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5.33).Let M(t) be the mass of the rocket (including fuel) at time t, V (t) its velocityat time t, and ω the speed (relative to the rocket) with which fuel flows out of thenozzle of the rocket as it burns.We wish to establish the connection among these quantities.2885Differential CalculusUnder these assumptions, we can regard the rocket with fuel as a closed systemwhose momentum (quantity of motion) remains constant over time.At time t the momentum of the system is M(t)V (t).At time t + h the momentum of the rocket with the remaining fuel is M(t +h)V (f + h) and the momentum ΔI of the mass of fuel ejected over that time|ΔM| = |M(t + h) − M(t)| = −(M(t + h) − M(t)) lies between the boundsV (t) − ω |ΔM| < ΔI < V (t + h) − ω |ΔM|,that is, ΔI = (V (t) − ω)|ΔM| + α(h)|ΔM, and it follows from the continuity ofV (t) that α(h) → 0 as h → 0.Equating the momenta of the system at times t and t + h, we haveM(t)V (t) = M(t + h)V (t + h) + V (t) − ω |ΔM| + α(h)|ΔM|,or, after substituting |ΔM| = −(M(t + h) − M(t)) and simplifying,M(t + h) V (t + h) − V (t) == −ω M(t + h) − M(t) + α(h) M(t + h) − M(t) .Dividing this last equation by h and passing to the limit as h → 0, we obtainM(t)V (t) = −ωM (t).(5.136)This is the relation we were seeking between the functions we were interested in,V (t), M(t), and their derivatives.We now must find the relation between the functions V (t) and M(t) themselves,using the relation between their derivatives.
In general a problem of this type is moredifficult than the problem of finding the relations between the derivatives knowingthe relation between the functions. However, in the present case, this problem has acompletely elementary solution.Indeed, after dividing Eq. (5.136) by M(t), we can rewrite it in the formV (t) = (−ω ln M) (t).(5.137)But if the derivatives of two functions are equal on an interval, then the functionsthemselves differ by at most a constant on that interval.Thus it follows from (5.137) thatV (t) = −ω ln M(t) + c.(5.138)If it is known, for example, that V (0) = V0 , this initial condition determines theconstant c completely. Indeed, from (5.138) we findc = V0 + ω ln M(0),5.6 Examples of Differential Calculus in Natural Science289and we then find the formula we were seeking27V (t) = V0 + ω lnM(0).M(t)(5.139)It is useful to remark that if mR is the mass of the body of the rocket and mF isthe mass of the fuel and V is the terminal velocity achieved by the rocket when allthe fuel is expended, substituting M(0) = mR + mF and M(t) = mR , we findmF.V = V0 + ω ln 1 +mRThis last formula shows very clearly that the terminal velocity is affected not somuch by the ratio mF /mR inside the logarithm as by the outflow speed ω, whichdepends on the type of fuel used.
It follows in particular from this formula that ifV0 = 0, then in order to impart a velocity V to a rocket whose own mass is mR onemust have the following initial supply of fuel:mF = mR eV /ω − 1 .5.6.2 The Barometric FormulaThis is the name given to the formula that exhibits the dependence of atmosphericpressure on elevation above sea level.Let p(h) be the pressure at elevation h.
Since p(h) is the weight of the column ofair above an area of 1 cm2 at elevation h, it follows that p(h + Δ) differs from p(h)by the weight of the portion of the gas lying in the parallelepiped whose base is theoriginal area of 1 cm2 and the same area at elevation h + Δ. Let ρ(h) be the densityof air at elevation h. Since ρ(h) depends continuously on h, one may assume thatthe mass of this portion of air is calculated from the formulaρ(ξ ) g/cm3 · 1 cm2 · Δ cm = p(ξ )Δg,where ξ is some height between h and h + Δ. Hence the weight of that mass28 isg · ρ(ξ )Δ.Thus,p(h + Δ) − p(h) = −gρ(ξ )Δ.27 This formula is sometimes connected with the name of K.E.
Tsiolkovskii (1857–1935), a Russianscientist and the founder of the theory of space flight. But it seems to have been first obtained bythe Russian specialist in theoretical mechanics I.V. Meshcherskii (1859–1935) in an 1897 paperdevoted to the dynamics of a point of variable mass.28 Withinthe region where the atmosphere is noticeable, g may be regarded as constant.2905Differential CalculusAfter dividing this equality by Δ and passing to the limit as Δ → 0, taking account of the relation ξ → h, we obtainp (h) = −gρ(h).(5.140)Thus the rate of variation in atmospheric pressure has turned out to be proportional to the density of the air at the corresponding elevation.To obtain an equation for the function p(h), we eliminate the function ρ(h) from(5.140).
By Clapeyron’s law29 (the ideal gas law) the pressure p, molar volume V ,and temperature T of the gas (on the Kelvin30 scale) are connected by the relationpV= R,T(5.141)where R is the so-called universal gas constant. If M is the mass of one mole of airand V its volume, then ρ = MV , so that from (5.141) we findp=Setting λ =RMT,1M RR·R·T =·· T = ρ · T.VV MMwe thus havep = λ(T )ρ.(5.142)If we now assume that the temperature of the layer of air we are describing is constant, we finally obtain from (5.140) and (5.142)gp (h) = − p(h).λThis differential equation can be rewritten asp (h)g=−p(h)λorg (ln p) (h) = − h ,λfrom which we derivegln p(h) = − h + c,λorp(h) = ec · e−(g/λ)h .29 B.P.E.30 W.Clapeyron (1799–1864) – French physicist who studied thermodynamics.Thomson (Lord Kelvin) (1824–1907) – famous British physicist.(5.143)5.6 Examples of Differential Calculus in Natural Science291The factor ec can be determined from the known initial condition p(0) = p0 ,from which it follows that ec = p0 .Thus, we have found the following dependence of pressure on elevation:p = p0 e−(g/λ)h .(5.144)For the air at sea level (at zero altitude) and at zero temperature (273 K = 0 ◦ C),by the above formula λ = RTM , it would be possible to obtain for λ the value427.8 × 10 (m/s) .
We set g = 10 m/s2 . The formula (5.144) acquires a completelyfinished form after these numerical values are substituted. In particular, for such values λ and g, the formula (5.144) shows that the pressure drops by a factor of e (≈ 3)at the height h = gλ = 7.8 × 103 m = 7.8 km. It will increase by the same amount,if one descends to a mine at a depth of order 7.8 km.315.6.3 Radioactive Decay, Chain Reactions, and Nuclear ReactorsIt is known that the nuclei of heavy elements are subject to sporadic (spontaneous)decay.
This phenomenon is called natural radioactivity.The main statistical law of radioactivity (which is consequently valid for amountsand concentrations of a substance that are not too small) is that the number of decayevents over a small interval of time h starting at time t is proportional to h and to thenumber N(t) of atoms of the substance that have not decayed up to time t, that is,N (t + h) − N (t) ≈ −λN (t)h,where λ > 0 is a numerical coefficient that is characteristic of the chemical element.Thus the function N (t) satisfies the now familiar differential equationN (t) = −λN (t),(5.145)from which it follows thatN (t) = N0 e−λt ,where N0 = N (0) is the initial number of atoms of the substance.The time T required for half of the initial number of atoms to decay is calledthe half-life of the substance.
The quantity T can thus be found from the equa210 ) thetion e−λT = 12 , that is, T = lnλ2 ≈ 0.69λ . For example, for polonium-210 (Po226half-life T is approximately 138 days, for radium-226 (Ra ), T ≈ 1600 years,for uranium-235 (U235 ), T ≈ 7.1 × 108 years, and for its isotope U238 , T ≈ 4.5 ×109 years.31 It should be noted that this barometric formula corresponds only approximately to the real distribution of pressure and density of the air in Earth’s atmosphere. Some corrections and additionsare given in Problem 2 at the end of this section.2925Differential CalculusA nuclear reaction is an interaction of nuclei or of a nucleus with elementaryparticles resulting in the appearance of a nucleus of a new type.
This may be nuclear fusion, in which the coalescence of the nuclei of lighter elements leads to theformation of nuclei of a heavier element (for example, two nuclei of heavy hydrogen– deuterium – yield a helium nucleus along with a release of energy); or it may bethe decay of a nucleus and the formation of one or more nuclei of lighter elements.In particular, such decay occurs in approximately half of the cases when a neutroncollides with a U235 nucleus. The breakup of the uranium nucleus leads to the formation of 2 or 3 new neutrons, which may then participate in further interactionswith nuclei, causing them to split and thereby leading to further multiplication ofthe number of neutrons.
A nuclear reaction of this type is called a chain reaction.We shall describe a theoretical mathematical model of a chain reaction in a radioactive element and obtain the law of variation in the number N (t) of neutrons asa function of time.We take the substance to have the shape of a sphere of radius r.
If r is not toosmall, on the one hand new neutrons will be generated over the time interval hmeasured from some time t in a number proportional to h and N (t), while on theother hand some of the neutrons will be lost, having moved outside the sphere.If v is the velocity of a neutron, then the only ones that can leave the sphere intime h are those lying within vh of its boundary, and of those only the ones whosedirection of motion is approximately along a radius. Assuming that those neutronsconstitute a fixed proportion of the ones lying in this zone, and that neutrons aredistributed approximately uniformly throughout the sphere, one can say that thenumber of neutrons lost over the time interval h is proportional to N (t) and theratio of the volume of this boundary layer to the volume of the sphere.What has just been said leads to the equalityN (t + h) − N (t) ≈ αN (t)h −βN (t)hr(5.146)(since the volume of the boundary layer is approximately 4πr 2 vh, and the volumeof the sphere is 43 πr 3 ).
Here the coefficients α and β depend only on the particularradioactive substance.After dividing by h and passing to the limit in (5.146) as h → 0, we obtainβN (t) = α −N (t),(5.147)rfrom whichβN (t) = N0 exp α −t .rIt can be seen from this formula that when (α − βr ) > 0, the number of neutronswill increase exponentially with time. The nature of this increase, independently ofthe initial condition N0 , is such that practically total decay of the substance occurs5.6 Examples of Differential Calculus in Natural Science293over a very short time interval, releasing a colossal amount of energy – that is anexplosion.If (α − βr ) < 0, the reaction ceases very quickly since more neutrons are beinglost than are being generated.If the boundary condition between the two conditions just considered holds, thatis, α − βr = 0, an equilibrium occurs between the generation of neutrons and theirexit from the reaction, as a result of which the number of neutrons remains approximately constant.The value of r at which α − βr = 0 is called the critical radius, and the mass ofthe substance in a sphere of that volume is called the critical mass of the substance.For U235 the critical radius is approximately 8.5 cm, and the critical mass approximately 50 kg.In nuclear reactors, where steam is produced by a chain reaction in a radioactivesubstance there is an artificial source of neutrons, providing the fissionable matter with a certain number n of neutrons per unit time.
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