**Mecflu - din?mica de um sistema de part?culas**

respostas do Marion

CHAPTER 5 Gravitation

5-1. a) Two identical masses: The lines of force (dashed lines) and the equipotential surfaces (solid lines) are as follows:

b) Two masses, +M and –M:

In this case the lines of force do not continue outward to infinity, as in a), but originate on the “negative” mass and terminate on the positive mass. This situation is similar to that for two electrical charges, +q and –q; the difference is that the electrical lines of force run from +q to –q.

Since ρ(r) is spherically symmetric, φ is also spherically symmetric. Thus,

The field vector is independent of the radial distance. This fact implies

Therefore, (2) becomes

or,

5-3. In order to remove a particle from the surface of the Earth and transport it infinitely far away, the initial kinetic energy must equal the work required to move the particle from erR= to r = ∞ against the attractive gravitational force:

Mm Gd r m

where eM and are the mass and the radius of the Earth, respectively, and is the initial velocity of the particle at . eR 0v er R=

Solving (1), we have the expression for : 0v

5-4. The potential energy corresponding to the force is dx mk

The central force is conservative and so the total energy is constant and equal to the potential energy at the initial position, x = d:

GRAVITATION 151

x m xd

Rewriting this equation in integrable form, 0 dx d x dx

k dxk where the choice of the negative sign for the radical insures that x decreases as t increases. Using Eq. (E.9), Appendix E, we find

or t k =

5-5. The equation of motion is

2Mm mx G x

Using conservation of energy, we find

2 xG M E GM

112dx GM

where is some fixed large distance. Therefore, the time for the particle to travel from x∞x∞ to x is xxdx xxGM

GM x ∞ ∞

∫∫td x

Making the change of variable, , and using Eq. (E.7), Appendix E, we obtain 2xy→ x x x x x

GM x

If we set x = 0 and 2xx∞= in (4), we can obtain the time for the particle to travel the total distance and the first half of the distance.

x Td t

x Td t

Hence,

Evaluating the expression,

or

5-6.

y r s P rdrd(cos θ)dφ

Since the problem has symmetry around the z-axis, the force at the point P has only a z-component. The contribution to the force from a small volume element is

()2 cos coszdg G r dr d d s where ρ is the density. Using cos cos zr s θα−= and integrating over the entire sphere, we have cos cos

2c os az zr r dr d d

Now, we can obtain the integral of cos θ as follows:

GRAVITATION 153 cos cos

2c os

2c os cos zr Id rz rz rz rz d z

Using Eq. (E.5), Appendix E, we find

2c os22

Ir z rz zr z z z

Therefore, substituting (3) into (2) and performing the integral with respect to r and φ, we have z a gG

Ga z

But 34 3 aπρ is equal to the mass of the sphere. Thus,

Thus, as we expect, the force is the same as that due to a point mass M located at the center of the sphere.

5-7. dx

PR sx

The contribution to the potential at P from a small line element is dG s dxρΦ=−A (1) where Mρ=A is the linear mass density. Integrating over the whole rod, we find the potential

Using Eq. (E.6), Appendix E, we have

24ln ln

RMG M Gx x R

4 ln

4 GM R

5-8.

z z

x r a rdrdθdz

Since the system is symmetric about the z-axis, the x and y components of the force vanish and we need to consider only the z-component of the force. The contribution to the force from a small element of volume at the point (r,θ,z) for a unit mass at (0,0,0z) is cosz rdrd dz dg G rz z z zr drd dz G

rz z where ρ is the density of the cylinder and where we have used cos z

. We can find the net gravitational force by integrating (1) over the entire volume of the cylinder. We find az z rdr d dz

Changing the variable to , we have 0xzz=−

z z xdx gG rdr rx πρ −

Using the standard integral,

GRAVITATION 155

we obtain

2 az r dr

rzrz πρgG

Next, using Eq. (E.9), Appendix E, we obtain

Now, let us find the force by first computing the potential. The contribution from a small element of volume is rdrd dz dG rz z θρΦ=− +−

Integrating over the entire volume, we have a r dG dz d dr

rz z

Using Eq. (E.9), Appendix E, again, we find

Now, we use Eqs. (E.1) and (E.8a), Appendix E, and obtain z a Ga z z z

z a az z z a z

Thus, the force is zaz a gG a z z az z z a

zaz a az az z z a

A (9) or,

and we obtain the same result as in (5).

In this case, it is clear that it is considerably easier to compute the force directly. (See the remarks in Section 5.4.)

5-9.

θ PrR

The contribution to the potential at the point P from a small line element dA is d G

where ρA is the linear mass density which is expressed as 2Maρπ=A. Using

222cosrRaaRθ=+− and dA = adθ, we can write (1) as

GM d

This is the general expression for the potential. If R is much greater than a, we can expand the integrand in (2) using the binomial expansion:

2c os

GRAVITATION 157

If we neglect terms of order

3 1c os cos

GM a a a d

or,

We notice that the first term in (5) is the potential when mass M is concentrated in the center of the ring. Of course this is a very rough approximation and the first correction term is 2 34GMaR −.

5-10.

xr a

R sin θ

R cos θθ φ

2Maρπ=A (the linear mass density), (3) the potential is expressed by 2 d GM d G

rR a

(4)

If we expand the integrand and neglect terms of order ()3 aR and higher, we have

2 213 1 2 sin cos 1 sin cos sin cos θφ (5)

Then, (4) becomes

Thus,

1 sin

5-1.

P ar zdm θ

The potential at P due to a small mass element dm inside the body is

2 2c os dm dm dG G

Integrating (1) over the entire volume and dividing the result by the surface area of the sphere, we can find the average field on the surface of the sphere due to dm:

2s in1

4 2c os ave ad dm a za za

Making the variable change cos θ = x, we have

ave

G dx d m

Using Eq. (E.5), Appendix E, we find ave G d m z a za z a za za

za z aG dm

G dm zaz

=− (4)

This is the same potential as at the center of the sphere. Since the average value of the potential is equal to the value at the center of the sphere at any arbitrary element dm, we have the same relation even if we integrate over the entire body.

GRAVITATION 159 5-12.

Rr r'

Let P be a point on the spherical surface. The potential dΦ due to a small amount of mass dm inside the surface at P is

Gdm

The average value over the entire surface due to dm is the integral of (1) over dΩ divided by 4π. Writing this out with the help of the figure, we have

4 2c os ave dGdm

Rr Rr πd π θθ

Making the obvious change of variable and performing the integration, we obtain 1

Gdm du Gdm

We can now integrate over all of the mass and get aveGmRΦ=−. This is a mathematical statement equivalent to the problem’s assertion.

5-13.

0R= position of particle. For , we calculate the force by assuming that all mass for which is at r = 0, and neglect mass for which . The force is in the radially inward

direction (−).

The magnitude of the force is

GMm where M = mass for which 0rR<

MRRR3πρπ=+−ρ

R R R πρρρ=−+−Fe

Gm R

Fe

5-14. Think of assembling the sphere a shell at a time (r = 0 to r = R).

For a shell of radius r, the incremental energy is dU = dm φ where φ is the potential due to the mass already assembled, and dm is the mass of the shell.

So r dr dm

Gmr φ=− where 33r mM R

So

r R

Ud u

Mrd r GMr R

GM rd r

5-15. When the mass is at a distance r from the center of the Earth, the force is in the inward radial direction and has magnitude rF:

m r

3r Gm F r rπρ = where ρ is the mass density of the Earth. The equation of motion is

GRAVITATION 161

3r Gm Fm r r or

This is the equation for simple harmonic motion. The period is

Substituting in values gives a period of about 84 minutes.

5-16.

x M h

For points external to the sphere, we may consider the sphere to be a point mass of mass M. Put the sheet in the x-y plane.

Consider force on M due to the sheet. By symmetry, 0xyFF==

coszz r

GMdm

Fd F rh θ

With 2sdmrdrρπ= and 2 s hrh θ=+co we have zs r

zs zs rdr FG Mh

FG Mh rh

Th e sphere attracts the sheet in the -direction with a force of magnitude 2sz GMπρ

5-17. y

Earth moon (not to scale) water

Start with the hint given to us. The expression for and xgyg are given by

GM x GM x

DR g=−; 3my

GM y GM y

where the first terms come from Equations (5.54) and the second terms come from the standard assumption of an Earth of uniform density. The origin of the coordinate system is at the center of the Earth. Evaluating the integrals:

2 max mex GM GM x mey yGM GM

To connect this result with Example 5.5, let us write (1) in the following way m ey GM x x

The right-hand side can be factored as

2R (4) If we make the approximation on the left-hand side of (3) that , we get exactly

Equation (5.5). Turning to the exact solution of (3), we obtain em e

em e

RD R DhR M M M

Upon substitution of the proper values, the answer is 0.54 m, the same as for Example 5.5. Inclusion of the centrifugal term in does not change this answer significantly. xg

GRAVITATION 163

5-18. From Equation (5.5), we have with the appropriate substitutions m m es

s s es

GM r gDhM

GM rhM gR

Substitution of the known values gives

5-19.

Because the moon’s orbit about the Earth is in the same sense as the Earth’s rotation, the difference of their frequencies will be half the observed frequency at which we see high tides. Thus

tides earth moon which gives T 12 hours, 27 minutes. tides

5-20. The differential potential created by a thin loop of thickness dr at the point (0,0,z) is

Gr drM GM d r GM dz z d z z R z

Then one can find the gravity acceleration, dG M z R k k

where is the unit vector in the z-direction. k

5-21. (We assume the convention that D > 0 means m is not sitting on the rod.)

The differential force dF acting on point mass m from the element of thickness dx of the rod, which is situated at a distance x from m, is

GM L mdx GMm dx GMm dF F dF

xL x

And that is the total gravitational force acting on m by the rod.