Problem Statement
A rocket ejects mass at rate $\mu = 50\,\text{kg/s}$ with exhaust speed $u = 3000\,\text{m/s}$ relative to the rocket. Find the thrust force.
Given Information
- $\mu = 50\,\text{kg/s}$
- $u = 3000\,\text{m/s}$
Physical Concepts & Formulas
The Tsiolkovsky rocket equation describes the motion of a rocket as it expels propellant. As the rocket ejects mass at exhaust velocity $u$ relative to the rocket, the thrust force is $F = -u\,dm/dt$. The key insight is that Newton’s second law must account for both the changing velocity and the changing mass.
- $M\,dv = -u\,dM$ — rocket equation (M is current mass)
- $\Delta v = u\ln(M_0/M_f)$ — Tsiolkovsky: velocity gain from exhaust
- $F_{\text{thrust}} = u\,|\dot M|$ — thrust force
Step-by-Step Solution
Step 1 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 2 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Step 3 — Verify the result: Check units, limiting cases, and order of magnitude to confirm the answer is physically reasonable.
Worked Calculation
$$m(t)v(t) = [m(t)-\mu\,dt][v+dv] + (\mu\,dt)(v-u)$$
$$m\,dv = \mu u\,dt$$
$$F = \frac{dp}{dt} = \mu u$$
Answer
$$\boxed{F_{\text{thrust}} = \mu u = 1.5\times10^5\,\text{N}}$$
Physical Interpretation
Rocket propulsion is Newton’s third law in action: ejecting mass backward imparts forward momentum to the rocket. The logarithmic dependence on mass ratio means that a rocket must carry enormous amounts of fuel to achieve modest velocity gains.
Leave a Reply