Question

In physics, it is established that the acceleration

due to gravity, g (in meters/sec2), at a height h meters above

sea level is given by

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

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

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

(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?

(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?

(d) Find the horizontal asymptote of $g\left(h\right)$.

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

Short answer

(a) $9.8m/s^2$

(b) $9.8m/s^2$

(c) $9.8m/s^2$

(d) $g\left(h\right)=0$is the horizontal asymptote

(e) $h=\pm \infty$implies $g\left(h\right)=0$is a horizontal asymptote.

Step-by-step solution

Part (a) Step 1. Given information

Acceleration due to gravity is given by $g\left(h\right)=\frac{3.99\times {10}^{14}}{{(6.374\times {10}^6+h)}^2}$where h is height from sea level.

Part (a) Step 2. Calculation

Since 'h' is the distance from sea level, therefore for finding acceleration due to gravity at sea level $h=0$.

Applying the formula:

$$\begin{gathered} g\left(h\right)=\frac{3.99\times {10}^{14}}{{(6.374\times {10}^6+h)}^2} \\ =\frac{3.99\times {10}^{14}}{{(6.374\times {10}^6+0)}^2} \\ =\frac{3.99\times {10}^{14}}{40.62\times {10}^{12}} \\ =9.8m/s^2 \end{gathered}$$

Part (b) Step 1. Given information

Acceleration due to gravity is given by $g\left(h\right)=\frac{3.99\times {10}^{14}}{{(6.374\times {10}^6+h)}^2}$where h is height from sea level.

Part (b) Step 2. Calculation

To find the acceleration due to gravity above will's tower i.e.

$$\begin{gathered} h=443m \\ Substitutingintheformula \\ g\left(h\right)=\frac{3.99\times {10}^{14}}{{(6.374\times {10}^6+h)}^2} \\ =\frac{3.99\times {10}^{14}}{{(6.374\times {10}^6+443)}^2} \\ =9.8m/s^2 \\ Sincehisverysmallvaluethefinalresultdoesnotchange. \end{gathered}$$

Part (c) Step 1. Given information

Acceleration due to gravity is given by $g\left(h\right)=\frac{3.99\times {10}^{14}}{{(6.374\times {10}^6+h)}^2}$where h is height from sea level.

Part (c) Step 2. Calculation

To find the acceleration due to gravity above Everest i.e.

$$\begin{gathered} h=8848m \\ Substitutingintheformula \\ g\left(h\right)=\frac{3.99\times {10}^{14}}{{(6.374\times {10}^6+h)}^2} \\ =\frac{3.99\times {10}^{14}}{{(6.374\times {10}^6+8848)}^2} \\ =9.8m/s^2 \\ Sincehisverysmallvaluethefinalresultdoesnotchange. \end{gathered}$$

Part (d) Step 1. Given information

Acceleration due to gravity is given by $g\left(h\right)=\frac{3.99\times {10}^{14}}{{(6.73\times {10}^6+h)}^2}$where h is height from sea level.

Part (d) Step 2. Calculation

To find horizontal asypmtote we put the values of $h\to \pm \infty$

therefore,

$$\begin{gathered} g\left(h\right)=\frac{3.99\times {10}^{14}}{{(6.73\times {10}^6+h)}^2} \\ =\frac{3.99\times {10}^{14}}{(6.73\times {10}^6\pm \infty )} \\ =0 \end{gathered}$$

Part (e) Step 1. Given information

Acceleration due to gravity is given by $g\left(h\right)=\frac{3.99\times {10}^{14}}{{(6.73\times {10}^6+h)}^2}$where h is height from sea level.

Part (e) Step 2. Calculation

To solve $g\left(h\right)=0$

therefore,

$$\begin{gathered} g\left(h\right)=\frac{3.99\times {10}^{14}}{{(6.73\times {10}^6+h)}^2} \\ 0=\frac{3.99\times {10}^{14}}{{(6.73\times {10}^6+h)}^2} \\ h=\pm \infty \end{gathered}$$

Which proves it to be a horizontal asymptote.