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A declared rocket check: initial mass $3 k g$ , loss rate $1 k g s^{- 1}$ and relative exhaust speed $2 m s^{- 1}$ give $v = u_{e} \ln (m_{0} / m) n$ and show that $(m v)^{'} = F$ fails

Physical scope

A declared-data check of the open-body material accounting: a body of initial mass $m_{0} = 3 k g$ losing mass at $r_{0} = 1 k g s^{- 1}$ has $m (t) = m_{0} - r_{0} t$ . With no external force and ejecta of relative speed $u_{e} = 2 m s^{- 1}$ opposite a fixed unit direction $n$ , its velocity from rest is $v (t) = u_{e} \ln (m_{0} / m (t)) n$ . It satisfies the material balance identities and violates the naive closed-system law $(m v)^{'} = F$ ; the example verifies the identities, the thrust $2 N$ and the failure of the naive law on declared model data. It reports no measurement, no engine data and no empirical comparison.

Example

Declared model data. Adopt the open-body material accounting of Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (iii) with, on the open interval $I = (- \frac{1}{2} s, \frac{5}{2} s)$ containing the initial time $0$ , set $m_{0} : = 3 k g$ , $r_{0} : = 1 k g s^{- 1}$ , and $m (t) : = m_{0} - r_{0} t, F \equiv 0, r_{i n} \equiv 0, r_{o u t} \equiv r_{0}, w_{o u t} (t) : = v (t) - u_{e} n,$ with a fixed unit vector $n$ , $u_{e} : = 2 m s^{- 1}$ , and initial velocity $v (0) = 0$ . The ejecta therefore leave with the body's velocity reduced by the constant exhaust speed $u_{e}$ along $n$ . Set $q : = t / (1 s)$ , the dimensionless numerical time, so $m (t) = (3 - q) k g$ . Primes denote differentiation with respect to physical frame time $t$ , read in seconds.

Assertions to check. The balance identities hold with $v (t) = u_{e} \ln \frac{m_{0}}{m (t)} n = 2 \ln \frac{3}{3 - q} n m s^{- 1} (t \in I),$ the thrust is $r_{o u t} u_{e} = 2 N$ along $n$ , and the naive closed-system law $(m v)^{'} = F$ fails: it would force $v \equiv 0$ on $I$ .

Facts & Assumptions

Given: The declared maps $m, F, r_{i n}, r_{o u t}, w_{o u t}$ and the vocabulary of Dimensions and SI unit conventions of the non-relativistic classical mechanics framework.

[L1]

Material accounting and derived law (Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (iii), Variable-mass momentum balance: from $m^{'} = r_{i n} - r_{o u t}$ and $(m v)^{'} = F + r_{i n} w_{i n} - r_{o u t} w_{o u t}$ to $m v^{'} = F + r_{i n} (w_{i n} - v) - r_{o u t} (w_{o u t} - v)$ , Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with $Δ K = - \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ , and the open-body and rocket laws clause 6): the model requires $m^{'} = r_{i n} - r_{o u t}$ and $(m v)^{'} = F + r_{i n} w_{i n} - r_{o u t} w_{o u t}$ , equivalently $m v^{'} = F + r_{i n} (w_{i n} - v) - r_{o u t} (w_{o u t} - v)$ .

[L2]

Calculus (The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ): for real-valued differentiable functions, $(\ln)^{'} = 1 / x$ on $(0, \infty)$ , the chain rule and product rule hold, and a curve with zero derivative on an interval is constant.

Verification

technique · direct

1.1givenL1L2algebra

Mass identity: $m^{'} = - r_{0} = r_{i n} - r_{o u t}$ on $I$ , in $k g s^{- 1}$ . Since $- 1 / 2 < q < 5 / 2$ , $m (t) = (3 - q) k g > 0$ , so the first balance identity and the positive-mass condition hold.

2.1step 1.1givenL2algebra

Velocity candidate: define $v (t) : = u_{e} \ln (m_{0} / m (t)) n$ . By [L2] and $m^{'} = - r_{0}$ , $\frac{d}{d t} \ln (m_{0} / m (t)) = r_{0} / m (t)$ , so $v^{'} (t) = u_{e} r_{0} n / m (t)$ , with acceleration units $m s^{- 2}$ , and $v (0) = u_{e} \ln 1 \cdot n = 0$ . The logarithm acts on the positive dimensionless ratio $m_{0} / m (t)$ .

3.1step 1.1step 2.1L1algebra

Strong form: $m v^{'} = r_{0} u_{e} n$ , while $F + r_{i n} (w_{i n} - v) - r_{o u t} (w_{o u t} - v) = - r_{0} ((v - u_{e} n) - v) = r_{0} u_{e} n$ . Thus the derived law holds with thrust $r_{0} u_{e} = 2 N$ along $n$ .

4.1step 1.1step 3.1L1L2algebra

Material identities: the product rule gives $(m v)^{'} = m^{'} v + m v^{'} = - r_{0} v + r_{0} u_{e} n = - r_{0} w_{o u t} = F + r_{i n} w_{i n} - r_{o u t} w_{o u t}$ . Along with step 1.1 this verifies both balance identities of [L1], with every term in this momentum-rate identity having force units.

5.1step 4.1L1L2algebra

Failure of the naive law: $(m v)^{'} = F \equiv 0$ would make $m v$ constant and, since $m (0) v (0) = 0$ and $m > 0$ , would force $v \equiv 0$ by [L2]. But the computed $v (t)$ is nonzero for $0 < t < 2 s$ ; at $t = 2 s$ , $(m v)^{'} = (2 - 2 \ln 3) n N \approx - 0.197 n N \neq F$ . Hence the naive law fails for this open body, as Consequences of the adopted impact and open-mass models: continuity of total momentum, the smooth-force exclusion, the restitution solution with $Δ K = - \frac{1 - e^{2}}{2} μ ∥ v_{2}^{-} - v_{1}^{-} ∥_{2}^{2}$ , and the open-body and rocket laws clause 6(b) records.

6.1step 5.1L1L2algebra∎

Tsiolkovsky value: at $t = 1 s$ , $q = 1$ , $m = 2 k g$ and $v = 2 \ln (3 / 2) n m s^{- 1} \approx 0.8109 n m s^{- 1}$ . The momentum is approximately $1.6219 n k g m s^{- 1}$ , whereas the naive law gives zero. These values come from the exact logarithm before rounding; all are declared model computations, with no measurement or empirical inference.

Remarks

Signs and conventions. The ejecta velocity $w_{o u t} = v - u_{e} n$ is expressed in the same inertial frame as $v$ , as Adopted impact and open-mass extensions of the force branch: the isolated reciprocal impulse model, the collinear restitution contact law and the open-body material accounting clause (iii) requires; $u_{e} > 0$ is the exhaust speed relative to the body, and the thrust $r_{o u t} (v - w_{o u t}) = r_{o u t} u_{e} n$ points along $+ n$ .
Units. $[m] = k g$ , $[r_{o u t}] = k g s^{- 1}$ , $[u_{e}] = m s^{- 1}$ , $[F] = N = k g m s^{- 2}$ (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework); $m v^{'}$ and $r_{o u t} u_{e}$ both have the dimension of force.

A declared rocket check: initial mass $3 k g$ , loss rate $1 k g s^{- 1}$ and relative exhaust speed $2 m s^{- 1}$ give $v = u_{e} \ln (m_{0} / m) n$ and show that $(m v)^{'} = F$ fails

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A declared rocket check: initial mass 3 kg3kg, loss rate 1 kg s−11kgs−1 and relative exhaust speed 2 m s−12ms−1 give v=ueln⁡(m0/m)nv=ue​ln(m0​/m)n and show that (mv)′=F(mv)′=F fails

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources

A declared rocket check: initial mass $3 k g$ , loss rate $1 k g s^{- 1}$ and relative exhaust speed $2 m s^{- 1}$ give $v = u_{e} \ln (m_{0} / m) n$ and show that $(m v)^{'} = F$ fails