Solutions to problems in Leighton and Vogt's "Exercises in Introductory Physics" textbook by Pier Franco Nali

Solutions to problems in Leighton and Vogtâ&#x20AC;&#x2122;s textbook "Exercises in Introductory Physics" (Addison-Wesley, 1969) Pier F. Nali (Revised April 10, 2016)

CHAPTER 1 Atoms in Motion B-3

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

nG ∼

N0 6.025 × 1023 molecules / mole ∼ ∼ 1019 molecules/cm3 . 3 3 Molar_vol._at_STP 22.41×10 cm / mole

nL liquid _ air _ density 1.0 g cm −3 ∼ ∼ ∼ 103 ⇒ nL ∼ 1022 molecules/cm3 . −3 nG air _ density 0.001 g cm b)

m∼

air _ density 0.001 g cm −3 ∼ ∼ 10 −23 g/molecule . nG 2.7 × 1019 molecules / cm3

c) Said σ the (approx.) diameter of an air molecule the inverse of the mean free path is 1 1 ∼ σ 2 ⋅ nG , where σ ∼ 3 ∼ 10−7 cm , assuming for a rough estimate that the molecules are condensed nL in liquid air in such a way that each molecule occupies a cubic cell of side σ . Thus 2 1 ∼ (10 −7 cm ) × 1019 molecules/cm3 ∼ 105 cm −1 or, calculating the reciprocal, the mean free path ∼ 10−5 cm . From P : P (

STP )

∼ nG : nG( STP ) ∼ ( STP ) : ,

P ( STP ) ∼ 1 atm, ( STP ) ∼ 10−5 cm, ∼ 1 m, one gets P ∼ 10−7 atm .

B-4

Said P the pressure of argon (A) gas spread in a layer of volume V (at constant temperature), we have P P(

STP )

Molar_vol._at_STP P ( STP ) × Molar_vol._at_STP 760 mm Hg × 22.41×103cm 3 /mole ⇒V = = = V P 6.0 × 10−4 mm Hg

= 2.839 ×1010 cm3 /mole

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

and

(

)

number _ density A _ atoms _ per _ cm3 =

N 0 6.025 × 1023 atoms/mole = = 2.123 × 1013 atoms / cm 3 . V 2.839 × 1010 cm3 /mole

Said A a constant of proportionality representing the effective target area per A atom and expressing (on experimental basis) the percent intensity reduction of the atomic beam of K atoms as A × number _ density × thickness = 3.0 % one gets

3.0 % 3.0 × 10−2 = ≈ 1.4 × 10−14 cm 2 /atom . number _ density × thickness 2.123 × 1013 atoms/cm3 × 10−1 cm

B-5

2 x interatomic spacing s

Cl….

Cl Na

Na….

( N − 1) × ( N − 1) × ( N − 1) spacings between nearest 3 neighbours (Cl and Na ions in alternance). Then lattice _ vol. = ( N − 1) ⋅ s 3 ≈ N 3 s 3 (for large N ) and The cubic lattice has N × N × N points and

° 3  N3 1 1 1 number _ density  atoms / A  ≈ 3 3 = 3 = = cm −3 . 3 −24 ° N s s 22.4258 × 10      2.820 A    Since the NaCl molecule is diatomic 1 number _ density × molecular _ weight _ of _ NaCl density _ of _ NaCl = 2 , N0

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

where: density _ of _ NaCl = 2.165 g/cm3 , molecular _ weight _ of _ NaCl = 58.454 g/mole, and hence 1

N0 =

number _ density × molecular _ weight _ of _ NaCl = 6.02 × 1023 molecules/mole . density _ of _ NaCl

B-6

The number of helium atoms expelled per second by one gram of radium in equilibrium with its disintegration products is 13.6 ×1010 atoms g −1 s −1 = 117.504 × 1014 atoms g −1 day −1 . The corresponding volume of helium produced per day by one gram of radium is −3 3 −1 0.0824 × 10 cm day = 4.291×10−4 cm3 g −1 day −1 and thus the number density of helium atoms at STP −3 192 ×10 g is a)

( STP ) nHe =

117.504 × 1014 atoms g −1 day −1 ≈ 2.7 × 1019 atoms/cm3 4.291×10−4 cm3 g −1 day −1

and the calculated Avogadro’s number b) ( STP ) N 0 = nHe × Molar_vol._at_STP ≈ 2.7 × 1019 atoms cm −3 × 22.41× 103 cm3 /mole ≈ 6.1× 1023 atoms/mole .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

C-1

The molecular weight of H(CH2)18COOH is approximately 298 g/mole and the molar volume is

molecular _ weight 298 g mole −1 ≈ ≈ 372.5 cm 3 /mole . −3 density 0.8 g cm The volume of oil is deduced by dividing 0.81 mg by its density, i.e.

0.81×10−3 g ≈ 1.0 × 10−3 cm 3 . 0.8 g cm −3

The surface area over which this volume of oil is spread is 14 π × 842 cm 2 ≈ 5.542 × 103 cm 2 , so the thickness of the oil film (calculated as if its density were the same as in non spread state) is 0.81×10 −3 g 0.8 g cm −3 1.0 ×10 −3 cm3 l≈ ≈ ≈ 0.2 × 10−6 cm . 2 2 3 2 1 5.542 × 10 cm 4 π × 84 cm If the oil molecules were arranged in linear chains 20 atoms high ( ≈ thickness l of the oil film) × 4 atoms wide ( ≈ 15 × l ) so that a molecule occupies a volume roughly estimated (assuming in liquid state a rotational degree of motion around the main axis of the chain) as

(

0.2 × 10−6 cm l3 1 1 l× 5l× 5l = ≈ 25 25

)

≈ 3.2 × 10 −22 cm3 .

N0 . So Molar _ vol. N 0 ≈ 0.3 × 1022 molecules/cm3 × Molar _ vol. ≈ 0.3 × 1022 mol./cm3 × 372.5 cm 3 /mole ∼ 10 24 molecules/mole.

The reciprocal of the latter ( 0.3 × 1022 molecules/cm3 ) is the number density

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

C-2

. Said m the mass of a gas molecule the gas density is ρ ( gas ) = m ⋅ N g . If one assume for sake of simplicity that in a solid the molecules are tightly packed in cells of side σ then ρ ( solid ) ≈ m σ 3 (1). Thus

ρ ( gas ) 1 ρ ( gas ) ≈ N g ⋅σ 3 ⇒ σ ≈ 3 ⋅ ⇒ N gσ 2 ≈ 3 N g N g ρ ( solid ) ρ ( solid )

 ρ ( gas )  ⋅  .  ρ ( solid )  3

(1)

For our purposes here we may ignore that in closed packaging configuration the proportion of space occupied by spherical nr _ of _ particles _ per _ cell × V( particle ) π particles is ∆ = = ≈ 74, 048% and therefore the number of particles per V( unitcell ) 3 2 unit cell is ∆ ⋅

V( unitcell ) V( particle )

σ = 2 particles/cell rather than 1 particle/cell, ρ ( solid ) is 3 3 2 π6 σ ⋅

m σ and so on, so the subsequent results should contain

2 ⋅ m σ rather than

in locus of N g . This is the densest packing of equal spheres 2 in three dimensions, as proved by Gauss in 1831. See Gauss, C. F. "Besprechung des Buchs von L. A. Seeber: Intersuchungen über die Eigenschaften der positiven ternären quadratischen Formen usw." Göttingsche Gelehrte Anzeigen (1831, July 9) 2, 188-196, 1876 = Werke, II (1876), 188-196. For a proof of this result see, e.g., J. W. S. Cassels, —An Introduction to the Geometry of Numbers, Springer-Verlag, 1971 [Chap. II, Th. III]

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics Comparing = 1

N gσ 2 = 1

(

)

2π N gσ 2 and η =

)

2π = ρ ( gas ) ⋅ v

(

ρ ( gas ) ⋅ v ⋅ one also gets

)

2π ⋅ 3η .

Comparing again the right-hand sides of latter two expressions for N gσ 2 and cubing the second members one gets 2

 ρ ( gas )  3 3 Ng ⋅   ≈  ρ ( gas )  ⋅ v  ρ ( solid ) 

(54

2π 3 ⋅η 3

)

and finally – assuming on experimental basis(2) ρ ( solid ) ≈ ρ ( liquid ) -

Ng ≈

(2)

ρ ( gas )  ρ ( liquid )  ⋅ v3 3

54 2 ⋅ π ⋅η

∼ 0.7 × 1019 molecules/cm3 .

This assumption is valid where we are dealing only whit orders of magnitude and is based on the observed condensation behaviour of many substances. Unfortunately, at the times of Loschimdt’s work (1865) was not yet known how to liquefy the air, so he could not directly observe condensation behaviour for air. However, the density of a substance in the condensed state can be estimated by its chemical composition, as Loschmidt did. See Porterfield, William W., and Walter Kruse. "Loschmidt and the Discovery of the Small.", J. Chem. Educ. , 1995, 72 (10), p 870

Solutions to problems in Leighton and Vogt’s textbook "Exercises in Introductory Physics" (Addison-Wesley, 1969) Pier F. Nali (Revised April 10, 2016)

CHAPTER 2 Conservation of Energy, Statics

According to the principle of virtual work (PVW) “the sum of the heights times the weights does not change”, said h1 and h2 the heights of the weights W1 and W2 respectively, then

∆ (W1h1 + W2 h2 ) = W1∆h1 + W2 ∆h2 = 0 . If we assume that the weight W1 goes down ∆h1 as the weight W2 ∆h1 = − 1 ∆h2 ⇒ W1 1 = W2 2 . 2

rises ∆h2 , then

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

The “sum of the heights times the weights” for an arbitrary number of weights becomes ∑ Wi ∆hi = 0 i

and considering the distances positive or negative depending on the side of the fulcrum, so that ∆h j ∆hi = i ∆h j (or ∆hi ∝ i for any i, being independent of j), it becomes thus ∑ Wi i = 0 . j j i

 n  ∑i  ∑j Fij i∆ri = 0 , where the sum over j is   extended to all the n forces acting on the mass i and the sum over i is extended to all the N masses.  n  In the present case (a single body) N = 1 , so the above formula reduces to  ∑ Fi i∆r = 0 .  i  a) For n = 1 is F i∆r = 0 ; so, for the arbitrariness of the choice of the vector ∆r , F = 0 . N

The PVW can be expressed in a more general form as

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

b) For n = 2 is ( F1 + F2 )i∆r = 0 and, again for the arbitrariness of the choice of the vector ∆r , is F1 + F2 = 0 , whence F1 = −F2 ; so, the two forces are equal in magnitude, opposite in direction and collinear. c) For n = 3 is, again, F1 + F2 + F3 = 0 . If we consider the plane on which lie the two vectors F1 and F2 (apart from the trivial case in which the two vectors are parallel, which degenerates to the case of

n = 2 ), F3 also lies in the same pane, being expressible as a linear combination of the other two

vectors. Let X 0 the vector joining the origin to the point of intersection of the lines of action of the two vectors F1 and F2 and X the vector joining the origin to an arbitrary point of one or the other line of action. Then the two lines of action are defined by the system of equations  ( X − X0 ) ∧ F1 = 0 .  ( X − X0 ) ∧ F2 = 0 By adding the latter two equation, immediately follows that X0 also lies on the line of action of F3 . X0 defines a single point (through which the three lines pass) as each line of action cannot intersect each other in more than one point (provided the lines are not parallel).  n   n  d) From Fi i∆r = Fi ∆ r⋅ cos ∆ i it follows that  ∑ Fi i∆r =  ∑ Fi cos ∆i  ⋅ ∆r = 0 , and – again for the  i =1   i =1  n

arbitrariness of the choice of ∆r -

∑ F cos ∆ i

i =1

= 0.

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

A-1

a’

c’

b’

a b

45°

∆θ

x W

Suppose the weight W goes down x as the centre of mass of the plank rises y. Therefore (refer to the above figure) y = L2 sin ∆θ , x = − ( c − c′ ) and a′ = a − L sin ∆θ , b′ = b + L (1 − cos ∆θ ) , c ′ = a ′ 2 + b′ 2 . L L +b So – assuming ∆θ small - c′ = c 1 − 2 aL sin ∆θ + 4 ( c2 ) sin 2 c2

∆θ 2

(

)

≅ c 1 − aL sin ∆θ and x ≅ − aLc sin ∆θ , c2

where we have used the trigonometric double-angle formula 1 − cos α = 2 sin 2 1+ x ≅ 1+

x for x small. 2

α 2

and the approx.

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

Being

also

c = 2 ⋅a

follows

x≅−

L 2

sin ∆θ = − 2 ⋅ y .

From

PVW

5 then

have

W( plank ) y 3 W ⋅ ( − x ) = W( plank ) ⋅ y , whence W = − W( plank ) = = kg-wts . x 2 2 (note: the length of the plank doesn't affect this derivation).

A-2

Let W the weight of the ball, FP and FW the forces exerted on the ball by the plane and by the wall n

respectively and made use of PVW in the form

∑ F cos ∆ i

= 0.

i =1

Thus, with respect to an horizontal line joining the point in which the ball touches the wall with the centre of mass of the ball, Fw cos ( 0 ) + FP cos ( π2 + α ) = 0 , that is Fw − FP sin α = 0 ; and, with respect to a vertical line lying in the plane of the sheet and passing through the centre of mass of the ball, FW cos π2 + W cos π + FP cos α = 0 , that is −W + FP cos α = 0 . So W FP = , cos α FW = W tan α , and the forces on the planes are 1 FP = kg-wt, cos α FW = tan α kg-wt.

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

A-3

Again make use of PVW, this time in the form

∑W h = 0 . i i

i =1

Without the loads: (WC + WAA′CD ) ⋅ AP = WBB′GH ⋅ PB , that is WC + WAA′CD = 2WBB′GH .

With the loads, assume the weight W1 goes down x as the weight W2 rises y, being x = −

Thus

(WC + WAA′CD + W1 ) ⋅ ( − x ) = (WBB′GH + W2 ) ⋅ y ,

that is WC + WAA′CD + W1 = 2 (WBB′GH

subtracting to this equation the one without the loads above one gets

1 y. 2 + W2 ) , and

W1 = 2W2 , so

1 2

W2 = W1 = 0.25 kg-wts .

A-4 L l1 xA

yA l2

−∆y A

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

The weight A can move both vertically and horizontally. Vertical displacement of A (assumed downwards): WA ⋅ ( −∆y A ) = W1 kg ⋅ ∆y1 kg + WB ⋅ ∆yB . -

Horizontal displacement of A (assuming B rises

(

∆yB

as A moves to the left):

)

W1 kg ⋅ −∆y1 kg = WB ⋅ ∆yB Case 1 – vertical displacement For the flattened triangle (l1, ∆y A , l1 + ∆l1 ), using Carnot’s theorem and assuming ∆y A small,

l1 + ∆l1 = l12 + 2l1∆y A cos ( π2 + ϕ ) + ∆y A2 = l12 + 2l1∆y A ( − sin ϕ ) + ∆y A2 = l1 1 − 2 Neglecting

the

l1 + ∆l1 ≅ l1 1 − 2 sin ϕ =

term

order

( ) sin ϕ ≅ l (1 − ∆y A l1

∆y A l1

∆y A2

and

)

and

sin ϕ

using

the

∆y A l1

∆y 2A l12

1 + x ≅ 1 + 12 x ,

approx.

(

( ) sin ϕ +

)

∆l1 ≅ l1 1 − ∆ly1 A sin ϕ − l1 ≅ −∆y A sin ϕ ,

then where

yA y y ≅ A . So ∆l1 ≅ −∆y A A . l1 + ∆l1 l1 l1

Similarly, for the triangle (not shown) (l2, ∆y A , l2 + ∆l2 ),

∆l2 ≅ −∆y A

yA . But ∆l1 = ∆y1 kg and l2

∆l2 = ∆y B , whence

WA ⋅ ( −∆y A ) ≅ W1 kg ⋅ ( −∆y A )

yA y y y + WB ⋅ ( −∆y A ) A → WA ≅ W1 kg ⋅ A + WB ⋅ A = l1 l2 l1 l2

= W1 kg ⋅ sin 30° + WB ⋅ sin 45° = 12 W1 kg +

1 2

WB = 12 +

1 2

WB .

Case 2 – horizontal displacement In the same way, for the triangle (not even shown as well) ( l1 − ∆l1

( ) cos 30° ≅ l (1 + ∆x A l1

∆x A l1

∆x A , l1 ) with ∆x A small,

( ) cos 30° + , and, with the above approximations, cos 30° ) → ∆l ≅ l − l (1 + cos 30° ) ≅ −∆x cos 30° .

l1 − ∆l1 = l12 + 2l1∆x A cos 30° + ∆x A2 = l1 1 + 2 l1 − ∆l1 ≅ l1 1 + 2

∆x A2

∆x A l1

l12

∆x A l1

xA x in the above expressions one gets ∆l1 ≅ −∆x A A , and similarly, for triangle l1 l1 L − xA ( l2 + ∆l2 , ∆x A , l2 ): ∆l2 ≅ ∆x A cos 45° = ∆x A . l2 So from PVW

Substituting cos 30° =

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

(

)

W1 kg ⋅ −∆y1 kg = WB ⋅ ∆yB → W1 kg ⋅ ( −∆l1 ) = WB ⋅ ∆l2 → W1 kg ⋅ ∆xA → W1 kg ⋅

xA L − xA = WB ⋅ → W1 kg ⋅ cos 30° = WB ⋅ cos 45° → W1 kg ⋅ l1 l2 3 2

→ WB =

WB =

(

1 2

3 2

xA L − xA = WB ⋅ ∆x A l1 l2 3 2

= WB ⋅

2 2

⋅ W1 kg .

Whence finally WA ≅ 12 +

WA =

1 2

WB = 12 +

3 2

W1 kg . In summary:

) kg-wts,

kg-wts.

A-5

From the PVW (in the form ∑ Fi cos ∆ i = 0 ): i =1

Case 1 – vertical fixed line:

T1 cos ϕ + T2 cos ϕ + W cos π = 0 → (T1 + T2 ) cos ϕ = W → (T1 + T2 ) 1 − sin 2 ϕ = W →

(T1 + T2 )

1−

( ) AB

2 AC

=W →

1 2

(T1 + T2 ) = W .

Case 2 – horizontal fixed line: −T1 cos ( π2 − ϕ ) + T2 cos ( π2 − ϕ ) + W cos π2 = 0 → T1 = T2 . Thus T1 = T2 =

W 50 = lb-wts 2 2

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

A-7

ϕ x

Let V the reaction force of the edge in contact with the wheel and assume null the reaction force of the n

wall at the ground soon as the force F is applied to the axle. Then from PVW

∑ F cos ∆ i

= 0:

i =1

R−h =W . R

Case 1 – vertical fixed line: V cos ϕ + W cos π = 0 → V cos ϕ = W → V

Case 2 – horizontal fixed line: F cos ( 0 ) + V cos ( π2 + ϕ ) + W cos π2 = 0 → F = V sin ϕ = V 2

where x = R 2 − ( R − h ) = h ( 2 R − h ) . Thus F = W

x =W R−h

h ( 2 R − h) ) R−h

x , R

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

A-9

Using

∑ F cos ∆ i

= 0 we have:

i =1

horizontal line AB : FO cos (π + 45° ) + FP cos π2 + FQ cos 0° + Fx cos (π + ϕ ) = 0 ⇒ − F cos 45° + F − Fx cos ϕ = 0 , vertical line OP : FO cos ( π2 + 45° ) + FP cos 0° + FQ cos π2 + Fx cos ( π2 + ϕ ) = 0 ⇒ − F sin 45° + F − Fx sin ϕ = 0 ; and confronting: cos ϕ = sin ϕ , whence ϕ = 45° or ϕ = 225° . Fx is positive when ϕ = 45° therefore it is parallel to, and has the same direction of, FO . Its magnitude is Fx = F

(

)

2 − 1 = 20.7 N , 45° down from AB .

When the plate is in equilibrium the torque, with respect to the corner O, is Fx ⋅ x ⋅ sin (π − ϕ ) + OP ⋅ FQ + 0 ⋅ FP + 0 ⋅ FO = 0 ⇒ Fx ⋅ x ⋅ sin ϕ + F ⋅ OP = 0 ⇒

x=−

F ⋅ OP 50 N × 0,100 m =− = −0, 34 m, Fx sin 45° 20.7 N × 0.7071

i.e. .34 m left of O.

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

A-10

Assume W1 > W2 . Then as W1 goes down − D sin θ the weight W2 rises + D sin θ . From the conservation of energy Thus v = 2 gD

1 2

W1 + W2  2  g  ⋅ v + W1 ⋅ ( − D sin θ ) + W2 ⋅ D sin θ = 0 .  

W1 − W2 sin θ . W1 + W2

A-11

See the previous problem. In the present case, 1 2W ⋅ v 2 + W ⋅ D sin θ + W ⋅ ( − D sin φ ) = 0 ⇒ v = gD ( sin φ − sin ϑ ) . 2⋅ g

Solutions to problems in Leighton and Vogtâ&#x20AC;&#x2122;s Exercises in Introductory Physics

B-1

At t = 0 ( K .E.)1 = ( K .E.)2 = 0, ( P.E.)1 = ( P.E.)2 = Mg H

, being M 1 = M 2 = M

At t > 0 ( K .E.)1 = ( K .E.)2 = 12 Mv 2 , ( P.E.)1 = Mgy, ( P.E.)2 = Mgx

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

H 1+ 2 2 Length of the cord on the side of M2: H − x H Length of the cord on the side of M1: 2 −1 + x 2 Height of M2: x H 1  x Height of M1: y = 1 − − 2 2 2

Length of the cord:

(

)

(

)

H  1  x  Then at t > 0 ( P.E.)1 = Mgy = Mg  1 −  and from the conservation of energy − 2 2 2 ( K .E.)1,t =0 + ( K .E.)2,t =0 + ( P.E.)1,t =0 + ( P.E.)2,t =0 = ( K .E.)1,t >0 + ( K .E.)2,t >0 + ( P.E.)1,t >0 + ( P.E.)2,t >0 ⇒ H H 1 1 x  H + Mg = Mv 2 + Mv 2 + Mg  1 − 1  −  + Mgx ⇒ 2 2 2 2 2 2 2 H MgH = Mv 2 + Mgx 1 − 1  + Mg 1 − 1  . 2 2 2 

0 + 0 + Mg

Taking the derivative with respect to t of each member of the above expression dx dv 2 dv 0 = 2 Mva + Mgv 1 − 1  + 0 , where we used v = and = 2v = 2va . 2 dt dt dt  1 1  a) So the vertical acceleration of M2 at t > 0 is a = − 1 − g . 2 2 b)

Being a < 0 the mass M2 will move downward. The eight of M2 is x =

v ( 0 ) = 0 at t = 0 whereas x = time t1 = − H

H and its velocity 2

1 2 H at + at t > 0 ; then it will strike the ground ( x = 0 ) at 2 2

. 1   g 1 −  2  H 1  x H 1  Being y = 1 − − the height of M1, then for x = 0 is y = 1 −   < H ; so the 2 2 2 2 2 other mass (M1) will no strike the pulley. a

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

B-2

When the boom rotates an angle ∆θ (assumed small, i.e. neglecting terms of the order if ( ∆θ ) or higher) around the pivot the length y of the horizontal cable will change of ∆y x cos θ∆θ . This can be seen using Carnot’s theorem and the approx. the angle θ :

1 + x ≅ 1 + 12 x to find the length of the side opposite to

∆y = y′ − y = x 2 + x 2 cos 2 θ − 2 x 2 cos θ cos (θ + ∆θ ) − x sin θ ≅ x sin θ

(

)

1 + 2 ctg θ∆θ − 1 x sin θ ⋅ ctg θ∆θ = x cos θ∆θ .

Or, in a more straightforward ′ y ≅ x sin (θ + ∆θ ) = x sin θ cos ∆θ + x cos θ sin ∆θ ≅ x sin θ + x cos θ∆θ ⇒

way,

∆y = y′ − y ≅ x cos θ . The height of the weight W changes of ∆Z = L cos (θ + ∆θ ) − L cos θ ≅ − L sin θ∆θ and the weight w of the boom (assumed applied to the centre of mass of the boom) moves of ∆z = L2 cos (θ + ∆θ ) − L2 cos θ ≅ − L2 sin θ∆θ . Then from PVW: T ∆y + W ∆Z + w∆z = 0 ⇒ T ⋅ x cos θ∆θ − W ⋅ L sin θ∆θ − w ⋅ L2 sin θ∆θ ≅ 0 ⇒ T=

L w  W +  tan θ . x 2

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

B-3

Let L = 10 ft the length of the ladder, r = 8 ft the initial height of the rollers at the top end and suppose that, slightly tilting the ladder leaning against the vertical wall, the rollers would go down ∆r . Then (assuming ∆r small) 2 2 L2 = ( r + ∆r ) + ( s + ∆s ) ⇒ L2 ≅ r 2 + 2r ∆r + s 2 + 2s∆s = r 2 + s 2 + 2 ( r ∆r + s∆s ) = L2 + 2 ( r ∆r + s∆s ) ⇒

(

)

⇒ r ∆r + s∆s = 0. r Thus ∆s = − ∆r . s Furthermore, said hW , w the heights of the weights W and w, hangings from rungs at LW , w feet from the bottom end, then hW , w =

LW ,w L

r and ∆hW , w =

LW , w L

∆r .

W is hung l = 2.5 feet from the top end and therefore LW = L − l whereas Lw =

L . 2

Let R, H and V, respectively, the force of the roller on the wall, the horizontal and vertical forces of the ladder on the ground, then, from PVW, L L  r  H ∆s + W ∆hW + w∆hw = 0 ⇒ H  − ∆r  + W W ∆r + w w ∆r = 0 ⇒ L L  s  −H

L L r s s L +W W + w w = 0 ⇒ H = (WLW + wLw ) = W ( L − l ) + w  . s L L Lr Lr  2

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

Now, from PVW in the form

∑ F cos ∆ i

= 0 one gets R = H and V = W + w , whence

i =1

R = 45 lb-wts

H = 45 lb-wts, V = 90 lb-wts

B-4

θ α −θ

Let L = 3R the length of the plank. Then we have sin α =

L 3 = ⇒ α = 60° . 2R 2

( 2)

For the heights of the weights W and w one gets that hW = R − R 2 − L

cos θ ,

hw = R − R cos (α − θ ) and (for a small tilt ∆θ )

( 2)

∆hW ≅ R 2 − L

(

sin θ∆θ = R 1 − L

2R )

sin θ∆θ = R cos α sin θ∆θ =

R sin θ∆θ , 2

being cos α = 12 , ∆hw ≅ R sin (α − θ ) ∆θ . From PVW, remembering that ∆hW and ∆hw lie in opposite directions, we have W ∆hW = w∆hw , whence R W α W sin θ∆θ = wR sin (α − θ ) ∆θ = R sin (α − θ ) ∆θ ⇒ sin θ = sin (α − θ ) ⇒ θ = α − θ ⇒ θ = = 30°. 2 2 2

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

B-5

y x

From PVW we have W ∆x sin 30° + W ∆y sin 60° = 0 . From the rectangular triangle shown (see problem B-3) x∆x + y∆y = 0 ⇒ ∆y = −

x ∆x = − tan α∆x . y

Then W ∆x sin 30° + W ∆y sin 60° = 0 ⇒ ∆x sin 30° + ∆y sin 60° = 0 ⇒ ∆x sin 30° − tan α∆x sin 60° = 0 ⇒ sin 30° − tan α sin 60° = 0 ⇒ sin 30° − tan α cos 30° = 0 ⇒ tan α =

sin 30° cos30°

= tan 30°,

and therefore α = 30° .

B-6

L R r

Being L the length of the string and r the radius of the sphere, then, from Carnot’s theorem,

( L + r ) + ∆L =

R 2 + ( R − r ) − 2 R ( R − r ) cos (α + ∆α ) .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

If ∆α is small ( L + r ) + ∆L ≅ R 2 + ( R − r ) − 2 R ( R − r ) cos α + 2 R ( R − r ) sin α∆α . 2

But R 2 + ( R − r ) − 2 R ( R − r ) cos α = ( L + r ) , thus

( L + r ) + ∆L ≅ ( L + r ) (L + r) +

R(R − r)

(L + r)

+ 2 R ( R − r ) sin α∆α ≅ ( L + r ) 1 +

sin α∆α ⇒ ∆L

R(R − r)

(L + r)

2R ( R − r )

(L + r)

sin α∆α

sin α∆α .

Furthermore, being h the height of ∆h = ( R − r ) sin (α + ∆α ) − ( R − r ) sin α ≅ ( R − r ) cos α∆α .

the

Let F the strength of the string, from PVW: F ∆L = W ∆h ⇒ F = W

centre

the

sphere,

∆h L+r ≅W cot α . One also gets ∆L R

R − r = 49 − 4.5 cm=44.5cm L + r = 40 + 4.5 cm=44.5cm L+r So the triangle is isosceles so that cot α = 1 ⇒ F ≅ W . Let h the height of the isosceles, then 2 2h 2 L+r 44.5 22.25 2 =W = W 0.6W . h = (L + r) − R = 44.52 − 24.52 = 1380 and F ≅ W 2 2h 2 × 1380 1380

( )

B-7

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

30°

The centres of the spheres lie on the vertices of a regular tetrahedron of edge a and height

6a , then 3

6 . 3 Let wa the component of the weight w along each edge a. From symmetry considerations w w 3 × wa cos θ = w , whence wa = = = 2 ton-wts . 3cos θ 6 cos θ =

3 and the two 3 welds at the contact points of each sphere with the two other exert along the same direction a force 2T cos 30° = T 3 , being T the tension of each weld. Along the horizontal surface of the tetrahedron wa has a component wa sin θ = wa ⋅

Then from PVW (in the form

∑ F cos ∆ i

= 0 ) and fixing the horizontal direction, one gets

i =1

T 3 = wa ⋅

w 3 ⇒ T = a , and thus, for a factor of safety of 3, T(3) = 3 × T = wa = 2 ton-wts . 3 3

B-8

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

Let l the length of the cord, h1 , h2 the heights of the two masses when they lie at distances c1 , c2 from the top in downwards direction along the two legs of the right triangle. Then from PVW T ∆l = m1∆h1 + m2 ∆h2 where: ∆h1 = ∆c1 sin 30° = 12 ∆c1 ∆h2 = ∆c2 sin 60° =

3 2

∆c2

Being c1 = l cos α c2 = l sin α one gets ∆c1 = −l sin α∆α + cos α∆l ⇒ ∆h1 = − 2l sin α∆α + 12 cos α∆l ,

∆c2 = l cos α∆α + sin α∆l ⇒ ∆h2 =

3 2

l cos α∆α +

3 2

sin α∆ l

Thus T ∆l = m1∆h1 + m2 ∆h2 =

(

1 2

m1 cos α +

3 2

)

m2 sin α ∆l −

(

1 2

m1 sin α −

3 2

)

m2 cos α l ∆α .

∆l , ∆α being independent of one another, tacking ∆l = 0 one finds that the system is in static equilibrium when m1 sin α − 3m2 cos α = 0 ⇒ tan α = 3 3 . Reading the 1 T = 2 m1 cos 79°.1 +

3 2

tables one gets m2 sin 79°.1 = 265 g-wts .

α = 79°.1

and,

taking

∆α = 0 ,

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

B-9

r’

L ∆y

r ∆s ∆x

x ∆x = − cot 30°∆x . y But ∆x = r ′ cos 30° + ∆s − r cos 30° = ∆s − ( r − r ′ ) cos 30° . Assuming ∆s small we have r − r ′ ≅ ∆s cos 30° . Thus ∆x ≅ ∆s (1 − cos 2 30° ) = ∆s sin 2 30° . Being ∆L = 0 we have x∆x + y∆y = 0 ⇒ ∆y = −

from PVW T ∆s + W ∆y = 0 ⇒ T = −W T = W cos 30° sin 30° =

 cot 30° ⋅ sin 2 30°∆s  ∆y  − cot 30°∆x  = −W  = W  ⇒  ∆s ∆s ∆s    

3 W. 4

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics b) normal

N T horizontal

Using the method of the force components (FC) on cart we have: 3 T cos180° + N cos 300° = 0 ⇒ T = N (horizontal). 2 1 Using FC on W: N cos 0° + W cos 240° = 0 ⇒ N = W . 2 3 W. So T = 4

B-10

Assume the centre of the bobbin goes down ∆H and m rises ∆h as the bobbin rolls downwards. Then from PVW: M ∆H = m∆h .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

For a rolling of the bobbin of ∆L = R∆α then M goes down ∆H = ∆L sin θ , whereas m rises (assuming ∆α small) ∆l ≅ r ∆α =

r ∆L and goes down ∆H . R

Thus r  r  ∆h = ∆l − ∆H = ∆L  − sin θ  ⇒ M ∆L sin θ = m∆L  − sin θ  ⇒ R  R  r r m r M sin θ = m − m sin θ ⇒ ( M + m ) sin θ = m ⇒ sin θ = = 0, 5 ⇒ R R M +m R θ = 30°.

B-11

∆h ∆x

If the chain rises ∆h resting in a horizontal circle, then its radius varies of ∆x = of ∆l = 2π∆x .

From PVW: W ∆h = T ∆l ⇒ W ∆h = T ⋅ 2π∆x = T ⋅ 2π

r Wh ∆h ⇒ T = . h 2π r

r ∆h and its length h

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

B-12

lW3

lW2 lW1

lw2 l w1

Assuming w rises ∆hw whereas W goes down ∆hW , being:

(

)

(

)

∆ lW1 + lW2 + lW3 = 0 , ∆lW1 = ∆lW2 ⇒ ∆lW1 = − 12 ∆lW3 , ∆ lw1 + lw2 = 0 , ∆lW3 = −∆lw1 = ∆lw2 ,

then ∆hw = ∆lw2 + ∆lW3 = 2∆lw2 , ∆hW = ∆lW1 sin θ = − 12 ∆lW3 sin θ = − 12 ∆lw2 sin θ = − 14 ∆hw sin θ .

From PVW: w∆hw + W ∆hW = 0 . So w∆hw − 14 sin θ ⋅ W ∆hw = 0 ⇒ W =

4w . sin θ

B-13

FG DF

x EG

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

From PVW: F1 x = F2 y , but x : 2 = y :1 ⇒ F2 = 2 F1 .

Using

∑ F cos ∆ i

= 0 then, with respect to the vertical direction, one gets

F1 − W + F2 = 0 ⇒ F1 + F2 = W ⇒ 3F1 = W ⇒ F1 =

W 2W , F2 = . 3 3

Using FC at the joint G:

vertical components: F2 cos 0° + TFG cos150° = 0 ⇒ F2 − TFG

3 2F = 0 ⇒ TFG = 2 . 2 3

Using FC at the joint F: -

horizontal components: TFG cos120° + TEF cos 240° + F cos 0° = 0 ⇒ F =

vertical components: TFG cos 30° + TEF cos150° = 0 ⇒

whence F = TFG =

2 F2 4W = . 3 3 3

1 2

(TFG + TEF )

3 (TFG − TEF ) 00 ⇒ TEF = TFG 2

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

B-14

a) members AB, BD, DF, FG should be rigid in order that A remains in fixed position; CD also rigid to fix C and D. The elements that could be flexible are therefore AC, CE, EG, BC, EF, ED.

ˆ = b) meanwhile note that cosA F1 =

3 5

ˆ 4 . From the result of the preceding problem is then , sin A= 5

W 2W , F2 = . 3 3

Using FC at the joint A: -

horizontal:

ˆ ABcos A=AC

vertical:

ˆ F = W ⇒ AB = 5W ABsin A= 1 3 12

Using FC at the joint B: -

horizontal:

ˆ ( AB+BC ) cos A=BD

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

vertical:

ˆ W AB=BC ⇒ BD=2ABcosA= 2

Using FC at the joint C: -

vertical:

BC = CD

Using FC at the joint D: -

vertical:

So BD=

DE = CD ⇒ DE=CD=BC=AB=

5W 12

W 5W , DE= . 2 12

B-15

At its maximum swing the pendulum has zero velocity. Then from the conservation of energy ∆h W ∆hW + w∆hw = 0 ⇒ W = − w w . At the maximum swing (pendulum at the ceiling) ∆hw = 3 , ∆hW 3 ∆hW = −4 . Thus W = 4 w .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

B-16

As the third mass reaches the table top, the two equal masses rise 50

(

)

2 − 1 cm ; when the third mass

2    ∆h   has descended a distance ∆h the two equal masses have risen by 50  1 +  . − 1     50   

Then, said V and v the final velocities (in module) of the third mass and of the two equal masses respectively, v =

V . 2

The kinetic energy varies of 2 ( mg ) ⋅ 50

(

1 2

( 2m )V 2 + 2 ⋅ 12 mv 2 = m (V 2 + v 2 ) = 32 mV 2

)

(

and the potential energy of

)

2 − 1 − ( 2mg ) ⋅ 50 = −2mg ⋅ 50 2 − 2 .

From the conservation of energy:

3 2

(

)

mV 2 = 2mg ⋅ 50 2 − 2 ⇒ V =

With g = 981 cm s −2 thus one obtains V = 196 cm s −1 .

(

200 g 2 − 2 3

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

B-17

Let’s consider a small layer of liquid of thickness ∆z a distance h up from the hole. Its potential energy of position is ρ gA∆z ⋅ h . The internal tension force due to the weight of the overlying liquid, ρ gA ( H − h ) , corresponds to a pressure p = ρ g ( H − h ) and an internal energy pA∆z = ρ gA ( H − h ) ∆z .

The total potential energy is the sum of the position energy and of the internal energy:

ρ gA∆z ⋅ h + ρ gA ( H − h ) ∆z = ρ gAH ∆z (independent of h, and hence also valid for

h=0

corresponding to the hole position). Soon

the

hole

open

the

ρ gAH ∆z = 12 ( ρ A∆z ) v 2 ⇒ v = 2 gH .

above

energy

converted

kinetic

energy,

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

C-1

W W

X W

α T

β β + 60° + θ = 90° ⇒ β = 30° − θ α = 60° − β = 30° + θ

60°

From FC: T cos α + X cos β = W

vert.

horiz. T sin α = X sin β

obliq. W sin θ = T cos 60° =

T 2

(Note that in the shown arrangement T( limit ) = 2W sin θ ). In terms of cot β : X cos β = X sin β cot β = T sin α cot β ⇒ T cos α + T sin α cot β = W .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

But cos α = cos ( 30° + θ ) = cos 30° cos θ − sin 30° sin θ =

3 2

cos θ − 12 sin θ

sin α = sin ( 30° + θ ) = sin 30° cos θ + cos 30° sin θ = 12 cos θ + cot β = ( cot 30° cot θ + 1) ( cot θ − cot 30° ) =

(

3 2

sin θ

) ( cot θ − 3 )

3 cot θ + 1

Thus we have  T ( cos α + sin α cot β ) = W ⇒ T  

(

3 2

cos θ − 12 sin θ +

)

⇒ 3 cos θ sin θ − sin 2 θ + cos θ sin θ + 3 sin 2 θ ⋅

(

1 2

cos θ +

3 2

)

sin θ ⋅

3 cot θ + 1  T  =W = 2sin θ cot θ − 3 

3 cot θ + 1 = 1. cot θ − 3

In terms of cot θ : cot θ 1 + cot 2 θ 1 sin 2 θ = 1 + cot 2 θ cos θ sin θ =

Thus

3 cot θ − 1 cot θ + 3 3 cot θ + 1 2 3 1 + ⋅ =1⇒ = 1 ⇒ cot θ = 3 3 ⇒ tan θ = 2 2 1 + cot θ 1 + cot θ cot θ − 3 cot θ − 3 3 3

finally θ = tan −1

1 3 3

= 10°,9 .

and

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

C-2

∆hw and W moves in the same r direction by the same distance ∆hw due to the vertical displacement of the spool, and in the opposite direction of a distance ∆hW ≅ R∆α due to the rotation of the spool.

As w moves of a (small) distance ∆hw the spool rotates an angle ∆α ≅

Then from PVW:

w∆hw = W ( ∆hW − ∆hw ) = W ( R∆α − r ∆α ) = W ( R − r ) ∆α ⇒ wr ∆α = W ( R − r ) ∆α , whence W =

wr . R−r

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

C-3

θ′

T′

T ′′

θ ′′ C

Define t ′ = tan θ ′ =

B-A A-C 4.8 ×103 kg-wts , t ′′ = tan θ ′′ = and w = . 9m 9m 6

From FC at the ends of B, A, C: B) Horiz. 1.

T 1 =T′ 2 1 + t ′2

Vert. 2.

T t′ =T′ +w 2 1 + t ′2

T ′′′

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

Horiz. 3.

T′

Vert. 4.

T′

1 1 + t ′2 t′ 1 + t ′2

= T ′′

1 1 + t ′′2 t ′′ 1 + t ′′2

C) Horiz. 5.

T ′′

Vert. 6.

T ′′

1 1 + t ′′2 t ′′ 1 + t ′′2

From 3. and 4.: T ′′

t′

1 + t ′′2

From 6. and the latter: T ′′

Thus t ′′ = t ′ − t ′′ ⇒ t ′′ =

From 6. and 4.: T ′

= T ′′′

= T ′′

t ′′

1 + t ′′2

t ′′

1 + t ′′2

= T ′′

+ w ⇒ T ′′

( t ′ − t ′′ ) 1 + t ′′2

1 + t ′2

= 2w , whence from 2.

But C = 2.00 m, so A=5 m, B=11 m. To determine Tmax let’s return back to 1. – 6.:

2’.

T 2T ′ = +w 2 13

3’.

3T ′ 3T ′′ = , 13 10

= w.

Finally, from 1., 3wt ′ = 2 w ⇒ t ′ = 23 ⇒ t ′′ = 13 . Thus

T 3T ′ = , 2 13

1 + t ′′2

t′ . 2

t′

1’.

( t ′ − t ′′ )

whence T > T ′

whence T ′ > T ′′

T = 3w . 2

B-A 2 = , 9m 3

A-C 1 = . 9m 3

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

4’.

2T ′ T ′′ = + w = 2 w using 6’. below, from which T ′ = 13w 13 10

5’.

3T ′′ = T ′′′ , 10

6’.

T ′′ =w 10

Finally, therefore, Tmax = T =

whence T ′′ > T ′′′

3 2 4.8 ×103 kg-wts T ′ = 3 2w = ≈ 34 × 103 kg-wts . 13 2

C-4

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

a) Given the condition FB = 0 , from PVW (in the form

∑W

i i

= 0 ), with respect to A, one gets

Fh = WL ⇒ F = W

L . h

b) From FC:

h. F = FA cos θ v. W = FA sin θ

whence tan θ =

W h h = ⇒ θ = tan −1   and F L L

2 2 L 1 L L h L 2 1 + tan θ = W 1+   = W 1+   . FA = W ⋅ =W h cos θ h h L h

Solutions to problems in Leighton and Vogt’s textbook "Exercises in Introductory Physics" (Addison-Wesley, 1969) Pier F. Nali (Revised April 10, 2016)

CHAPTER 3 Kepler’s Laws and Gravitation

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

In general, for a curve having parametric equations x = ϕ ( t ) , in the figure below

y = ψ ( t ) , the curvilinear trapezoid area

where a = ϕ ( t1 ) , b = ϕ ( t2 ) and ψ ( t ) ≥ 0 in the interval ( t1 , t2 ) , is A = ∫ψ ( t ) ⋅ ϕ ′ ( t ) dt . t1

We can use this formula to calculate the area of an ellipse having parametric equations x = a cos θ , y = b sin θ . From the symmetry of the ellipse we can calculate the area of a quarter ellipse between the limits x = 0 (ϑ = π2 ) , x = a (ϑ = 0 ) . b 0

Hence

2 A = ∫ b sin θ ⋅ ( − a sin θ ) dθ = ab ∫ sin 2 θ dθ = π4 ab ⇒ A=π ab . 4 π 0 2

A-1

x=0

x=a

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

For the moon a =

1 2

+ rp

) = 405, 500 km 2+ 363,300 km = 384,500 km

For the Earth sat.

a =

1 2

+ rp ) =

1 2

( r⊕ + 710 km + r⊕ + 225 km ) = 12 ( d ⊕ + 935 km ) =

12, 756 km + 935 km = 6,845.5 km 2

 a 2  6,845.5 km  2 d d Thus t =   t =   × 27 .322 ≈ 0 .065 ≈ 1.6 hr a 384,500 km    

A-2

From the Kepler’s second law ra va = rp v p ⇒

vp va

ra a + c 1 + e⊕ 1 + 0.0167 1.0167 = = = = = 1.033 . rp a − c 1 − e⊕ 1 − 0.0167 0.9833

A-3

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics 2

4 2

M M g M  R⊕  M  R⊕  1.000  6,378 km  g ⊕ = G 2⊕ , g = G 2 ⇒ = × × 9.80 m s −2 =   ⇒g =   g⊕ =  R⊕ R g⊕ M ⊕  R  M⊕  R  81.3  1, 738 km  g 1.6 m s −2 ⊕ 6

A-4

 Tcomet   acomet   Tcomet  3 From the Kepler’s third law   =  ⇒ acomet =   ⋅ a⊕ . Thus  T⊕   a⊕   T⊕  2

acomet

2  (1986 − 1456 ) / 7  3 3 =  × 1 A.U. ≈ ( 75.7 ) A.U. ≈ 17.9 A.U. and 1 yr  

racomet = 2acomet − rpcomet ≈ 2 × 17.9 A.U. - 0.60 A.U. ≈ 35.8 A.U. - 0.60 A.U. ∼ 35.2 A.U.

From the Kepler’s second law vmax rpcomet = vmin racomet ⇒

vmax racomet 35.2 A.U. = = ∼ 59 . vmin rpcomet 0.60 A.U.

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

A-5

From the Kepler’s third law 2

2  T24 h   R24 h   T24 h  3  24 × 60 min  3 3 = ⇒ R = ⋅ R =       24 h E   ⋅ RE = (14.4 ) ⋅ RE ∼ 5.9 RE .  100 min   T100 min   RE   T100 min 

B-1

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

a) The geographic latitude of P is restricted to λ = 0 (equator). Although a synchronous orbit doesn’t need to be equatorial, a body in a synchronous non equatorial orbit could not remain in a fixed position with respect to P: it would appear to oscillate between the north and south of the Earth's equator. Moreover, a body in a synchronous non circular (elliptical) orbit would appear to oscillate East-and-West. The combination of the two latter (synchronous non-circular nonequatorial) would appear as a path in the shape of eight. 3

 r 2 1  1 3 b) From the Kepler’s third law ts (= 1 day)=  s  T (= 27 days) ⇒ rs =   r⊕ = r⊕ . 9  27   r⊕ 

B-2

T2 4π 2 a) From the Kepler’s third law = , a⊕ 3 GM ⊕

From Earth and Moon data: rP⊕ = 1.47 ×1011 m, rA⊕ = 1.52 × 1011 m, T⊕ = 365d .242 rP = 3.63 × 108 m,

T  a  M T⊕ 2 4π 2 = ⇒   ⋅  ⊕  = . 3 a ⊕ GM  T⊕   a⊕ 3 M ⊕

rA = 4.06 × 1011 m, T = 27 d .322

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics a⊕ = a =

1 2 1 2

(r (r

P⊕

) ) = 3.84 ×10

+ rA⊕ = 1.50 × 1011 m + rA

M  27 d .322  Thus =   M ⊕  365d .242 

T 2 4π 2 , = a 3 GM

 1.50 × 1011 m  5 ⋅  = 3.33 × 10 8  3.84 × 10 m  2

T   a  T2 4π 2 M = ⇒   ⋅  = 3 a⊕ GM ⊕  T   a⊕ 3 M ⊕

a = 421,800 km, a⊕ = 384, 000km, T = 1d .769

 27 d .322  M Thus = d  M ⊕  1 .769 

 421,800 km  ⋅  = 318 .  384, 000 km 

B-3

For elliptic orbits ma + mb =

4π 2 R 3 . GT 2

For the Earth-Sun system m + m⊕ ( ≅ m ) =

4π 2 (1 A.U.) G (1 yr )

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

Thus ma + mb =

R3 m with R expressed in A.U. and T in years. T2

B-4

II: unchanged. The Kepler’s second low says that the areal velocity is constant, i.e., said d the arc length of the R

trajectory of m and R the distance from M :

dA 1 d 1 1 L = R = Rv = R ( mv ) = = const. , being L the orbital angular momentum of m . dt 2 dt 2 2m 2m

But L remains constant for central forces, so the areal velocity also doesn’t change. Indeed, d L  GMm  GMm = R × F = R ×  − (3+ a ) R  = − (3+ a ) R × R = 0 , being R × R ≡ 0 . dt R  R 

(

Thus also L cost. and

)

dA so well. dt

2π III: For a circular orbit F = − mω 2 R , where ω = . Hence T −

GMm GM 4π 2 GM 4π 2 ( 3+ a ) 2 2 2 ω ω R m R T R = − ⇒ = ⇒ = ⇒ = . 3+ a 3+ a 3+ a T2 GM R( ) R( ) R( )

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

C-1

Ignoring the moon’s gravitation one can assume g =

Moon at zenith: g z = g −

( R⊕ − R⊕ )

GM ⊕ . R⊕2

, at nadir: g n = g +

( R⊕ + R⊕ )

    ∆g g n − g z M  1 1  = = + . Thus 2 2  g g M⊕  R    R ⊕  ⊕ +1 − 1      R⊕ R   ⊕  

R⊕ 3.84 × 108 m But = , R⊕ 6.38 × 106 m

R⊕ 3.90 × 108 m +1 = , 6.38 × 106 m R⊕

R⊕ 3.78 × 108 m M 7.34 × 1022 kg −1 = , = , so 6.38 × 106 m M ⊕ 5.98 × 1024 kg R⊕

2 2 ∆g 7.34  6.38   6.38   −6 −6 −6 = ×   +   ×10 = 1.23 × ( 2.68 + 2.85 ) × 10 = 1.23 × 5.53 ×10 = g 5.98  3.90   3.78   = (approx.) 7 × 10−6 .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

C-2

4π 2 a ⊕3  a   T⊕  2 4π 2 a 3 M= , M = ⇒M =  ⋅  . GT 2 GT⊕2  a⊕   T  Being T in days and V in km s-1 ,

2π a [ km ]   seconds   T [s ] = T days ×  day    

(

Thus M = 86, 400 km

2π a [ km ] T × 86, 400 [s ]

2π × 1.50 ×10

⇒a=

86, 400 TV . 2π

× ( 365 days ) × TV ⋅ M km ) 2

= 1.02 ×10 −7 TV 3 M ⊕ .

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

C-3

For the ellipse the radius of curvature is Rc

(R R ) = p

3 2

where Ra = a (1 + e ) > R p = a (1 − e ) , b = a 1 − e 2 .

Thus Rc = R p (1 + e ) = Ra (1 − e ) and

1 1 = (1 − e cos ϕ ) . r Rc

Solutions to problems in Leighton and Vogt’s Exercises in Introductory Physics

1 2 GM 1 2 GM v − = va − (simplifying the comet mass), 2 2 Rp Ra

a) From the conservation of energy:

where v is the velocity at perihelion, and from the Kepler’s second law vR p = va Ra .  1 GM 1 2 2 1  , v − va = GM  −  = 2e   2 Rc  R p Ra 

(

Thus

va = v

Rp Ra

)

e 1− e 1 ⇒ v − va = 2v , v + va = 2v , 1+ e 1+ e 1+ e

GM 1 2 2 e v − va = 2v 2 , whence = 2e 2 2 Rc (1 + e )

(

)

GM Rc = 2 v

R ⋅ c  Rp 

(

500.0 km s −1 ×1.00 × 106 km vR p )  ( =  ⇒ Rc = GM 1.33 ×1011 km 3 s −2 

)

= 1.88 × 106 km .

Rp Rp Rc Rc 1.00 × 106 km 106 km b) 1 + e = ⇒e= − 1, a = = = = = 8.33 × 106 R 1.88 Rp Rp 1− e 2 − c 0.12 2− 1.00 Rp

c) From the Kepler’s third law Tc2 =

km .

4π 2 a 3 2π a a ⇒ Tc = . GM GM

From a) GM =

vR p Rc

⇒ Tc =

2π a aRc vRp

2π × 8.33 × 106 km 8.33 × 106 km × 1.88 ×106 km = 414, 000 s 500.0 Km s −1 × 1.00 × 106 km −1

 seconds  ( ) 414, 000 s ≈ 4.8 days .  days  Tc [ days ] = Tc [s ] ×  = Tc [s ] ×  =   86, 400 s day −1  s   day 