Chapter 1, Problem 12P

In our example of the free-falling parachutist, weassumed that the acceleration due to gravity was a constant value. Although this is a decent approximation when we are examining falling objects near the surface of the earth, the gravitational force decreases as we move above sea level. A more general representation based on Newton's inverse square law of gravitational attraction can be written as

$g\left(x\right)=g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}$

where $g\left(x\right)=$ gravitational acceleration at altitude x (in m) measured upward from the earth's surface $\left({\text{m/s}}^{\text{2}}\right),g\left(0\right)=$ gravitational acceleration at the earth's surface $\left(\cong 9.81{\text{ m/s}}^2\right)$, and $R=$ the earth's radius $\left(\cong 6.37\times {10}^6\text{ m}\right)$.

(a) In a fashion similar to the derivation of Eq. (1.9) use a force balance to derive a differential equation for velocity as a function of time that utilizes this more complete representation of gravitation. However, for this derivation, assume that upward velocity is positive.

(b) For the case where drag is negligible, use the chain rule to express the differential equation as a function of altitude rather than time. Recall that the chain rule is

$\frac{dv}{dt}=\frac{dv}{dx}\frac{dx}{dt}$

(c) Use calculus to obtain the closed form solution where $v=v_0\text{ at }x=0$.

(d) Use Euler's method to obtain a numerical solution from $x=0$ to 100,000 m using a step of 10,000 m where the initial velocity is 1500 m/s upward. Compare your result with the analytical solution.

To determine

The differential equation for velocity v if the gravitational force is not constant and is given by the function $g\left(x\right)=g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}$.

Answer to Problem 12P

Solution:

The differential equation for velocity v is $\frac{dv}{dt}=-g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}+\frac{C_{d}v\left|v\right|}{m}$.

Explanation of Solution

Given Information:

The function $g\left(x\right)=g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}$.

Where, $g\left(x\right)$ is the gravitational acceleration at an altitude x, $g\left(0\right)$ is the acceleration due to gravity at the surface of earth which is $9.81{\text{ m/sec}}^2$, R is the earth’s radius which is $6.37\times {10}^6\text{ m}$.

Assume the upward velocity is positive. Therefore, the force balance is given as,

$F=-F_{\text{gravity}}+F_{\text{drag}}$

Here,

$\begin{array}{c}F=ma\\ =m\frac{dv}{dt}\end{array}$

And,

$\begin{array}{c}{F}_{\text{gravity}}=mg\left(x\right)\\ =mg\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}\end{array}$

And,

$F_{\text{drag}}=C_{d}v\left|v\right|$

Thus, the force balance is,

$m\frac{dv}{dt}=-mg\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}+C_{d}v\left|v\right|$

Divide the both sides of the above equation by m,

$\begin{array}{c}\frac{dv}{dt}=-\frac{mg\left(0\right)}{m}\frac{R^2}{{\left(R+x\right)}^2}+\frac{C_{d}v\left|v\right|}{m}\\ =-g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}+\frac{C_{d}v\left|v\right|}{m}\end{array}$

Hence, the differential equation for the velocity is, $\frac{dv}{dt}=-g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}+\frac{C_{d}v\left|v\right|}{m}$.

To determine

The differential equation of for velocity as a function of altitude if the differential equation for velocity as a function of time is, $\frac{dv}{dt}=-g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}+\frac{C_{d}v\left|v\right|}{m}$.

Answer to Problem 12P

Solution:

The differential equation of $\frac{dv}{dt}$ for the special case of a sphere is $\frac{dv}{dt}=g\left(1-\frac{\rho }{{\rho }_s}\right)-\frac{3\rho C_{d}v\left|v\right|}{4{\rho }_{s}d}$.

Explanation of Solution

Given Information:

The differential equation $\frac{dv}{dt}=-g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}+\frac{C_{d}v\left|v\right|}{m}$.

The drag force is negligible.

And, the chain rule is given as,

$\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}$

Consider the chain rule,

$\frac{dv}{dt}=\frac{dv}{dx}\cdot \frac{dx}{dt}$

Here, $\frac{dx}{dt}=v$ is the velocity. Therefore,

$\frac{dv}{dt}=\frac{dv}{dx}\cdot v$

Now, consider the equation,

$\frac{dv}{dt}=-g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}+\frac{C_{d}v\left|v\right|}{m}$

Since, drag force is negligible. Therefore,

$\begin{array}{c}\frac{dv}{dt}=-g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}+0\\ =-g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}\end{array}$

Thus, from the chain rule,

$\begin{array}{c}\frac{dv}{dt}=\frac{dv}{dx}\cdot v\\ -g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}=\frac{dv}{dx}\cdot v\\ \frac{dv}{dx}=-\frac{g\left(0\right)}{v}\cdot \frac{R^2}{{\left(R+x\right)}^2}\end{array}$

Hence, differential equation of velocity as a function of altitude x is $\frac{dv}{dx}=-\frac{g\left(0\right)}{v}\frac{R^2}{{\left(R+x\right)}^2}$.

To determine

To calculate: The solution for the velocity by the use of calculus if the differential equation of velocity as a function of altitude x is $\frac{dv}{dx}=-\frac{g\left(0\right)}{v}\frac{R^2}{{\left(R+x\right)}^2}$.

Answer to Problem 12P

Solution:

The solution for the velocity is, $v=\pm \sqrt{-2g\left(0\right)\left(\frac{Rx}{R+x}\right)+v_0{}^2}$.

Explanation of Solution

Given Information:

The differential equation $\frac{dv}{dx}=-\frac{g\left(0\right)}{v}\frac{R^2}{{\left(R+x\right)}^2}$.

The initial condition, at $x=0$, $v=v_0$.

Formula used:

Integration formula,

${\displaystyle \int x^{n}dx}=\frac{x^{n+1}}{n+1}+C$

Calculation:

Consider the differential equation,

$\frac{dv}{dx}=-\frac{g\left(0\right)}{v}\frac{R^2}{{\left(R+x\right)}^2}$

Separate the variable as below,

$vdv=-g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}dx$

Integrate both the sides of the above equation,

$\begin{array}{c}{\displaystyle \int vdv}={\displaystyle \int -g\left(0\right)\frac{R^2}{{\left(R+x\right)}^2}dx}\\ \frac{v^2}{2}=-g\left(0\right)R^2{\displaystyle \int \frac{dx}{{\left(R+x\right)}^2}}\\ \frac{v^2}{2}=-g\left(0\right)R^2\left(-\frac{1}{\left(R+x\right)}\right)+C\\ \frac{v^2}{2}=g\left(0\right)\left(\frac{R^2}{\left(R+x\right)}\right)+C\end{array}$

Now, for $x=0$, $v=v_0$. Therefore, the above equation is,

$\begin{array}{c}\frac{v_0{}^2}{2}=g\left(0\right)\left(\frac{R^2}{\left(R+0\right)}\right)+C\\ \frac{v_0{}^2}{2}=g\left(0\right)R+C\\ C=\frac{v_0{}^2}{2}-g\left(0\right)R\end{array}$

Substitute the value of C in the equation $\frac{v^2}{2}=g\left(0\right)\left(\frac{R^2}{\left(R+x\right)}\right)+C$ as below,

$\begin{array}{l}\frac{v^2}{2}=g\left(0\right)\left(\frac{R^2}{\left(R+x\right)}\right)+\frac{v_0{}^2}{2}-g\left(0\right)R\\ \frac{v^2}{2}=g\left(0\right)\left(\frac{R^2}{\left(R+x\right)}-R\right)+\frac{v_0{}^2}{2}\\ v^2=2g\left(0\right)\left(\frac{R^2-R^2-Rx}{R+x}\right)+v_0{}^2\\ v=\pm \sqrt{-2g\left(0\right)\left(\frac{Rx}{R+x}\right)+v_0{}^2}\end{array}$

Hence, the solution for the velocity is $v=\pm \sqrt{-2g\left(0\right)\left(\frac{Rx}{R+x}\right)+v_0{}^2}$. Here, the positive sign represent upward direction and negative sign indicates downward direction.

To determine

To calculate: The velocity from $x=0\text{ to }100,000\text{ m}$ by the Euler’s method with the step size $\Delta x=10,000\text{ m}$ and the initial condition $v\left(0\right)=1500\text{ m/sec}$ if the differential equation of velocity as a function of altitude x is $\frac{dv}{dx}=-\frac{g\left(0\right)}{v}\frac{R^2}{{\left(R+x\right)}^2}$. Also, compare the result with the analytical solution.

Answer to Problem 12P

Solution:

The velocity from $x=0\text{ to }100,000\text{ m}$ by the Euler’s method is,

x	v- Euler	v- analytical
0	1500	1500
10000	1434.518	1433.216
20000	1366.261	1363.388
30000	1294.818	1290.023
40000	1219.669	1212.476
50000	1140.138	1129.885
60000	1055.324	1041.05
70000	963.9789	944.2077
80000	864.2883	836.5811
90000	753.4434	713.3028
100000	626.6846	564.2026

The velocity by the Euler’s method is approximately same as the analytical solution.

Explanation of Solution

Given Information:

The differential equation $\frac{dv}{dx}=-\frac{g\left(0\right)}{v}\frac{R^2}{{\left(R+x\right)}^2}$.

Where, $g\left(x\right)$ is the gravitational acceleration at an altitude x, $g\left(0\right)$ is the acceleration due to gravity at the surface of earth which is $9.81{\text{ m/sec}}^2$, R is the earth’s radius which is $6.37\times {10}^6\text{ m}$.

The initial condition $v\left(0\right)=1500\text{ m/sec}$.

Formula used:

Euler’s method for $\frac{dy}{dx}=f\left(x,y\right)$ is,

$y\left(x+h\right)=y\left(x\right)+hf\left(x,y\right)$

Where, h is the step size.

Calculation:

Consider the differential equation,

$\frac{dv}{dx}=-\frac{g\left(0\right)}{v}\frac{R^2}{{\left(R+x\right)}^2}$

Substitute $g\left(0\right)=9.81{\text{ m/sec}}^2$ and $R=6.37\times {10}^6\text{ m}$ in the above equation and solve as below,

$\begin{array}{c}\frac{dv}{dx}=-\frac{9.81}{v}\frac{{\left(6.37\times {10}^6\right)}^2}{{\left(6.37\times {10}^6+x\right)}^2}\\ =-\frac{398.56\times {10}^{12}}{{\left(6.37\times {10}^6+x\right)}^{2}v}\end{array}$

The iteration formula for Euler’s method with step size $\Delta x=10,000\text{ m}$ for the above equation is,

$v\left(x+10,000\right)=v\left(x\right)+10,000\left(-\frac{398.56\times {10}^{12}}{{\left(6.37\times {10}^6+x\right)}^{2}v}\right)$

From part (c), the analytical solution for the velocity is,

$v=\pm \sqrt{-2g\left(0\right)\left(\frac{Rx}{R+x}\right)+v_0{}^2}$

Substitute $g\left(0\right)=9.81{\text{ m/sec}}^2$, and $R=6.37\times {10}^6\text{ m}$ in the above equation and solve as below,

$v=\pm \sqrt{-2\times 9.81\times \left(\frac{6.37\times {10}^{6}x}{6.37\times {10}^6+x}\right)+{\left(1500\right)}^2}$

Use excel to find all the iteration with step size $\Delta x=10,000\text{ m}$ from $x=0\text{ to }100000\text{ m}$ as below,

Step 1: Name the column A as x and go to column A2 and put 0 then go to column A3and write the formula as,

=A2+10000

Then, Press enter and drag the column up to the $x=100000\text{ m}$.

Step 2: Now name the column B as v-Euler and go to column B2 and write 1500 and then go to the column B3 and write the formula as,

=B2+10000*(-398.56*10^12/((6.37*10^6+A2)^2*B2))

Step 3: Press enter and drag the column up to the $x=100000\text{ m}$.

Step 4. Now name the column C as v-analytical and go to column C2 and write 1500 and then go to the column C3 and write the formula as,

=(-2*9.81*((6.37*10^6*A3)/(6.37*10^6+A3))+1500^2)^(1/2)

Step 5. Press enter and drag the column up to the $x=100000\text{ m}$.

Thus, all the iterations are as shown below,

x	v- Euler	v- analytical
0	1500	1500
10000	1434.518	1433.216
20000	1366.261	1363.388
30000	1294.818	1290.023
40000	1219.669	1212.476
50000	1140.138	1129.885
60000	1055.324	1041.05
70000	963.9789	944.2077
80000	864.2883	836.5811
90000	753.4434	713.3028
100000	626.6846	564.2026

To draw the graph of the above results, follow the steps as given below,

Step 6:Select the column A and column B. Then, go to the Insert and select the scatter (X, Y) from the chart.

Step 7: Select the column A and column C. Then, go to the Insert and select the scatter (X, Y) from the chart.

Step 8: Select one of the graphs and paste it on another graph to Merge the graphs.

The graph obtained is,

From the graph, it is observed that both the graphs of velocity by analytical method and by Euler’s method is approximately same.