Mecflu - movimento de for?a central

Mecflu - movimento de for?a central

(Parte 1 de 2)

CHAPTER 8 Central-Force Motion

8-1. x m m r r

In a uniform gravitational field, the gravitational acceleration is everywhere constant. Suppose the gravitational field vector is in the x direction; then the masses and have the

UF x m x

introducting the relative coordinate r and the center of mass coordinate R according to r r

we can express r and r in terms of r and R by 12 m m r R R

which differ from Eq. (8.3) in the text by R. The Lagrangian of the two-particle system can now be expressed in terms of r and R:

ggLm m U U U

m m U r m m m mx X m x m m αα

(4) where x and X are the components of r and R, respectively. Then, (4) becomes 1x m Lm m m m m m m

m m

Hence, we can write the Lagrangian in the form

where µ is the reduced mass:

Therefore, this case is reducible to an equivalent one-body problem.

8-2. Setting 1u=r, Eq. (8.38) can be rewritten as

2 du Ek

12 sin const.

dx ax b

we have const. sin k r

CENTRAL-FORCE MOTION 235 or, equivalently, sin const.

k r

We can choose the point from which θ is measured so that the constant in (4) is 2π−. Then, 2


E k which is the desired expression.

8-3. When 2kk→, the potential energy will decrease to half its former value; but the kinetic energy will remain the same. Since the original orbit is circular, the instantaneous values of T and U are equal to the average values, T andU. For a 2r1 force, the virial theorem states


Now, consider the energy diagram E D


CBE= original total energy CAU= original potential energy

CCDU= original centrifugal energy

Hence, if U suddenly is halved, the total energy is raised from B by an amount equal to ()12CA or by CB. Thus, the total energy point is raised from B to C; i.e., E(final) = 0 and the orbit is parabolic.

8-4. Since the particle moves in a central, inverse-square law force field, the potential energy is

so that the time average is

from which

Therefore, (2) becomes

1 kr k U d

k U

where a is the semimajor axis of the ellipse. Using the standard integral [see Eq. (E.15), Appendix E],

and the relation,

(6) becomes

The kinetic energy is


and the time average is k T dt Tr

Part of this integral is trivial, k T r d

To evaluate the integral above, substitute the expression for r and make a change of variable dx r d

The reader is invited to evaluate this integral in either form. The solution presented here is to integrate by parts twice, which gives a third integral that can be looked up in a table:

x x d x xεε ε−−

x x xε εε ε−−


1π ε ε

Substituting this into (13) and then into (12), we obtain the desired answer,

This explicitly verifies the virial theorem, which states that for an inverse-square law force,

8-5. a

Suppose two particles with masses and move around one another in a circular orbit with radius a. We can consider this motion as the motion of one particle with the reduced mass µ moving under the influence of a central force

212Gmma. Therefore, the equation of motion before the particles are stopped is


The radius of circular motion is

24 Gm ma τπµ

After the circular motion is stopped, the particle with reduced mass µ starts to move toward the force center. We can find the equation of motion from the conservation of energy:


Therefore, the time elapsed before the collision is 0 dx td t

Gm m

xaµ where the negative sign is due to the fact that the time increases as the distance decreases. Rearranging the integrand, we can write a x t

yx Id x


Using Eq. (E.7), Appendix E, we find

12s in ya y yaa

Therefore, a t

Gm m µπ= or,


m m x xO r rxx=−

When two particles are initially at rest separated by a distance , the system has the total energy 0r

The coordinates of the particles, and , are measured from the position of the center of mass. At any time the total energy is 1x 2x

From the conservation of energy we have E0E=, or m m

Using (3) in (4), we find

G xvm

Mr r

G xv m

M r

8-7. Since ()Frkr=− is a central force, angular momentum is conserved and the areal velocity, 2dAdtµ=A, is trivially constant (see Section 8.3). In order to compute U, we start with dr dt

EU rµµ =


The time average of the potential energy becomes

r r

U dt kr dr kr Er


2 r x dr dxr

(4) becomes xd xk U

k Ex x

Using the integrals in Eqs. (E.9) and (E.8c), Appendix E, xd x b ax ax bx c

(5) becomes kE E kr k Ur

k k E

A 2Er (7)

But and were originally defined as the roots of maxrminr2 22EU rµ−− A

. Hence, the second term vanishes at both limits of integration. On the other hand,


r r

EU r rdr k rE r or, using (5), r r k xE x

Ek r k k E

Using (9) to substitute for τ in (7), we have


The virial theorem states:

In our case n = 1, therefore,

8-8. The general expression for θ (r) is [see Eq. (8.17)] rd r r

EU r

dx r kE x x θ µ=

Using Eq. (E.10b), Appendix E, dx bx c

and expressing again in terms of r, we find

E r

Ek r or,


In order to interpret this result, we set 2


or, yx α α

Since a′ > 0, ε′ > 1 from the definition, (10) is equivalent to


which is the equation of a hyperbola.

8-9. (a) By the virial theorem, 2U=−T for a circular orbit.

The firing of the rocket doesn’t change U, so fiUU= But


0fiEE =

The firing of the rocket doesn’t change the angular momentum since it fires in a radial direction.

(b) E = 0 means the orbit is parabolic. The satellite will be lost.

() ()0 esGM m Er U r r

() esGM m Tr E U r esGM m Vr U r

Behavior of V(r) is determined by

222for small

for large esrr GM m r r

dVdr = at 0rr=

0 esr GM mµ=− A

8-10. For circular motion


GM m

U r =−

We can get 2ω by equating the gravitational force to the centripetal force

22se e

GM m mrr ω= or

So ses e


If the sun’s mass suddenly goes to 1

2 its original value, T remains unchanged but U is halved.


T or he energy is 0, so the orbit is a parabola. For a parabolic bic, the earth will escape the solar system.

8-1. For central-force motion the equation of orbit is [Eq. (8.21)]

11d r F r

force center r a

Therefore, (1) becomes

cos cos cos d Fr

But we have sin cos


2s in1 cos cos d θθθθ θ

Therefore, we have

cos cos cos aFrθµθθθθ++=−A (5) or,

so that

246 CHAPTER 8 8-12.

βr r

The orbit of the comet is a parabola (ε = 1), so that the equation of the orbit is

We choose to measure θ from perihelion; hence

Therefore, 2

2 Erk α βµ==A

Since the total energy is zero (the orbit is parabolic) and the potential energy is Uk=−r, the time spent within the orbit of the Earth is r r

r E dr T rd r k r r r k β β from which

Now, the period and the radius of the Earth are related by



Substituting (7) into (5), we find k T

where skGMµ= and sEkGMµ=′. Therefore,

where 1 yearEτ=. Now, Mercury0.387Errβ==. Therefore,

so that

8-13. Setting 1≡u we can write the force as r

Then, the equation of orbit becomes [cf. Eq. (8.20)]

1du u ku

from which ud 2 µλµθ

or, du k u

If we make the change of variable, kvu µ

we have


where 221βµλ=−A. This equation gives different solutions according to the value of λ. Let us consider the following three cases:

i) 2λµ<A:

By proper choice of the position θ = 0, the integration constant δ can be made to equal zero. Therefore, we can write cos k A

When β = 1 (λ = 0), this equation describes a conic section. Since we do not know the value of the constant A, we need to use what we have learned from Kepler’s problem to describe the motion. We know that for λ = 0, kr µεθ=+A

and that we have an ellipse or circle (0 ≤ ε < 1) when E < 1, a parabola (ε = 1) when E = 0, and a hyperbola otherwise. It is clear that for this problem, if E ≥ 0, we will have some sort of parabolic or hyperbolic orbit. An ellipse should result when E < 0, this being the only bound orbit. When β ≠ 1, the orbit, whatever it is, precesses. This is most easily seen in the case of the ellipse, where the two turning points do not have an angular separation of π. One may obtain most constants of integration (in particular A) by using Equation (8.17) as a starting point, a more formal approach that confirms the statements made here.

i) 2λµ=A

so that k u

from which we see that r continuously decreases as θ increases; that is, the particle spirals in toward the force center.


cosh k A

Again, the particle spirals in toward the force center.

8-14. The orbit equation for the central-force field is [see Eq. (8.17)] dr r EU dr µ θµ 2

But we are given the orbit equation:

from which

Substituting (2) into (3), we have

From (1) and (4), we find the equation for the potential U:

r kr E U rµ µ 2

from which

8-15. AB

Let us denote by v the velocity of the particle when it is infinitely far from P and traveling along the line AB. The angular momentum is

where we have used m = 1. Therefore,

The total energy E of the particle is equal to the initial kinetic energy:

dr d

r k E

Substituting for A and E from (1) and (3), we have kd r d

br k k br b r dr b

rb r b dr b


where we have taken the negative square root because r decreases as θ increases (see the diagram). We can now use the integral [see Eq. (E.4b), Appendix E]


from which we obtain


Notice that r is always greater than b (because coth φ → 1 as φ → ∞), so that the denominator in

(5) never equals zero nor changes sign. Thus, r always decreases as θ increases. This is, the particle spirals in toward P but never approaches closer than a distance b.

a principle that by no means pushes the philosophical envelope of physical interpretation. The impulse that causes v → v + δv changes the kinetic energy, not the potential energy. We therefore have

By the virial theorem, for a nearly circular orbit we have

so that

where we have written –E since E < 0. The major and minor axes of the orbit are given by k a b

Now let us compute the changes in these quantities. For a we have k E E a

and for b we have

E b

E E E b

A A A A (7)

Easily enough, we can show that vδδ=A and therefore

8-17. The equation of the orbit is

from which

ε =+ A . Therefore, the radial distance r can vary from the

Thus, the maximum and minimum values of ω become r ω µ αµ ε

Thus, 2

maxmin 1 1 from which we find

8-18. Kepler’s second law states that the areal velocity is constant, and this implies that the angular momentum L is conserved. If a body is acted upon by a force and if the angular momentum of the body is not altered, then the force has imparted no torque to the body; thus,

CENTRAL-FORCE MOTION 253 the force must have acted only along the line connecting the force center and the body. That is, the force is central.

Kepler’s first law states that planets move in elliptical orbits with the sun at one focus. This means the orbit can be described by Eq. (8.41):

On the other hand, for central forces, Eq. (8.21) holds:

11dr F r

Substituting 1 from (1) into the left-hand side of (2), we find r

which implies, that

8-19. The semimajor axis of an orbit is defined as one-half the sum of the two apsidal distances, and [see Eq. (8.4)], so maxrminr

This is the same as the semimajor axis defined by Eq. (8.42). Therefore, by using Kepler’s Third Law, we can find the semimajor axis of Ceres in astronomical units:

k a ka τπµ where cskMmcγ=, and

Here, sM and are the masses of the sun and Ceres, respectively. Therefore, (2) becomes cm

Es e E aM m from which

so that

The period of Jupiter can also be calculated using Kepler’s Third Law:


EEs E a ka Mm

Mm a a from which


8-20. Using Eqs. (8.42) and (8.41) for a and r, we have

1c os1 cos cos a dt r τ εθθ θτε

From Kepler’s Second Law, we can find the relation between t and θ:

dt dA d

1 cos cos 1 cos a a d

It is easily shown that the value of the integral is 2πε. Therefore,


After substituting a and b in terms of ε and α [see Eqs. (8.42) and (8.43)], we obtain

8-21. If we denote the total energy and the potential of the family of orbits by E and U(r), we have the relation

from which

Here, E and U(r) are same for all orbits, and the different values of A result from different values

circular motions, and A is smaller than for the circular case. That is, the angular momentum of the circular orbit is the largest among the family. 2 0r >

8-2. For the given force, ()3Frkr=−, the potential is

and the effective potential is

2 Vr k rµ

The equation of the orbit is [cf. Eq. (8.20)]

or, du k u

Let us consider the motion for various values of A.

i) 2kµ=A:

In this case the effective potential V(r) vanishes and the orbit equation is

with the solution

and the particle spirals toward the force center. i) 2kµ>A:

In this case the effective potential is positive and decreases monotonically with increasing r. For any value of the total energy E, the particle will approach the force center and will undergo a reversal of its motion at ; the particle will then proceed again to an infinite distance. 0rr= r r

with the solution

Since the minimum value of u is zero, this solution corresponds to unbounded motion, as expected from the form of the effective potential V(r).

i) 2kµ<A:

with the solution

so that the particle spirals in toward the force center.



Since 0rpr== , Eq. (8.87) shows that 2kµ=A. Therefore, (12) reduces to so that the perturbation x increases uniformly with the time. The circular orbit is therefore not stable.

We can also reach the same conclusion by examining the basic criterion for stability, namely, that

20 and 0 r r

The first of these relations requires 2kµ=A while the second requires 2kµ>A. Since these requirements cannot be met simultaneously, no stable circular orbits are allowed.

8-23. Start with the equation of the orbit:

and take its time derivative

2 sin sin r

Now from Equation (8.45) and (8.43) we have

so that from (2) ar ε πε

as desired.

r (b) (a) θ b r a) With the center of the earth as the origin, the equation for the orbit is

Also we know

Substituting (2) gives ε = 0.193. When θ = 0,


T he satellite is 1590 km above the earth. b) a – r β θ


Using tan 101 b ar a ar

1900 km above the earth

8-25. Let us obtain the major axis a by exploiting its relationship to the total energy. In the following, let M be the mass of the Earth and m be the mass of the satellite.


where pr and pv are the radius and velocity of the satellite’s orbit at perigee. We can solve for a and use it to determine the radius at apogee by

ra r r rv 1

we obtain , or 288 km above the earth’s surface. We may get the

aappmrvmrv= (7)

Substitution of the value of a found from (1) gives τ = 1.49 hours.


r r r

First, consider a velocity kick applied along the direction of travel at an arbitrary place in the orbit. We seek the optimum location to apply the kick. v∆ initial energy

final energy

GMm mv

GMm mv v

We seek to maximize the energy gain EE21−:

For a given , this quantity is clearly a maximum when v is a maximum; i.e., at perigee. v∆ Now consider a velocity kick applied at perigee in an arbitrary direction: V∆ v v

The final energy is

GMm mv r =

This will be a maximum for a maximum 2v; which clearly occurs when and are along the same direction. 1v ∆v

Thus, the most efficient way to change the energy of elliptical orbit (for a single engine thrust) is by firing along the direction of travel at perigee. an


8-27. By conservation of angular momentum a p mr v mr v or v r

(Parte 1 de 2)