Goldstein Classical Mechanics Notes Michael Good May 30, 2004 1 Chapter 1: Elementary Principles 1.1 Mechanics of a Single Particle Classical mechanics incorporates special relativity. ‘Classical’ refers to the con- tradistinction to ‘quantum’ mechanics. Velocity: v = dr dt. Linear momentum: p = mv. Force: F = dp dt.

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In most cases, mass is constant and force is simplified: F = d dt (mv) = m dv dt = ma. Acceleration: a = d2r dt2. Newton’s second law of motion holds in a reference frame that is inertial or Galilean. Angular Momentum: L = r× p.

Torque: T = r× F. Torque is the time derivative of angular momentum: 1. Work: W12 = ∫ 2 1 F dr. In most cases, mass is constant and work simplifies to: W12 = m ∫ 2 1 dv dt vdt = m ∫ 2 1 v dv dt dt = m ∫ 2 1 v dv W12 = m 2 (v22 − v21) = T2 − T1 Kinetic Energy: T = mv2 2 The work is the change in kinetic energy.

A force is considered conservative if the work is the same for any physically possible path. Independence of W12 on the particular path implies that the work done around a closed ciruit is zero:∮ F dr = 0 If friction is present, a system is non-conservative.

Potential Energy: F = −∇V (r). The capacity to do work that a body or system has by viture of is position is called its potential energy. V above is the potential energy. To express work in a way that is independent of the path taken, a change in a quantity that depends on only the end points is needed.

This quantity is potential energy. Work is now V1 − V2. The change is -V.

Energy Conservation Theorem for a Particle: If forces acting on a particle are conservative, then the total energy of the particle, T + V, is conserved. The Conservation Theorem for the Linear Momentum of a Particle states that linear momentum, p, is conserved if the total force F, is zero. The Conservation Theorem for the Angular Momentum of a Particle states that angular momentum, L, is conserved if the total torque T, is zero. 1.2 Mechanics of Many Particles Newton’s third law of motion, equal and opposite forces, does not hold for all forces. It is called the weak law of action and reaction. Center of mass: R = ∑ miri∑ mi = ∑ miri M. Center of mass moves as if the total external force were acting on the entire mass of the system concentrated at the center of mass.

Internal forces that obey Newton’s third law, have no effect on the motion of the center of mass. F(e) ≡M d 2R dt2 = ∑ i F(e)i. Motion of center of mass is unaffected. This is how rockets work in space. Total linear momentum: P = ∑ i mi dri dt =M dR dt. Conservation Theorem for the Linear Momentum of a System of Particles: If the total external force is zero, the total linear momentum is conserved. The strong law of action and reaction is the condition that the internal forces between two particles, in addition to being equal and opposite, also lie along the line joining the particles.

Then the time derivative of angular momentum is the total external torque: dL dt = N(e). Torque is also called the moment of the external force about the given point.

Conservation Theorem for Total Angular Momentum: L is constant in time if the applied torque is zero. Linear Momentum Conservation requires weak law of action and reaction. Angular Momentum Conservation requires strong law of action and reaction. Total Angular Momentum: L = ∑ i ri × pi = R×Mv+ ∑ i r′i × p′i.

Total angular momentum about a point O is the angular momentum of mo- tion concentrated at the center of mass, plus the angular momentum of motion about the center of mass. If the center of mass is at rest wrt the origin then the angular momentum is independent of the point of reference. Total Work: W12 = T2 − T1 where T is the total kinetic energy of the system: T = 12 ∑ i miv 2 i. Total kinetic energy: T = 1 2 ∑ i miv 2 i = 1 2 Mv2 + 1 2 ∑ i miv ′2 i. Kinetic energy, like angular momentum, has two parts: the K.E. Obtained if all the mass were concentrated at the center of mass, plus the K.E. Of motion about the center of mass.

Total potential energy: V = ∑ i Vi + 1 2 ∑ i,j i 6=j Vij. If the external and internal forces are both derivable from potentials it is possible to define a total potential energy such that the total energy T + V is conserved. The term on the right is called the internal potential energy. For rigid bodies the internal potential energy will be constant. For a rigid body the internal forces do no work and the internal potential energy remains constant. 1.3 Constraints.

holonomic constraints: think rigid body, think f(r1, r2, r3., t) = 0, think a particle constrained to move along any curve or on a given surface. nonholonomic constraints: think walls of a gas container, think particle placed on surface of a sphere because it will eventually slide down part of the way but will fall off, not moving along the curve of the sphere. Rheonomous constraints: time is an explicit variable.example: bead on moving wire 2. Scleronomous constraints: equations of contraint are NOT explicitly de- pendent on time.example: bead on rigid curved wire fixed in space Difficulties with constraints: 4.

Lagrange’s Equations come from this principle. If you remember the indi- vidual coefficients vanish, and allow the forces derivable from a scaler potential function, and forgive me for skipping some steps, the result is: d dt ( ∂L ∂q̇j )− ∂L ∂qj = 0 1.5 Velocity-Dependent Potentials and The Dissipation Function The velocity dependent potential is important for the electromagnetic forces on moving charges, the electromagnetic field. L = T − U where U is the generalized potential or velocity-dependent potential. For a charge mvoing in an electric and magnetic field, the Lorentz force dictates: F = qE+ (v×B).

The equation of motion can be dervied for the x-dirction, and notice they are identical component wise: mẍ = qEx + (v×B)x. If frictional forces are present(not all the forces acting on the system are derivable from a potential), Lagrange’s equations can always be written: d dt ( ∂L ∂q̇j )− ∂L ∂qj = Qj. Where Qj represents the forces not arising from a potential, and L contains the potential of the conservative forces as before.

Friction is commonly, Ffx = −kxvx. Rayleigh’s dissipation function: Fdis = 1 2 ∑ i (kxv2ix + kyv 2 iy + kzv 2 iz). The total frictional force is: Ff = −∇vFdis Work done by system against friction: dWf = −2Fdisdt 6.

The rate of energy dissipation due to friction is 2Fdis and the component of the generalized force resulting from the force of friction is: Qj = − ∂Fdis ∂q̇j. In use, both L and Fdis must be specified to obtain the equations of motion: d dt ( ∂L ∂q̇j )− ∂L ∂qj = −∂Fdis ∂q̇j. 1.6 Applications of the Lagrangian Formulation The Lagrangian method allows us to eliminate the forces of constraint from the equations of motion. Scalar functions T and V are much easier to deal with instead of vector forces and accelerations. Procedure: 1. Write T and V in generalized coordinates. Form L from them.

Put L into Lagrange’s Equations 4. Solve for the equations of motion. Simple examples are: 1. A single particle is space(Cartesian coordinates, Plane polar coordinates) 2.

Atwood’s machine 3. A bead sliding on a rotating wire(time-dependent constraint).

Forces of contstraint, do not appear in the Lagrangian formulation. They also cannot be directly derived.

Goldstein Chapter 1 Derivations Michael Good June 27, 2004 1 Derivations 1. Show that for a single particle with constant mass the equation of motion implies the follwing differential equation for the kinetic energy: dT dt = F v while if the mass varies with time the corresponding equation is d(mT ) dt = F p. Answer: dT dt = d( 12mv 2) dt = mv v̇ = ma v = F v with time variable mass, d(mT ) dt = d dt ( p2 2 ) = p ṗ = F p. Prove that the magnitude R of the position vector for the center of mass from an arbitrary origin is given by the equation: M2R2 =M ∑ i mir 2 i − 1 2 ∑ i,j mimjr 2 ij.

Answer: MR = ∑ miri 1. M2R2 = ∑ i,j mimjri rj Solving for ri rj realize that rij = ri − rj. Square ri − rj and you get r2ij = r 2 i − 2ri rj + r2j Plug in for ri rj ri rj = 1 2 (r2i + r 2 j − r2ij) M2R2 = 1 2 ∑ i,j mimjr 2 i + 1 2 ∑ i,j mimjr 2 j − 1 2 ∑ i,j mimjr 2 ij M2R2 = 1 2 M ∑ i mir 2 i + 1 2 M ∑ j mjr 2 j − 1 2 ∑ i,j mimjr 2 ij M2R2 =M ∑ i mir 2 i − 1 2 ∑ i,j mimjr 2 ij 3. Suppose a system of two particles is known to obey the equations of mo- tions, M d2R dt2 = ∑ i F(e)i ≡ F (e) dL dt = N(e) From the equations of the motion of the individual particles show that the in- ternal forces between particles satisfy both the weak and the strong laws of ac- tion and reaction. The argument may be generalized to a system with arbitrary number of particles, thus proving the converse of the arguments leading to the equations above. Answer: First, if the particles satisfy the strong law of action and reaction then they will automatically satisfy the weak law.

The weak law demands that only the forces be equal and opposite. The strong law demands they be equal and oppo- site and lie along the line joining the particles. The first equation of motion tells us that internal forces have no effect. The equations governing the individual particles are ṗ1 = F (e) 1 + F21 ṗ2 = F (e) 2 + F12 2. Assuming the equation of motion to be true, then ṗ1 + ṗ2 = F (e) 1 + F21 + F (e) 2 + F12 must give F12 + F21 = 0 Thus F12 = −F21 and they are equal and opposite and satisfy the weak law of action and reaction.

If the particles obey dL dt = N(e) then the time rate of change of the total angular momentum is only equal to the total external torque; that is, the internal torque contribution is null. For two particles, the internal torque contribution is r1×F21 + r2×F12 = r1×F21 + r2× (−F21) = (r1− r2)×F21 = r12×F21 = 0 Now the only way for r12 × F21 to equal zero is for both r12 and F21 to lie on the line joining the two particles, so that the angle between them is zero, ie the magnitude of their cross product is zero. A×B = ABsinθ 4. The equations of constraint for the rolling disk, dx− a sin θdψ = 0 dy + a cos θdψ = 0 are special cases of general linear differential equations of constraint of the form n∑ i=1 gi(x1,., xn)dxi = 0. A constraint condition of this type is holonomic only if an integrating function f(x1,., xn) can be found that turns it into an exact differential.

Clearly the function must be such that ∂(fgi) ∂xj = ∂(fgj) ∂xi for all i 6= j. Show that no such integrating factor can be found for either of the equations of constraint for the rolling disk. First attempt to find the integrating factor for the first equation. Note it is in the form: Pdx+Qdφ+Wdθ = 0 where P is 1, Q is −a sin θ and W is 0. The equations that are equivalent to ∂(fgi) ∂xj = ∂(fgj) ∂xi are ∂(fP ) ∂φ = ∂(fQ) ∂x ∂(fP ) ∂θ = ∂(fW ) ∂x ∂(fQ) ∂θ = ∂(fW ) ∂φ These are explicitly: ∂(f) ∂φ = ∂(−fa sin θ) ∂x ∂(f) ∂θ = 0 ∂(−fa sin θ) ∂θ = 0 Simplfying the last two equations yields: f cos θ = 0 Since y is not even in this first equation, the integrating factor does not depend on y and because of ∂f∂θ = 0 it does not depend on θ either. Thus f = f(x, φ) The only way for f to satisfy this equation is if f is constant and thus appar- ently there is no integrating function to make these equations exact. Performing the same procedure on the second equation you can find ∂(fa cos θ) ∂y = ∂f ∂φ a cos θ ∂f ∂y = ∂f ∂φ and f sin θ = 0 4.

∂f ∂θ = 0 leading to f = f(y, φ) and making it impossible for f to satsify the equations unless as a constant. If this question was confusing to you, it was confusing to me too. Mary Boas says it is ‘not usually worth while to spend much time searching for an integrating factor’ anyways. That makes me feel better. Two wheels of radius a are mounted on the ends of a common axle of length b such that the wheels rotate independently. The whole combination rolls with- out slipping on a palne. Show that there are two nonholonomic equations of constraint, cos θdx+ sin θdy = 0 sin θdx− cos θdy = 1 2 a(dφ+ dφ′) (where θ,φ, and φ′ have meanings similar to those in the problem of a single vertical disk, and (x,y) are the corrdinates of a point on the axle midway between the two wheels) and one holonomic equation of constraint, θ = C − a b (φ− φ′) where C is a constant.

Answer: The trick to this problem is carefully looking at the angles and getting the signs right. I think the fastest way to solve this is to follow the same procedure that was used for the single disk in the book, that is, find the speed of the disk, find the point of contact, and take the derivative of the x component, and y component of position, and solve for the equations of motion. Here the steps are taken a bit further because a holonomic relationship can be found that relates θ, φ and φ′. Once you have the equations of motion, from there its just slightly tricky algebra. Here goes: We have two speeds, one for each disk v′ = aφ̇′ v = aφ̇ and two contact points, (x± b 2 cos θ, y ± b 2 sin θ) 5. The contact points come from the length of the axis being b as well as x and y being the center of the axis. The components of the distance are cos and sin for x and y repectively.

So now that we’ve found the speeds, and the points of contact, we want to take the derivatives of the x and y parts of their contact positions. This will give us the components of the velocity. Make sure you get the angles right, they were tricky for me. D dt (x+ b 2 cos θ) = vx ẋ− b 2 sin θθ̇ = v cos(180− θ − 90) = v cos(90− θ) = v cos(−90 + θ) = v sin θ ẋ− b 2 sin θθ̇ = aφ̇ sin θ Do this for the next one, and get: ẋ+ b 2 sin θθ̇ = aφ̇′ sin θ The plus sign is there because of the derivative of cos multiplied with the negative for the primed wheel distance from the center of the axis. For the y parts: d dt (y + b 2 sin θ) = vy ẏ + b 2 cos θθ̇ = −v cos θ = −aφ̇ cos θ It is negative because I decided to have axis in the first quadrent heading south-east. I also have the primed wheel south-west of the non-primed wheel. A picture would help, but I can’t do that on latex yet.

So just think about it. Do it for the next one and get: ẏ − b 2 cos θθ̇ = −aφ̇′ cos θ All of the derivatives together so you aren’t confused what I just did: ẋ− b 2 sin θθ̇ = aφ̇ sin θ ẋ+ b 2 sin θθ̇ = aφ̇′ sin θ ẏ + b 2 cos θθ̇ = −aφ̇ cos θ ẏ − b 2 cos θθ̇ = −aφ̇′ cos θ Now simplify them by cancelling the dt′s and leaving the x and y’s on one side: 6.

Dx = sin θ b 2 dθ + adφ (1) dx = sin θ− b 2 dθ + adφ′ (2) dy = − cos θ b 2 dθ + adφ (3) dy = − cos θ− b 2 dθ + adφ′ (4) Now we are done with the physics. The rest is manipulation of these equa- tions of motion to come up with the constraints.

For the holonomic equation use (1)-(2). (1)− (2) = 0 = bdθ + a(dφ− dφ′) dθ = −a b (dφ− dφ′) θ = −a b (φ− φ′) + C For the other two equations, I started with (1) cos θ + (3) sin θ = cos θ sin θ b 2 dθ + adφ− sin θ cos θ b 2 dθ + adφ cos θdx+ sin θdy = 0 and (1) + (2) = 2dx = sin θadφ+ dφ′ (3) + (4) = 2dy = − cos θadφ+ dφ′ multiply dy by − cos θ and multiply dx by sin θ to yield yourself − cos θdy = cos2 θa 2 dφ+ dφ′ sin θdx = sin2 θ a 2 dφ+ dφ′ Add them together and presto! Sin θdx− cos θdy = a 2 dφ+ dφ′ 6. A particle moves in the xy plane under the constraint that its velocity vector is always directed towards a point on the x axis whose abscissa is some given function of time f(t).

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Show that for f(t) differentiable, but otherwise arbitrary, 7. The constraint is nonholonomic. Answer: The abscissa is the x-axis distance from the origin to the point on the x-axis that the velocity vector is aimed at. It has the distance f(t).

I claim that the ratio of the velocity vector components must be equal to the ratio of the vector components of the vector that connects the particle to the point on the x-axis. The directions are the same.

The velocity vector components are: vy = dy dt vx = dx dt The vector components of the vector that connects the particle to the point on the x-axis are: Vy = y(t) Vx = x(t)− f(t) For these to be the same, then vy vx = Vy Vx dy dx = y(t) x(t)− f(t) dy y(t) = dx x(t)− f(t) This cannot be integrated with f(t) being arbituary. Thus the constraint is nonholonomic. It’s nice to write the constraint in this way because it’s frequently the type of setup Goldstein has: ydx+ (f(t)− x)dy = 0 There can be no integrating factor for this equation.

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The Lagrangian equations can be written in the form of the Nielsen’s equa- tions. ∂Ṫ ∂q̇ − 2∂T ∂q = Q Show this. Answer: I’m going to set the two forms equal and see if they match. That will show that they can be written as displayed above. Lagrangian Form = Nielsen’s Form d dt ( ∂T ∂q̇ )− ∂T ∂q = ∂Ṫ ∂q̇ − 2∂T ∂q d dt ( ∂T ∂q̇ ) + ∂T ∂q = ∂Ṫ ∂q̇ (5) What is ∂Ṫ∂q̇ you may ask? Well, lets solve for Ṫ first.

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Ṫ ≡ d dt T (q, q̇, t) Because ddt is a full derivative, you must not forget the chain rule. Ṫ ≡ d dt T (q, q̇, t) = ∂T ∂t + ∂T ∂q q̇ + ∂T ∂q̇ q̈ Now lets solve for ∂Ṫ∂q̇, not forgetting the product rule ∂Ṫ ∂q̇ = ∂ ∂q̇ ∂T ∂t + ∂T ∂q q̇ + ∂T ∂q̇ q̈ ∂Ṫ ∂q̇ = ∂ ∂q̇ ∂T ∂t + ∂ ∂q̇ ∂T ∂q q̇ + ∂T ∂q ∂q̇ ∂q̇ + ∂ ∂q̇ ∂T ∂q̇ q̈ ∂Ṫ ∂q̇ = ∂ ∂t ∂T ∂q̇ + ∂ ∂q ∂T ∂q̇ q̇ + ∂T ∂q + ∂ ∂q̇ ( ∂T ∂q̇ )q̈ Now we have ∂Ṫ∂q̇, so lets plug this into equation (5).

D dt ( ∂T ∂q̇ ) + ∂T ∂q = ∂ ∂t ∂T ∂q̇ + ∂ ∂q ∂T ∂q̇ q̇ + ∂T ∂q + ∂ ∂q̇ ( ∂T ∂q̇ )q̈ d dt ( ∂T ∂q̇ ) = ∂ ∂t ∂T ∂q̇ + ∂ ∂q ∂T ∂q̇ q̇ + ∂ ∂q̇ ( ∂T ∂q̇ )q̈ Notice that this is indeed true. D dt ( ∂T ∂q̇ ) = ∂ ∂t ( ∂T ∂q̇ ) + ∂ ∂q ( ∂T ∂q̇ )q̇ + ∂ ∂q̇ ( ∂T ∂q̇ )q̈ because T = T (q, q̇, t). If L is a Lagrangian for a system of n degrees of freedom satisfying Lagrange’s equations, show by direct substitution that L′ = L+ dF (q1., qn, t) dt also satisfies Lagrange’s equations where F is any arbitrary, but differentiable, function of its arguments. Answer: Let’s directly substitute L′ into Lagrange’s equations. D dt ∂L′ ∂q̇ − ∂L ′ ∂q = 0 d dt ∂ ∂q̇ (L+ dF dt )− ∂ ∂q (L+ dF dt ) = 0 d dt ∂L ∂q̇ + ∂ ∂q̇ dF dt − ∂L ∂q − ∂ ∂q dF dt = 0 d dt ∂L ∂q̇ − ∂L ∂q + d dt ∂ ∂q̇ dF dt − ∂ ∂q dF dt = 0 On the left we recognized Lagrange’s equations, which we know equal zero. Now to show the terms with F vanish. D dt ∂ ∂q̇ dF dt − ∂ ∂q dF dt = 0 d dt ∂Ḟ ∂q̇ = ∂Ḟ ∂q This is shown to be true because ∂Ḟ ∂q̇ = ∂F ∂q We have d dt ∂Ḟ ∂q̇ = d dt ∂F ∂q = ∂ ∂t ∂F ∂q + ∂ ∂q ∂F ∂q q̇ = ∂ ∂q ∂F ∂t + ∂F ∂q q̇ = ∂Ḟ ∂q 10.

Thus as Goldstein reminded us, L = T − V is a suitable Lagrangian, but it is not the only Lagrangian for a given system. The electromagnetic field is invariant under a gauge transformation of the scalar and vector potential given by A→ A+∇ψ(r, t) φ→ φ− 1 c ∂ψ ∂t where ψ is arbitrary (but differentiable). What effect does this gauge trans- formation have on the Lagrangian of a particle moving in the electromagnetic field? Is the motion affected? Answer: L = 1 2 mv2 − qφ+ q c A v Upon the gauge transformation: L′ = 1 2 mv2 − qφ− 1 c ∂ψ ∂t + q c A+∇ψ(r, t) v L′ = 1 2 mv2 − qφ+ q c A v+ q c ∂ψ ∂t + q c ∇ψ(r, t) v L′ = L+ q c ∂ψ ∂t +∇ψ(r, t) v L′ = L+ q c ψ̇ In the previous problem it was shown that: d dt ∂ψ̇ ∂q̇ = ∂ψ̇ ∂q For ψ differentiable but arbitrary. This is all that you need to show that the Lagrangian is changed but the motion is not. This problem is now in the same form as before: L′ = L+ dF (q1., qn, t) dt And if you understood the previous problem, you’ll know why there is no effect on the motion of the particle( i.e.

There are many Lagrangians that may describe the motion of a system, there is no unique Lagrangian). Let q1., qn be a set of independent generalized coordinates for a system 11. Of n degrees of freedom, with a Lagrangian L(q, q̇, t). Suppose we transform to another set of independent coordinates s1., sn by means of transformation equations qi = qi(s1., sn, t), i = 1., n. (Such a transformatin is called a point transformation.) Show that if the Lagrangian function is expressed as a function of sj, ṡj and t through the equa- tion of transformation, then L satisfies Lagrange’s equations with respect to the s coordinates d dt ∂L ∂ṡj − ∂L ∂sj = 0 In other words, the form of Lagrange’s equations is invariant under a point transformation. Answer: We know: d dt ∂L ∂q̇i − ∂L ∂qi = 0 and we want to prove: d dt ∂L ∂ṡj − ∂L ∂sj = 0 If we put ∂L∂ṡj and ∂L ∂sj in terms of the q coordinates, then they can be substitued back in and shown to still satisfy Lagrange’s equations.

∂L ∂sj = ∑ i ∂L ∂qi ∂qi ∂sj ∂L ∂ṡj = ∑ i ∂L ∂q̇i ∂q̇i ∂ṡj We know: ∂qi ∂sj = ∂q̇i ∂ṡj Thus, ∂L ∂ṡj = ∑ i ∂L ∂q̇i ∂qi ∂sj Plug ∂L∂ṡj and ∂L ∂sj into the Lagrangian equation and see if they satisfy it: d dt ∑ i ∂L ∂q̇i ∂qi ∂sj − ∑ i ∂L ∂qi ∂qi ∂sj = 0 12. Goldstein Chapter 1 Exercises Michael Good July 17, 2004 1 Exercises 11. Consider a uniform thin disk that rolls without slipping on a horizontal plane. A horizontal force is applied to the center of the disk and in a direction parallel to the plane of the disk. Derive Lagrange’s equations and find the generalized force.

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Discuss the motion if the force is not applied parallel to the plane of the disk. Answer: To find Lagrangian’s equations, we need to first find the Lagrangian. L = T − V T = 1 2 mv2 = 1 2 m(rω)2 V = 0 Therefore L = 1 2 m(rω)2 Plug into the Lagrange equations: d dt ∂L ∂ẋ − ∂L ∂x = Q d dt ∂ 12mr 2ω2 ∂(rω) − ∂ 12mr 2ω2 ∂x = Q d dt m(rω) = Q m(rω̈) = Q 1. If the motion is not applied parallel to the plane of the disk, then there might be some slipping, or another generalized coordinate would have to be introduced, such as θ to describe the y-axis motion. The velocity of the disk would not just be in the x-direction as it is here. The escape velocity of a particle on Earth is the minimum velocity re- quired at Earth’s surface in order that that particle can escape from Earth’s gravitational field. Neglecting the resistance of the atmosphere, the system is conservative.

From the conservation theorme for potential plus kinetic energy show that the escape veolcity for Earth, ingnoring the presence of the Moon, is 11.2 km/s. Answer: GMm r = 1 2 mv2 GM r = 1 2 v2 Lets plug in the numbers to this simple problem: (6.67× 10−11) (6× 1024) (6× 106) = 1 2 v2 This gives v = 1.118× 104 m/s which is 11.2 km/s. Rockets are propelled by the momentum reaction of the exhaust gases expelled from the tail. Since these gases arise from the raction of the fuels carried in the rocket, the mass of the rocket is not constant, but decreases as the fuel is expended. Show that the equation of motion for a rocket projected vertically upward in a uniform gravitational field, neglecting atmospheric friction, is: m dv dt = −v′ dm dt −mg where m is the mass of the rocket and v’ is the velocity of the escaping gases relative to the rocket. Integrate this equation to obtain v as a function of m, assuming a constant time rate of loss of mass.

Show, for a rocket starting initally from rest, with v’ equal to 2.1 km/s and a mass loss per second equal to 1/60th of the intial mass, that in order to reach the escape velocity the ratio of the wight of the fuel to the weight of the empty rocket must be almost 300! Answer: This problem can be tricky if you’re not very careful with the notation. But here is the best way to do it.

Defining me equal to the empty rocket mass, mf is the total fuel mass, m0 is the intitial rocket mass, that is, me +mf, and dm dt = − m0 60 as the loss rate of mass, and finally the goal is to find the ratio of 2. Mf/me to be about 300. The total force is just ma, as in Newton’s second law. The total force on the rocket will be equal to the force due to the gas escaping minus the weight of the rocket: ma = d dt −mv′−mg m dv dt = −v′ dm dt −mg The rate of lost mass is negative. The velocity is in the negative direction, so, with the two negative signs the term becomes positive. Use this: dv dm dm dt = dv dt Solve: m dv dm dm dt = −v′ dm dt −mg dv dm dm dt = − v ′ m dm dt − g dv dm = − v ′ m + 60g m0 Notice that the two negative signs cancelled out to give us a positive far right term. Dv = − v ′ m dm+ 60g m0 dm Integrating, ∫ dv = −v′ ∫ me m0 dm m + ∫ me m0 60g m0 dm v = −v′ ln me m0 + 60g m0 (me −m0) v = −v′ ln me me +mf + 60g me −me −mf me +mf v = v′ ln me +mf me − 60g mf me +mf Now watch this, I’m going to use my magic wand of approximation.

This is when I say that because I know that the ratio is so big, I can ignore the empty 3. Rocket mass as compared to the fuel mass.