# Mecflu - gravita??o, Exercícios de Meteorologia

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1-1

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.

149

150 CHAPTER 5

5-2. Inside the sphere the gravitational potential satisfies

( )2 4 G rφ π ρ∇ = (1)

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

( )22 1

4r G r r r

φ π ρ

∂ ∂  = ∂ ∂  r (2)

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

r φ∂ ∂

= constant ≡ C (3)

Therefore, (2) becomes

2

4 C

G r

π ρ= (4)

or,

2

C Gr

ρ π

= (5)

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 er R= to r = ∞ against the attractive gravitational force:

202 1 2e

e

R

M m G dr m

r

∞ =∫ v (1)

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

0 2 e

e

G M v

R = (2)

Substituting G , , , we have 11 3 26.67 10 m /kg s−= × ⋅ 245.98 10 kgeM = × 66.38 10 meR = ×

0 11.2 km/secv ≅ (3)

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

2

2 3 2

dx mk U F dx mk

x x = − = = −∫ ∫ 2 (1)

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

2 2

2 2

1 1 1 constant

2 2 2 k

x m x d

= = − = − 2 k

mE m (2)

Rewriting this equation in integrable form,

0 0

2 2 2

2 2

1 1d d

dx d x dx

k d xk x d

= − = − − −  

∫ ∫dt (3)

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

0

2 2

d

d t d x

k = − (4)

or

2d

t k

=

5-5. The equation of motion is

2 Mm

mx G x

= − (1)

Using conservation of energy, we find

2 1 1 2

x G M E G M x x

1

− = = − (2)

1 1

2 dx

GM dt x x

  = − − 

  (3)

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

( ) 1

21 1 2

x x

x x

xxdx x xGM

GM x x

∞ ∞

= − = − − 

−   

∫ ∫t d x

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

( ) 1sin 2

x

x

x x x x x

GM x t x

−∞ ∞ ∞

  = − − 

  (4)

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

152 CHAPTER 5

3 20

0 1

2x

x T dt

GM

∞ = =   ∫ (5)

2 3 2

1 2 1

1 2 2

x

x

x T dt

GM π∞

∞   = = +     ∫ (6)

Hence,

1 2

0

1 2T

T

π

π

+ =

Evaluating the expression,

1 2

0

0.818 T

T = (7)

or

1 2

0

9 11

T

T ≅ (8)

5-6.

α

θ

φ

z

x

y r

s

P

r2drd(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

( )22 cos coszdg G r dr d ds ρ

θ φ= − α (1)

where ρ is the density. Using cos

cos z r

s θ

α −

= and integrating over the entire sphere, we have

( ) ( )

1 2 2

3 22 2 0 1 0

cos cos

2 cos

a

z

z r r dr d d

r z rz

π θ ρ θ φ

θ

+

− = −

+ − ∫ ∫ ∫g G (2)

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

GRAVITATION 153

( )

( )

( ) ( )

1

3 22 2 1

1 1 22 2

1

cos cos

2 cos

2 cos cos

z r I d

r z rz

r z rz d z

θ θ

θ

θ θ

+

+

− =

+ −

∂ = − + − ∂

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

( ) 1

1 22 2

1

2

1 2 cos

2 2

I r z rz z rz

z z z

θ +

 ∂ = − − + − ∂  

∂  = − = ∂   (3)

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

3

2

3 2

2 2

3

4 1 3

z a

g G z

G a z

ρ π

π ρ

= − ⋅

= − (4)

But 3 4 3

a π

ρ is equal to the mass of the sphere. Thus,

2 1

zg GM z = − (5)

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

d G s

dx ρ

Φ = − (1)

where M

ρ = is the linear mass density. Integrating over the whole rod, we find the potential

2

2 22

1M G

x R− Φ = −

+ ∫ dx (2)

154 CHAPTER 5

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

2 2

2 2 2

22 2

2 4ln ln

2 4

RM GM G x x R

R

  + + 

   − + + = − Φ =   − + +   

2 2

2 2

4 ln

4

GM R

R

 + + Φ = −  

+ −   (3)

5-8.

z

z

y

x r

a

rdrdθdz

z0

θ

α r z z2 0 2+ −( )

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

( )

( ) ( )

22 0

0 3 222

0

cosz rdrd dz

dg G r z z

z z rdrd dz G

r z z

θ ρ α

θ ρ

= − + −

− = −

 + − 

(1)

where ρ is the density of the cylinder and where we have used ( ) ( )

0

22 0

cos z z

r z z α

− =

+ − . We can

find the net gravitational force by integrating (1) over the entire volume of the cylinder. We find

( )

2 0

3 222 0 0 0 0

a

z

z z rdr d dz

r z z

π

ρ θ −

= −g G  + − 

∫ ∫ ∫

Changing the variable to , we have 0x z z= −

0

0

3 22 2 0

2 za

z z

xdx g G rdr

r x π ρ

=  + 

∫ ∫ (2)

Using the standard integral,

GRAVITATION 155

( )3 22 2

1xdx

a xa x 2

− =

±± ∫ (3)

we obtain

( )2 2 220 00

2 a

z r r

dr r zr z

π ρg G    = − −  ++ − 

∫ (4)

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

( )22 202z a z a zπ ρ 20g G  = − + − − + +   (5)

Now, let us find the force by first computing the potential. The contribution from a small element of volume is

( )22 0

rdrd dz d G

r z z

θ ρΦ = −

+ − (6)

Integrating over the entire volume, we have

( )

2

22 0 0 0 0

a r d G dz d dr

r z z

π

ρ θΦ = − + −

∫ ∫ ∫ (7)

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

( ) (22 0 0 0

2 dz a z z zπ ρ )zd G  Φ = − + − − −  ∫ (8)

Now, we use Eqs. (E.11) and (E.8a), Appendix E, and obtain

( ) ( ) ( ) ( ) 2

2 20 2 2 0 0 0

2 2 2 2 2 20

0 0 0 0

2 ln 2 2 2

1 ln 2 2

2 2 2

z a G a z z z

z a a z z z a z

π ρ  −

2 a Φ = − − + − + − − + − +   

 + + − − + + − +   

Thus, the force is

156 CHAPTER 5

( ) ( ) ( )

( ) ( ) ( )

0 22 22

2 002 0 2 22 20 0 0 0

0 2 22 2 02 2 0

0 2 2 2 2 0 0 0

1 1 1

2 2 2 2

1 1 1 2 2 2

z

z

z az a g G a z

z a z z z a

z

z az a a z

a z z z a

π ρ

− − − +−∂Φ = − = − + − + +∂ + − − − + − +



− + − + − − + + − + +  

(9)

or,

( )22 202z a z a zπ ρ 20g G  = − + − − + +   (10

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 P

r

R

a

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

d

G r ρ

Φ = − ∫ (1)

where ρ is the linear mass density which is expressed as 2 M

a ρ

π = . Using

2 2 2 cosr R a aR θ= + − and d = adθ, we can write (1) as

2

2 2 02 2 cos

GM d

R a aR

π θ π θ

Φ = − + −

∫ (2)

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:

1 22

22 2

22 2

2 2

1 1 1 2 cos

2 cos

1 3 1 2 cos 2 cos

2 8

a a R R RR a aR

a a a a R R R R R

θ θ

θ θ

−   

= − −   + −  

     = + − + − +          

…1 (3)

GRAVITATION 157

If we neglect terms of order 3a

R     

and higher in (3), the potential becomes

2 2 2 2

2 2 0

2 2

2 2

3 1 cos cos

2 2 2

3 2

2 2

GM a a a d

R R R

GM a a R R

π

θ θ θ π

π π π π

  Φ = − + − + 

 

  − + 

 

= − (4)

or,

( ) 2

2

1 1

4 GM a

R R R  

Φ ≅ − +   

(5)

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

34 GMa

R − .

5-10.

P R

x

r

a R sin θ

R cos θ θ

φ

Using the relations

( )2 2sin 2 sin cosx R a aRθ θ= + − φ (1)

2 2 2 2 2cos 2 sin cosR R a aRr x θ θ= + = + − φ (2)

2 M

a ρ

π = (the linear mass density), (3)

the potential is expressed by

2

2 0

2

2 1 2 sin cos

d GM d G

r R a a R R

− − =Φ =

  − −   

∫ ∫ πρ φ

π θ φ

(4)

If we expand the integrand and neglect terms of order ( )3a R and higher, we have 1 22 2

2 2 2 2

1 3 1 2 sin cos 1 sin cos sin cos

2 2 a a a a a R R R R R

2

2θ φ θ φ −

   − − ≅ + − +     

θ φ (5)

Then, (4) becomes

158 CHAPTER 5

2 2

2 2 2

1 3 2 2 sin

2 2 2 GM a a

R R R π π π

π θ

  − − +Φ ≅  

 

Thus,

( ) 2

2 2

1 3 1 1 sin

2 2 GM a

R R R

θ   − − −Φ ≅     

(6)

5-11.

P ar

zdm θ

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

2 2 2 cos

dm dm d G G

r z a za θ Φ = − = −

+ − (1)

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:

2

2 2 2 0

2 sin1 4 2 cos

ave

a d dm

a z a za

π π θ θ π

d G θ

  Φ = − 

+ −  ∫ (2)

Making the variable change cos θ = x, we have

( )

1

2 2 12 2

ave

G dx d dm

z a zax

+

Φ = − + −

∫ (3)

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

( ) ( )

( ) ( )

2 2 2 21 12 2 2

2

ave G

d dm z a za z a za za

z a z aG dm

za

G dm

za

z

 Φ = − − + − + + +  

− − + +  = −   

= − (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.

O

P

dm

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

d r

Φ = − (1)

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

20 2

2 sin 4 2 cos

ave

dGdm

R r Rr

π d

π θ θ π θ

Φ = − + − ′′

∫ (2)

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

1

21 24 2 ave

Gdm du Gdm RR r Rurπ −

Φ = − = − + − ′′

d (3)

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

5-13.

R1

R2

R0

ρ2

ρ1

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 ( − ).

1 0R R R< < 2 0r R< e

0r R>

r

The magnitude of the force is

2 0

GMm F

R =

where M = mass for which 0r R<

160 CHAPTER 5

( )3 31 1 0 1 24 43 3M R R R 3π ρ π= + − ρ

So ( )3 3 31 1 2 0 2 12 0

4 3 r

Gm R R R

R π

ρ ρ ρ= − + −F e

( ) 31 2 1

2 02 0

4 3 r

R Gm R

R

ρ ρ π ρ

 − = − + 

  F e

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

2

2 2 3 3

3 3 4 4

4 M M

r dr r dr R R

ρ π π π

 = = =   r dr

dm

Gm

r φ = − where

3

3

r m M

R =

So

2 2

3 3 0

2 4

6 0

3

3

R

r

R

U du

Mr dr GMr R R

GM r dr

R

=

=

    = −       

= −

23

5 GM

U 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

3 2

4 3r

Gm F

r rπ ρ=     where ρ is the mass density of the Earth. The equation of motion is

GRAVITATION 161

32 4 3r

Gm F mr r

r π ρ = = −   

or

where 2 0r rω+ = 2 4

3 Gπ ρ

ω =

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

2 3

T G

π π ω ρ

= =

Substituting in values gives a period of about 84 minutes.

5-16.

z

y

x

M

h

r

θ r h2 2+

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, 0x yF F= =

( )2 20 cosz z

r

GMdm F dF

r h θ

=

= = +∫ ∫

With 2sdm rdrρ π= and 2 2s h

r h θ =

+ co

we have

( )

( )

3 22 2 0

1 22 2

0

2

1 2

2

z s r

z s

z s

rdr F GMh

r h

F GMh r h

F GM

πρ

πρ

πρ

=

= +

   = −  + 

=

Th

e sphere attracts the sheet in the -direction

with a force of magnitude 2 s z

GMπρ

162 CHAPTER 5

5-17.

y

x

Earth

moon (not to scale)

water

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

3 3 2 m e

x

GM x GM x D R

g = − ; 3 m

y

GM y GM y g

D R = − − 3

e (1)

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:

max

2 max

3 30

2 2

x m e

x

GM GM x g dx

D R  = −  ∫ ;

max 2 max

3 30 2

y m e

y

yGM GM g dy

D R = − − ∫

  (2)

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

( ) 2

2 2max max max max3 32 2

m ey GMx x D R

  + = − 

  2y

GM (3)

The right-hand side can be factored as

( ) ( ) ( ) ( )max max max max3 22 2 eGM GMx y x y R h g

R R + − = 3

e h=

2R

(4)

If we make the approximation on the left-hand side of (3) that , we get exactly Equation (5.55). Turning to the exact solution of (3), we obtain

2 2 max maxx y

3 3 3 3

3 3 3 3

2

2 2

e m e

e m e

M M M M R D R Dh R M M M M R D R D

+ − − =

+ + −

m

m

(5)

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 (5.55), we have with the appropriate substitutions

2

33 moon

2 sun

3

3 2

3 2

m

m es

s s

es

GM r gDh M

GM rh M gR

R D  = =   

(1)

Substitution of the known values gives

322 11

moon 30 8

sun

7 350 10 kg 1 495 10 m 2 2

1 993 10 kg 3 84 10 m h h

 . × . × =  . × . × 

. (2)

5-19.

ωearth ωmoon

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

1 1 1 2T T T

= − (1)

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

( ) ( ) 2

2 2 2 2 22 2 2 2

2 2 ( ) ( ) ( )

G rdrM GM d r GM d z z d z z R z

R R Rz r z r

π π

− − − Φ = = ⇒Φ = Φ = + −

+ + ∫

Then one can find the gravity acceleration,

2 2

2 2 2

2ˆ ˆ( ) d GM z R

k k dz R z R

 z g z

Φ + − = − = −  

 +

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

164 CHAPTER 5

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

( )

2 2 ( )

L D

D

G M L mdx GMm dx GMm dF F dF

x L x

+

= ⇒ = = = +∫ ∫ D L D

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

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