The data for a single stage rocket are as follows:

Launch mass: $m_{o}$ = 68,000 kg
Mass ratio: n = 15
Specific impulse: $I_{s p}$ = 390 s
Thrust: T = 933.91 kN

It is launched into a vertical trajectory, like a sounding rocket. Neglecting drag and assuming that the gravitational acceleration is constant at its sea-level value $g_{o}$ = 9.81 m/s², calculate
(a) The time until burnout.
(b) The burnout altitude.
(c) The burnout velocity.
(d) The maximum altitude reached.

Question

The data for a single stage rocket are as follows:

Launch mass: [latex]m_{o}[/latex] = 68,000 kg
Mass ratio: n = 15
Specific impulse: [latex]I_{s p}[/latex] = 390 s
Thrust: T = 933.91 kN

It is launched into a vertical trajectory, like a sounding rocket. Neglecting drag and assuming that the gravitational acceleration is constant at its sea-level value [latex]g_{o}[/latex] = 9.81 m/s², calculate
(a) The time until burnout.
(b) The burnout altitude.
(c) The burnout velocity.
(d) The maximum altitude reached.

Accepted Answer

(a) From Example 11.1(b) the burnout time [latex]t_{b o}[/latex] is

[latex]t_{b o}=\frac{m_{o}-m_{f}}{\dot{m}_{e}}[/latex] (a)

The burnout mass [latex]m_{f}[/latex] is obtained from Equation 11.24,

[latex]m_{f}=\frac{m_{o}}{n}=\frac{68,000}{15}=4533.3 kg[/latex] (b)

The propellant mass flow rate [latex]\dot{m}_{e}[/latex] is given by Equation 11.20,

[latex]I_{s p}=\frac{T}{\dot{m}_{e} g_{o}}[/latex] (11.20)

[latex]\dot{m}_{e}=\frac{T}{I_{s p} g_{o}}=\frac{933,913}{390 \cdot 9.81}=244.10 kg / s[/latex] (c)

Substituting (b), (c) and [latex]m_{o}=68,000 kg[/latex] into (a) yields the burnout time,

[latex]t_{b o}=\frac{68,000-4533.3}{244.10}=260.0 s[/latex]

(b) The burnout altitude is given by Example 11.1(e)

[latex]h_{b o}=\frac{c}{\dot{m}_{e}}\left\lgroup m_{f} \ln \frac{m_{f}}{m_{o}}+m_{o}-m_{f} ight group -\frac{1}{2}\left\lgroup \frac{m_{o}-m_{f}}{\dot{m}_{e}} ight group ^{2} g_{o}[/latex] (d)

The exhaust velocity c is found in Equation 11.21,

[latex]c=I_{s p} g_{o}=390 \cdot 9.81=3825.9 m / s[/latex] (e)

Substituting (b), (c) and (e), along with [latex]m_{o}=68,000 kg ext { and } g_{o}=9.81 m / s ^{2}[/latex] into (d), we get

[latex]h_{b o}=\frac{3825.9}{244.1}\left\lgroup 4533.3 \ln \frac{4533.3}{68,000}+68,000-4533.3 ight group -\frac{1}{2}\left\lgroup \frac{68,000-4533.3}{244.1} ight group ^{2} \cdot 9.81[/latex]
[latex]h_{b o}=470.74 km[/latex]

(c) From Example 11.1(f) we find

[latex]v_{b o}=c \ln \frac{m_{o}}{m_{f}}-\frac{g_{o}}{\dot{m}_{e}}\left(m_{o}-m_{f} ight)=3825.9 \ln \frac{68,000}{4533.3}-\frac{9.81}{244.1}(68,000-4533.3)[/latex]
[latex]v_{b o}=7.810 km / s[/latex]

(d) To find [latex]h_{\max }[/latex], where the speed of the rocket falls to zero, we use Example 11.1(i),

[latex]h_{\max }=\frac{1}{2} \frac{c^{2}}{g_{o}} \ln ^{2} n-\frac{c m_{o}}{\dot{m}_{e}} \frac{n \ln n-(n-1)}{n}=\frac{1}{2} \frac{3825.9^{2}}{9.81} \ln ^{2} 15-\frac{3825.9 \cdot 68,000}{244.1} \frac{15 \ln 15-(15-1)}{15}[/latex]
[latex]h_{\max }=3579.7 km[/latex]

Notice that the rocket coasts to a height more than seven times the burnout altitude.

Question 11.2: The data for a single stage rocket are as follows: Launch ma...

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