Chapter 4.4, Problem 57AYU

To determine

To find:

a.What is the acceleration due to gravity at sea level?

Answer to Problem 57AYU

a.$9.82\text{m}/{\text{sec}}^2$

Explanation of Solution

Given:

The acceleration due to gravity, $g(\text{in meters}{\text{/sec}}^2)$, at a height $h$ meters above sea level is given by $g\left(h\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+h\right)}^2}$ where $6.374\times {10}^6$ is the radius of Earth in meters.

a.Set $h=0\text{ and }g\left(0\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6\right)}^2}=9.82\text{m}/{\text{sec}}^2$

To determine

To find:

b.The Willis Tower in Chicago, Illinois, is 443 meters tall. What is the acceleration due to gravity at the top of the Willis Tower?

Answer to Problem 57AYU

b.$9.8195\text{ m}/{\text{sec}}^2$

Explanation of Solution

Given:

The acceleration due to gravity,$g(\text{in meters}{\text{/sec}}^2)$, at a height $h$ meters above sea level is given by $g\left(h\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+h\right)}^2}$ where $6.374\times {10}^6$ is the radius of Earth in meters.

b.Set $h=443m=0.000443\times {10}^6$

$g\left(h\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+0.000443\times {10}^6\right)}^2}=9.8195\text{m}/{\text{sec}}^2$

To determine

To find:

c.The peak of Mount Everest is 8848 meters above sea level. What is the acceleration due to gravity on the peak of Mount Everest?

Answer to Problem 57AYU

c.$9.7936\text{m}/{\text{sec}}^2$

Explanation of Solution

Given:

The acceleration due to gravity, $g(\text{in meters}{\text{/sec}}^2)$, at a height $h$ meters above sea level is given by $g\left(h\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+h\right)}^2}$ where $6.374\times {10}^6$ is the radius of Earth in meters.

c.Set $h=8848=0.008848\times {10}^6$

$g\left(h\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+0.008848\times {10}^6\right)}^2}=9.7936\text{m}/{\text{sec}}^2$

To determine

To find:

d.Find the end behavior of $g$. That is, find $\underset{h\to \infty }{\lim }g\left(h\right)$. What does the result suggest?

Answer to Problem 57AYU

d.0

Explanation of Solution

Given:

The acceleration due to gravity, $g(\text{in meters}{\text{/sec}}^2)$, at a height $h$ meters above sea level is given by $g\left(h\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+h\right)}^2}$ where $6.374\times {10}^6$ is the radius of Earth in meters.

d.$\underset{h\to \infty }{\lim }g\left(h\right)=\underset{h\to \infty }{\lim }\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+h\right)}^2}=0$

Therefore the result suggest, the acceleration due to gravity, $g\left({\text{in meters/sec}}^2\right)$, at a height $h\to \infty$ is zero.

To determine

To find:

e. Solve $g\left(h\right)= 0$. How do you interpret your answer?

Answer to Problem 57AYU

e.When $h\to \infty$

Explanation of Solution

Given:

The acceleration due to gravity,$g(\text{in meters}{\text{/sec}}^2)$, at a height $h$ meters above sea level is given by $g\left(h\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+h\right)}^2}$ where $6.374\times {10}^6$ is the radius of Earth in meters.

e.$\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+h\right)}^2}=0,g\left(h\right)=0$ is only possible when $h$ is so large, otherwise the result is not possible.