Systems with Varying Mass: A Rocket — Question 5

Section 9-11 · Physics pages 1360–1361

Question 5 Hovering requires a declining mass flow An ideal rocket hovers at fixed height in uniform gravity. Initially m0 = 100 kg, of which 60 kg is dry mass. Relative exhaust speed is u = 1000 m/s and g = 9.8 m/s2. Neglect air effects. hover uq mg a = 0, q = mg/u Exact hover requires thrust to track the decreasing weight. (a) Derive the mass as a function of time needed to hover. (b) Find maximum hover time and compare initial and final propellant flow rates.
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Question 5 - Solution Hovering requires a declining mass flow hover uq mg a = 0, q = mg/u (a) Derivation and calculation Zero acceleration requires uq = mg, with q = −˙m: dm dt = −g um ⇒ ∫m(t) 100 dm m = −g ut. m(t) = 100e−gt/u = 100e−0.0098t kg . (b) Result and interpretation The model applies until the dry-mass limit is reached: 60 = 100e−gtmax/u ⇒tmax = 1000 9.8 ln 100 60 = 52.13 s . qi = 100(9.8)/1000 = 0.980 , qf = 60(9.8)/1000 = 0.588 kg/s . A constant flow would exceed the required thrust as fuel is consumed. Physical check The hovering trajectory ends at fuel exhaustion; the exponential cannot continue below dry mass.

Question 5

Hovering requires a declining mass flow

An ideal rocket hovers at fixed height in uniform gravity. Initially m0=100kgm_0=100\,\mathrm{kg}, of which 60kg60\,\mathrm{kg} is dry mass. Relative exhaust speed is u=1000m/su=1000\,\mathrm{m/s} and g=9.8m/s2g=9.8\,\mathrm{m/s^2}. Neglect air effects.

Diagram shown in the original SVG worksheet.

  1. Derive the mass as a function of time needed to hover.

  2. Find maximum hover time and compare initial and final propellant flow rates.

Question 5 - Solution

Hovering requires a declining mass flow

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Zero acceleration requires uq=mguq=mg, with q=−ṁq=-\dot m: dmdt=−gum⇒∫100m(t)dmm=−gut.\frac{dm}{dt}=-\frac gu m\Rightarrow\int_{100}^{m(t)}\frac{dm}{m}=-\frac gu t. m(t)=100e−gt/u=100e−0.0098tkg.\boxed{m(t)=100e^{-gt/u}=100e^{-0.0098t}\ \mathrm{kg}}.

(b) Result and interpretation

The model applies until the dry-mass limit is reached: 60=100e−gtmax/u⇒tmax=10009.8ln⁡10060=52.13s.60=100e^{-gt_{\max}/u}\Rightarrow\boxed{t_{\max}=\frac{1000}{9.8}\ln\frac{100}{60}=52.13\ \mathrm s}. qi=100(9.8)/1000=0.980,qf=60(9.8)/1000=0.588kg/s.\boxed{q_i=100(9.8)/1000=0.980},\qquad\boxed{q_f=60(9.8)/1000=0.588\ \mathrm{kg/s}}. A constant flow would exceed the required thrust as fuel is consumed.

Physical check

The hovering trajectory ends at fuel exhaustion; the exponential cannot continue below dry mass.