Systems with Varying Mass: A Rocket — Question 7

Section 9-11 · Physics pages 1364–1365

Question 7 Discarding empty structure helps only subsequent propulsion An ideal two-stage rocket in force-free space starts at 1000 kg. The first burn expels 400 kg, then 100 kg of empty structure is detached with no relative speed. The second burn expels 300 kg, leaving 200 kg. Both stages have u = 2000 m/s. 1000 600 500 200 burn discard burn Masses are in kg; only the solid-arrow burn intervals contribute rocket velocity gain. (a) Find total velocity gain. (b) Compare with retaining the empty structure through the second burn. Does jettison itself create an instantaneous speed increase?
Show solutionHide solution
Question 7 - Solution Discarding empty structure helps only subsequent propulsion 1000 600 500 200 burn discard burn (a) Derivation and calculation Sum the two powered increments using their own mass endpoints: ∆v1 = 2000 ln(1000/600), ∆v2 = 2000 ln(500/200). ∆v = 2000 ln(25/6) = 2854.23 m/s . (b) Result and interpretation Retaining the empty structure gives ∆vretain = 2000[ln(1000/600) + ln(600/300)] = 2407.95 m/s. ∆vadvantage = 446.29 m/s . For equal-velocity detachment, 600v = 500v + 100v: jettison creates no instantaneous speed change, but reduces the mass accelerated afterward. Physical check The unpowered mass ratio 600/500 must not be inserted as a rocket thrust increment.

Question 7

Discarding empty structure helps only subsequent propulsion

An ideal two-stage rocket in force-free space starts at 1000kg1000\,\mathrm{kg}. The first burn expels 400kg400\,\mathrm{kg}, then 100kg100\,\mathrm{kg} of empty structure is detached with no relative speed. The second burn expels 300kg300\,\mathrm{kg}, leaving 200kg200\,\mathrm{kg}. Both stages have u=2000m/su=2000\,\mathrm{m/s}.

Diagram shown in the original SVG worksheet.

  1. Find total velocity gain.

  2. Compare with retaining the empty structure through the second burn. Does jettison itself create an instantaneous speed increase?

Question 7 - Solution

Discarding empty structure helps only subsequent propulsion

Diagram shown in the original SVG worksheet.

(a) Derivation and calculation

Sum the two powered increments using their own mass endpoints: Δv1=2000ln⁡(1000/600),Δv2=2000ln⁡(500/200).\Delta v_1=2000\ln(1000/600),\qquad\Delta v_2=2000\ln(500/200). Δv=2000ln⁡(25/6)=2854.23m/s.\boxed{\Delta v=2000\ln(25/6)=2854.23\ \mathrm{m/s}}.

(b) Result and interpretation

Retaining the empty structure gives Δvretain=2000[ln⁡(1000/600)+ln⁡(600/300)]=2407.95m/s.\Delta v_{\mathrm{retain}}=2000[\ln(1000/600)+\ln(600/300)]=2407.95\ \mathrm{m/s}. Δvadvantage=446.29m/s.\boxed{\Delta v_{\mathrm{advantage}}=446.29\ \mathrm{m/s}}. For equal-velocity detachment, 600v=500v+100v600v=500v+100v: jettison creates no instantaneous speed change, but reduces the mass accelerated afterward.

Physical check

The unpowered mass ratio 600/500600/500 must not be inserted as a rocket thrust increment.