gravitational pull of the earth.