Chapter 4.2, Problem 55AYU

(a)

To determine

To find: the acceleration due to gravity at sea level

Answer to Problem 55AYU

The acceleration due to gravity at sea level is $9.8208\text{m}/{\sec }^2$ .

Explanation of Solution

Given:

  $g\left(h\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+h\right)}^2}$

  $6.374\times {10}^6$ is the radius of Earth in meters.

Calculation:

The acceleration due to gravity at sea level is obtained when h = 0.

Substitute 0 for h .

  $\begin{array}{c}g\left(0\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+0\right)}^2}\\ =\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6\right)}^2}\\ =\frac{3.99\times {10}^{14}}{40.627876\times {10}^{12}}\\ =\frac{3.99}{40.627876}\times {10}^2\\ =9.8208\end{array}$

Therefore, the acceleration due to gravity at sea level is $9.8208\text{m}/{\sec }^2$ .

Conclusion:

Therefore, the acceleration due to gravity at sea level is $9.8208\text{m}/{\sec }^2$ .

(b)

To determine

To find: the acceleration due to gravity at the top of the Tower

Answer to Problem 55AYU

The acceleration due to gravity at the top of is $9.8195\text{m}/{\sec }^2$

Explanation of Solution

Given:

Tower height =443 meters

Calculation:

The acceleration due to gravity at the top of is obtained when

  $\text{h}=443\text{meters}$ .

Substitute 443 for h

  $\begin{array}{c}g\left(443\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+443\right)}^2}\\ =\frac{3.99\times {10}^{14}}{{\left(6374443\right)}^2}\\ =9.8195\end{array}$

So, the acceleration due to gravity at the top of is $9.8195\text{m}/{\sec }^2$

Conclusion:

The acceleration due to gravity at the top of is $9.8195\text{m}/{\sec }^2$

(c)

To determine

To find: the acceleration due to gravity on the peak of Mount Everest

Answer to Problem 55AYU

The acceleration due to gravity on the peak is $9.7936\text{m}/{\sec }^2$ .

Explanation of Solution

Given:

The peak of Mount Everest is 8848 meters

Calculation:

The acceleration due to gravity on the is obtained when

h = 8848meters.

Substitute 8848 for h

  $\begin{array}{c}g\left(8848\right)=\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+8848\right)}^2}\\ =\frac{3.99\times {10}^{14}}{{\left(6382848\right)}^2}\\ =9.7936\end{array}$

Therefore, the acceleration due to gravity on the peak is $9.7936\text{m}/{\sec }^2$ .

Conclusion:

Therefore, the acceleration due to gravity on the peak is $9.7936\text{m}/{\sec }^2$ .

(d)

To determine

To find: the horizontal asymptote of $g\left(h\right)$ .

Answer to Problem 55AYU

The horizontal asymptote for the rational function $g\left(h\right)$ is the h -axis.

Explanation of Solution

Calculation:

The given rational function $g\left(h\right)$ is in its lowest terms. So, the function is proper.

If a rational function R ( x ) is proper, then the x-axis is a horizontal asymptote of its graph.

So, the horizontal asymptote for the rational function $g\left(h\right)$ is the h -axis.

Conclusion:

The horizontal asymptote for the rational function $g\left(h\right)$ is the h -axis.

(e)

To determine

To solve and interpret: $g\left(h\right)=0$ .

Answer to Problem 55AYU

The graph of the rational function $g\left(h\right)$ has no x-intercept.

Explanation of Solution

Calculation:

If $g\left(h\right)=0$ , then $\frac{3.99\times {10}^{14}}{{\left(6.374\times {10}^6+h\right)}^2}=0$ .

If the value of the above fraction has to be zero, then the only possibility is that the numerator has to be zero. Since the numerator is a constant, no solution exists for $g\left(h\right)=0$ .

Therefore, the graph of the rational function $g\left(h\right)$ has no x-intercept.

Conclusion:

Therefore, the graph of the rational function $g\left(h\right)$ has no x-intercept.