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Physics 181

Assignment 4

Solutions

1. A sphere has within it a gravitational field given by g = !g

r

r, where g is constant

and r is the position vector of the field point relative to the center of the sphere. This field has constant magnitude and is directed toward the center.

a. Find the potential !(r) at points in the sphere, taking !(0) = !0 . Check

that your answer obeys !" = #g .

b. Find the mass density !(r) from Gauss’s law ! "g = #4$G% .

c. Let the sphere have radius R. Find its total mass. d. Find the field g(r) at a point outside the sphere ( r > R ).

e. Find the potential at a point outside the sphere.

f. Determine !

0 so that !(r)" 0 as r !" .

a. We have !(r) " !0 = " g #dr

0

r

$ == gr #dr

r0

r

$ . But r !dr = r dr , so we find

!(r) = !0 + gr .

b. ! "g = #igi = $g#i(ri /r) . But

!

i(r

i/r) = (!

iri) " (1/r) + r

i" !

i(1/r) , where

!

iri= 3 and

!

i(1/r) = "(1/r

2 ) # (!ir) = "(1/r

2 ) # (ri/r) . Using

riri= r

2 and putting the pieces together we find

! "g = #2g/r . Using Gauss’s law we

have ! =

g

2"G

1

r.

c. The mass is given by M = !dV" where the integral covers the volume of

the sphere. Since ! depends only on the distance from the center, we can calculate easily the mass in a thin spherical shell of radius r and thickness

dr: dM = ! "dV =

g

2#G

1

r" 4#r2dr =

2g

G" rdr . We integrate this over r to get

the total mass: M =

2g

Gr dr

0

R

! =gR2

G.

d. As with all spherically symmetric cases, at points outside all the mass the fields are just those of a point mass with the total mass, placed at the

center of symmetry. In this case, for r > R , g = !GM

r

r3= !gR2 r

r3.

2

e. Since we have fixed the potential at r = 0 to be !

0, we must start there in

calculating the potential elsewhere. First we find the value at the surface

of the sphere: !(R) = !0 " g #dr

0

R

$ = !0 + g dr0

R

$ = !0 + gR . (We could use the answer to (a) for this part.) Then for points outside the sphere we find

!(r) = !(R) " g #dr

R

r

$ = !0 + gR + gR2 dr

r2R

r

$ = !0 + 2gR " gR2 /r .

f. If !" 0 as r !" , then !

0= "2gR .

2. A sphere with uniform mass density ! and

radius R has a spherical cavity in it, of radius

R

0, centered at distance a from the

center of the massive sphere. We are interested in the field within the cavity. To get at it, we will calculate the potential in the cavity first. a. Consider first the sphere without the

cavity. Use the integral form of Gauss’s law,

g !ndS!" = #4$G %dV"

to find g(r) at points inside the sphere. [Choose for the surface a sphere of radius r about the center of the actual sphere. Be careful about directions and signs.]

b. Find the potential !(r) at a point inside the sphere, taking ! to vanish at infinite distance. [The integral is in two parts, one for r from ! to the surface ( r = R ), the other from R to the location of the field point.]

The fields in the cavity can be thought of as a superposition of two fields: (1) that of the solid sphere of density ! ; (2) a “solid” sphere of negative density !" . The densities cancel, giving the cavity. (It’s a trick, of course.)

c. The drawing shows a close-up view of the cavity and an arbitrary field point in it, which is at distance r from the center of the solid sphere and at distance !r from the center of the cavity. By the law of cosines we have

a

r r’ !

x

y

a

3

!r2= r

2+ a

2" 2ar cos# . Find the total potential at the field point, as

a function of r and ! , or of x and y. (You may lump all the additive constants in the potential into one.)

d. Use g = !"# to find the field at the field point. You should find that it is uniform, and entirely in the negative x-direction.

e. Obtain the same result directly by using the answer to (a) and adding the g fields of the positive and negative density spheres.

a. By the spherical symmetry, g has the same magnitude at all points on the sphere suggested, and is pointing directly inward, so

g !ndS = "g(r)dS .

We have

g !ndS!" = #g(r) dS!" = #g(r) ! 4$r2 . The other side of the equation is !4"G times the mass enclosed by the sphere of radius r, which is

4

3!r

3" . We find therefore g(r) = 4

3!G" # r . Since the field is toward the

center, the vector formula is g(r) = ! 4

3"G# $ r .

b. Since we chose the potential at infinity to be zero, we must start there to

find its value elsewhere. So we calculate !(r) = " g #dr

$

R

% " g #drR

r

% . For the first integral we use the field outside the sphere, which is

g(r > R) = !GM

r

r3= ! 4

3"GR

3#r

r3

. For the second integral we use the

answer to (a). The result is ! 2

3"G#(3R

2 ! r2 ) .

c. At the point shown the contribution of the sphere of positive density is

!1(r) = 2

3"G#r

2+ const. That of the sphere of negative density is

!2( "r ) = # 2

3$G% "r 2

+ const. The total is ! =

23"G#(r2 $ %r 2 ) + const. Using

the law of cosines we find r2! "r

2= 2racos# ! a

2 , and using the fact that

r cos! = x we find ! =

4

3"G#ax + const.

d. From g = !"# we find easily g = ! 4

3"G#a $ i . This is in the negative x-

direction (towad the center of the solid sphere) and uniform in magnitude. e. We use the same drawing, with

vectors indicating the directions of the contributions. We have

g = g1 + g2 =

43!G"(r + #r ) = $ 4

3!G"a .

Since a = !ai , this is the same as (d).

a

r r’ !

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3. A satellite has speed vp at perigee and

v

a at apogee.

a. Find the eccentricity ! of the orbit in terms of these speeds. b. It is desired to make a small increase in the energy of the orbit. To do this

the engine can be fired in a short burst providing a force F in any direction. Explain why the largest increase in the energy is accomplished by firing the engine when the satellite is at perigee, and directing the force parallel to the satellite’s velocity at that point.

c. Let the energy added be dE , a small fraction of the magnitude E of the

original orbit energy. Show that to first order in small quantities, the fractional change da/a of the semi-major axis is equal to

dE/ E . [Since

the total energy is negative, increasing it means making it closer to zero.] d. Show that the orbit becomes less circular, i.e., that ! increases. [Hint: there

is essentially no change in the perigee distance rp during the brief time

while the engine is fired.] e. Suppose that when this satellite reaches apogee it encounters a second

satellite moving in a circular orbit of radius equal to ra

. Which satellite will be moving faster? Explain. [You may assume they have the same mass, because the mass doesn’t matter in determining the speed at a certain point.]

a. By conservation of angular momentum, mrpvp = mrava , where m is the

satellite’s mass, rp is the perigee distance and

ra

is the apogee distance. Thus

vp /va = ra /rp . Now by definition

rp = a(1 ! ") and

ra= a(1 + !) ,

where a is the semi-major axis. So we find vp /va = (1 + !)/(1 " !) . Solving,

we find

! =

vp " va

vp + va

.

b. The power supplied by a force F acting on a particle that is moving with velocity v is P = F !v . For a given force this is clearly largest when v is largest (which is true at perigee) and when F is parallel to v.

c. Use the formula

E = !GMm

2a to find

dE = +GMm

2a2

da = E !da

a.

d. Easiest: rp = a(1 ! ") , so

! = 1 " rp /a . Since

rp is fixed while a increases,

! gets closer to 1. e. The value of a for the circular orbit is greater than for the elliptical one, so

its total energy is greater (closer to zero). Since the two are at the same point their potential energies are equal (if the masses are), so the kinetic energy (and thus the speed) of the one in the circular orbit is larger.

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4. The example worked out in the notes to Unit 3 can be solved using ordinary

Newtonian mechanics if one uses effective gravity.

Recall the situation. A small block of mass m slides without friction down a wedge of mass M, which is on a frictionless floor. The wedge also slides across the floor. The figure shows the arrangement.

a. Let the acceleration of the wedge be a

w. We will analyze the situation in a

reference frame moving with the wedge. In that frame, find the components of

geff in the x-y set of axes shown.

b. The motion of the block is more easily described in the x’-y’ set of axes shown. Find the components of

geff in that set. [This involves rotating the

axes; see the notes on linear algebra.] c. Draw the free body diagram showing the forces acting on the block (in the

reference frame we are using).

d. Find the acceleration of the block a

b (in terms of

a

w and known

quantities). e. Find the normal force exerted on the block by the wedge. f. If the frame we are using, the wedge is at rest. Relate the normal force in

(e) to the acceleration a

w. [Return to the x-y axes for this.]

g. Solve for a

w and

a

b in terms of given quantities. [You should get the same

answers as in the notes on Unit 3.]

a. In vectors geff = g ! aw , so we have

geff = !awi ! gj . In column form:

geff =!aw

!g

"

#$$

%

&''

.

b. The transformation matrix is

M =cos! sin!"sin! cos!

#

$%&

'(,

so we have

cos! sin!"sin! cos!

#

$%&

'("aw

"g

#

$%%

&

'((=

"aw cos! " gsin!

aw sin! " gcos!

#

$%%

&

'((

.

In vector notation geff = (!aw cos" ! gsin") #i + (aw sin" ! gcos") #j .

θ

x’

x

y

y’

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c. Drawing as shown. The component of geff along the slope is the block’s

acceleration. The component of mgeff perpendicular to

the slope cancels the normal force N. d. The magnitude of the block’s acceleration down the

slope is ab = gsin! + aw cos! .

e. We have from the free body diagram N = m(gcos! " aw sin!) .

f. The horizontal component of the normal force exerted by the block on the wedge (the reaction to the one in the free body diargram) is cancelled by the horizontal component of

Mgeff , so we have

N sin! = Ma

w, and

rearranging this we find (M + msin2

!)aw = mgsin! cos! .

g. We have from (f) aw =

mgsin! cos!

M + msin2!

. Then from (d) we find after some

algebra ab =

M + m

M + msin2!

gsin! . These are the same answers we got using

the Lagrangian approach.

5. Two questions about the Coriolis force.

a. In some locations on the earth’s equator there are demonstrations for tourists. Standing north of the line, water is observed to rotate counter-clockwise as it goes down a drain; a few steps away, south of the line, the water rotates clockwise going down the drain. Is it the Coriolis force that makes this happen? Explain.

b. There were two major battles between the German and British navies in World War I. The first took place near Jutland, about 55° N latitude, the last major battle between large battleships. The other took place near the Falkland Islands, about 50° S latitude.

Consider a shell fired due north in the Battle of Jutland, with initial speed

v

0= 800 m/s and elevation angle 45°. Use a coordinate system with origin

where the shell is fired from the surface of the earth; let the x-axis run north along the surface, and the y-axis be upward perpendicular to the surface. (The z-axis will run east along the surface.) 1. Write (in components) the earth’s angular velocity ! , the shell’s

velocity v

s(t) (neglecting the Coriolis effect), and the acceleration

a

c(t) produced by the Coriolis effect.

2. Find the contribution v

c(t) of the Coriolis acceleration to the shell’s

velocity.

N

mgeff

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3. Find the deflection rc(t) caused by the Coriolis effect.

4. How long (in s) is the shell in the air? (Neglect the Coriolis effect.) 5. Find the total deflection (in m) due to the Coriolis effect.

a. No. The drains are designed to produce this difference, but many tourists believe the scam. The Coriolis force is exactly zero at the equator, so it is negligbly small near it. In fact it is too small to produce any significant effect on water going down drains anywhere.

b. Shown is the situation. We see that without the Coriolis effect the motion would be in the x-y plane.

1. The shell’s velocity is

vs = v0 cos! " i + (v0 sin! # gt) " j ,

where ! = 45° . The earth’s angular velocity is

! = cos" # i + sin" # j . The acceleration produced by the Coriolis force is

ac = 2vs ! " = 2"[v0 sin(# $%) + cos# & gt)] &k .

2. We integrate a

c with respect to time to find

vc = 2![v0 sin(" #$) % t + 1

2cos" % gt2] %k .

3. Integrate again to find the displacement. (The time of flight is T.)

rc = vc dt

0

T

! ="[v0 sin(# $%) &T2+

13

gcos# &T3] &k .

4. The time of flight is twice the time to reach maximum height, or

T =2v

0sin!

g" 115 s.

5. Using ! = 2" /(24 #3600) $ 7.27 % 10&5 s&1 and the given data we find

rc! 340.7 m. This is a small angular deflection (the range is

about 65 km) but it is enough to make the shell miss its target.

x y

!

v

0

! = 55°