Classical Mechanics
Classical Mechanics
5th Edition
ISBN: 9781891389221
Author: John R. Taylor
Publisher: University Science Books
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Chapter 3, Problem 3.13P
To determine

To show the rocket’s height as a function of t as y(t)=vext12gt2mvexkln(mm0), and to estimate the space shuttle’s height after two minutes.

Expert Solution & Answer
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Answer to Problem 3.13P

The rocket’s height as a function of t is shown as y(t)=vext12gt2mvexkln(mm0)_, and the space shuttle’s height after two minutes is approximately 4.0×104m_.

Explanation of Solution

From Problem 11, the equation of motion of rocket is given by

    mv˙=m˙vextmg

Here, m is the mass, v˙ is the acceleration, m˙ is the variable mass, vext is the exhaust speed, and g is the acceleration due to gravity.

Rearrange the above equation.

    mv˙+mg=m˙vextdvdt+g=m˙mvext

Integrate the above equation.

    v+gt=m˙mvextdt

Since f(x)f(x)dx=lnf(x)+C, the above equation is reduced as

    v+gt=vext(ln(m)m0mC)v+gt=vext(ln(m)ln(m0))v+gt=vexln(mm0)        (I)

From problem 3.11, the value of m is obtained as m0kt.

Use the above equation in equation (I).

    v+gt=vexln(m0ktm0)v=vexln(m0ktm0)gt        (II)

Use velocity v as dydt in the above equation.

    dydt=vexln(m0ktm0)gtdy=(vexln(m0ktm0)gt)dt

Integrate the above equation, dy from 0to y(t), and dt from 0to t.

    0y(t)dy=0t(vexln(m0ktm0)gt)dty(t)=vex0t(ln(m0ktm0))dt0tgtdt=vex0t(ln(1ktm0)dt)12gt2=vex[(1ktm0)ln(1ktm0)(1ktm0)]0t(1km0)12gt2        (III)

Since ln(1ax)dx=(1ax)ln(1ax)(1ax)a, where a=km0.

Use the above condition in equation (III).

    y(t)=m0vexk[(1ktm0)ln(1ktm0)(1ktm0)]0t12gt2=m0vexk[{(1ktm0)ln(1ktm0)(1ktm0)}(1)]12gt2=m0vexk[{(1ktm0)ln(1ktm0)(1ktm0)}+1]12gt2        (IV)

Rearrange the equation m=m0kt, to substitute for mm0

    m=m0ktmm0=1ktm0        (V)

Use equation (V) in (IV).

    y(t)=m0vexk[mm0ln(mm0)(mm0)+1]12gt2=m0vexk[mm0ln(m0m)+1(mm0)]12gt2=m0vexk[mm0ln(m0m)+(m0mm0)]12gt2        (VI)

Rearrange the equation m=m0kt, for kt.

    m=m0ktkt=m0m

Use the above equation in equation (VI).

    y(t)=m0vexk[mm0ln(m0m)+(ktm0)]12gt2=vextmvexkln(m0m)12gt2=vext12gt2mvexkln(m0m)        (VII)

Substitute m0mt for k in the above equation.

    y(t)=vext12gt2mvex(m0mt)ln(m0m)=vext12gt2mvext(m0m)ln(m0m)        (VIII)

Conclusion:

Substitute 2min  for t, 3000m/s for vex, 9.8m/s2 for g, 2×106kg for m0, and 1×106kg for m in equation (VIII), to find y(t).

y=(3000m/s)(2min ×60s1min)12(9.8m/s2)(2min ×60s1min)2(1×106kg)(3000m/s)(2min ×60s1min)((2×106kg)(1×106kg))ln(2×106kg1×106kg)=(3000m/s)(120s)12(9.8m/s2)(120s)2(1×106kg)(3000m/s)(120s)((2×106kg)(1×106kg))ln(2×106kg1×106kg)=4.0×104m

Therefore, The rocket’s height as a function of t is shown as y(t)=vext12gt2mvexkln(mm0)_, and the space shuttle’s height after two minutes is approximately 4.0×104m_.

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