quiz samples for chapter 13 general physics i may 11, 2020...

General Physics IQuiz Samples for Chapter 13

GravitationMay 11, 2020

Name: Department: Student ID #:

Notice

� +2 (−1) points per correct (incorrect) answer.

� No penalty for an unanswered question.

� Fill the blank ( ) with � (8) if the statement iscorrect (incorrect).

� Textbook: Walker, Halliday, Resnick, Principlesof Physics, Tenth Edition, John Wiley & Sons(2014).

13-1 Newton’s Law of Gravitation

1. Any particle in the universe attracts any otherparticle with a gravitational force.

(a) (�) The magnitude of the gravitational forceis

F = Gm1m2

r2,

where m1 and m2 are the masses of theparticles, r is their separation, and thegravitational constant is

G = 6.67× 10−11N ·m2/kg2.

(b) (�) The gravitational force is central: thedirection of the force is along the line thatconnects the two interacting particles.

(c) (�) The units for the gravitational constantare

m3 · kg−1 · s−2.

(d) (�) The quantity G is a universal constant ofnature.

(e) (�) The gravitational force satisfies theinverse square law : the strength isattenuated as 1/r2 as the distance r increases.

(f) (�) The gravitational force between extendedbodies is found by adding (integrating) theindividual forces on individual particles withinthe bodies.

(g) (�) If a body is a uniform spherical shell or aspherically symmetric solid, the netgravitational force exerts on an externalobject may be computed as if all the massesof the shell or body were located at its center.It called the shell theorem.

2. (�) Consider a spherical shell has inner radius R1,outer radius R2, and mass M , distributeduniformly throughout the shell.

The magnitude of the gravitational force exertedon the shell by a point mass particle of m with adistance d from the center, outside the outerradius, is

F =GMm

d2.

3. (�) Let M denote the mass of Earth and let Rdenote its radius. The ratio g/G at Earth’s surfaceis

M

R2.

13-2 Gravitation and the Principle ofSuperposition

©2020 KPOPEEE All rights reserved. Korea University of 9



1. A mass m is located at the origin; a second massm is at x = d. A third mass m is above the firsttwo so the three masses form an equilateraltriangle on the xy plane.

(a) (�) The positions of the three particles aregiven by

r1 = 0,

r2 = di,

r3 =1

2di +

√3

2dj.

(b) (�) The gravitational forces on the particle 3by 1 and 2 are

F1→3 = −Gm2

d2

(1

2i +

√3

2j

),

F2→3 = −Gm2

d2

(−1

2i +

√3

2j

).

(c) (�) The net force on the third particle is

F3,net = −√

3Gm2

d2j.

2. Gravitational forces obey the principle ofsuperposition.

(a) (�) If n particles interact, the net force on aparticle labeled particle 1 is the sum of theforces on it from all the other particles takenone at a time:

F1,net =n∑

i=2

Fi→1,

where Fi→j is the gravitational force acting onparticle j by particle i.

(b) (�) The gravitational force on a particle froman extended body can be found by dividingthe body into units of differential mass dm,each of which produces a differential force dFon the particle, and then integrating over allthose units to find the sum of those forces:

F1 =

∫dF .

3. (�) Consider 99 identical particles placed at

rk = d

(i cos

kπ

50+ j sin

kπ

50

),

for k = 1, 2, · · · , 99. Each particle’s mass is m.The gravitational force acting on a particle massM placed at the origin is

F = −GMm

d2i.

4. Consider a uniform thin ring of mass M and radiusa placed on the xy plane. The center is at theorigin. A particle of mass m is at the point Pwhose position vector is

r = zk.

Q is a point on the ring.

(a) (�) The distance d between P and Q is

d =√z2 + a2.




(b) If θ is the angle between the z axis and theline segment PQ, then

cos θ =z

d.

(c) (�) For any mass element dM on the ring, thecomponent of the gravitational forceperpendicular to k cancels that for theidentical mass element in the opposite side ofthe diameter.

(d) (�) The magnitude of the gravitational forceparallel to k is the gravitational force betweenm and dM multiplied by cos θ which isconstant for all dM around the ring.

(e) (�) The gravitational force on the particle mis

F = −GMm

d2cos θk = −G Mmz

(z2 + a2)3/2k.

5. Consider a half circle of radius a on the xy plane inthe x > 0 region whose center is at the origin. Themass distribution is uniform on the circle and thetotal mass is M . At the origin, a particle of massm is fixed.

(a) (�) An infinitesimal arc of angle dθ has themass

dM = Mdθ

π.

(b) (�) The gravitational force on the particle ofmass m by the mass dM which has an angulardisplacement θ from x axis is

dF =Gm

a2dM (i cos θ + j sin θ)

=GmM

πa2dθ(i cos θ + j sin θ).

(c) (�) The net force is

Fnet =

∫ π2

−π2

dF

=GmM

πa2

∫ π2

−π2

dθ(i cos θ + j sin θ).

=2GmM

πa2i.

6. Consider a uniform rod of length L and mass M isplaced parallel to the y axis. The center of the rodis at ai. At the origin, a particle of mass m is fixed.

(a) (�) The position of a point of the rod can beexpressed as

y = ai + yj, y ∈ [−L2 ,

L2 ],

=√a2 + y2(i cos θ + j sin θ),

cos θ =a√

a2 + y2,

sin θ =y√

a2 + y2.

The infinitesimal element dy in the range[y, y + dy] has the mass

dM = Mdy

L, y ∈ [−L

2 ,L2 ].




(b) (�) The infinitesimal element dy has thecontribution dF to the gravitational force onthe particle of mass m as

dF =Gm

y2 + a2dM (i cos θ + j sin θ)

=GmM

L

(i

ady

(y2 + a2)32

+jydy

(y2 + a2)32

).

(c) (�) The net gravitational force on the particleof mass m is

Fnet =

∫Gm

y2 + a2dM (i cos θ + j sin θ)

=GmM

L

(i

∫ L2

−L2

ady

(y2 + a2)32

+j

∫ L2

−L2

ydy

(y2 + a2)32

).

(d) (�) The contribution parallel to the unitvector j vanishes because the integrand is anodd function of y.

(e) (�) If we make use of the integral table∫dx

(x2 + 1)32

=x√

x2 + 1,

then the following integral can be evaluated as∫ L2

−L2

ady

(y2 + a2)32

=1

a

∫ L2a

− L2a

dx

(x2 + 1)32

=1

a

x√x2 + 1

∣∣∣∣∣L2a

− L2a

=L

a2√

1 + L2

4a2

.

Here, we have made a replacement y = ax.

(f) (�) The net force is

Fnet =GmM

Li

∫ L2

−L2

ady

(y2 + a2)32

=GmM

a2√

1 +(L2a

)2 i.

13-3 Gravitation Near Earth’s Surface

1. The gravitational acceleration ag of a particle ofmass m is due solely to the gravitational forceacting on it.

(a) (�) When the particle is at distance r fromthe center of a uniform, spherical body ofmass M , the magnitude F of the gravitationalforce on the particle is given by

F = GmM

r2.

Thus, by Newton’s second law,

F = mag,

which gives

ag =GM

r2.

(b) (�) Due to the fact that Earth’s mass is notdistributed uniformly, that the planet is notperfectly spherical, and that it rotates, theactual free-fall acceleration of a particle nearEarth differs slightly for different locations,and the particle’s weight differs from thedifferent magnitude of the gravitational forceon it.

2. A rocket ship is coasting toward a planet withradius R. Its captain wishes to know the value of gat the surface of the planet.

(a) (�) The captain needs to measure the altitudeH from the surface, the acceleration ag(H) atH, and the radius of the planet.

(b) (�) The acceleration on the planet surfaceag(0) can be inferred as

ag(0) = ag(H)×(R+H

R

)2

.

13-4 Gravitation Inside Earth

1. (a) (�) A uniform shell of matter exerts no netgravitational force on a particle located insideit.




(b) (�) The gravitational force on a particle ofmass m inside a uniform solid sphere, at adistance r from the center, is due only to massMinside in an “inside sphere” with that radiusr:

Minside =4

3πr3ρ =

M

R3r3,

where ρ is the solid sphere’s density, R is itsradius, and M is its mass.

(c) (�) We can assign this inside mass to be thatof a particle at the center of the solid sphereand then apply Newton’s law of gravitationfor particles. We find that the magnitude ofthe force acting on mass m is

F =GmM

R3r.

2. A spherical shell has inner radius R1, outer radiusR2, and mass M , distributed uniformly throughoutthe shell. The center is placed at the origin. Aparticle of mass m is placed at r = xi and thegravitational force on the particle by the shell isF (x).

(a) (�) If x < R1, then

F (x) = 0.

(b) (�) If R1 < x < R2, then

F (x) = −GmMx2

(x3 −R3

1

R32 −R3

1

)i.

(c) (�) If x > R2, then

F (x) = −GmMx2

i.

3. The mass density of a certain planet has sphericalsymmetry but is a function of the distance fromthe center of the planet. Let M and R be the totalmass and radius of the planet.

(a) (�) If we define M(r) by the mass inside thespherical surface of radius r, which isproportional to r, then we find that

M(r) = αr, r ≤ R.

Because M(R) = M , the constant α is M/R .

M(r) = Mr

R.

(b) (�) If we define ρ(r) by the mass density atradius r, then we find that

M(r) =

∫ r

04πr2ρ(r)dr.

Thus,

ρ(r) =1

4πr2dM(r)

dr

=1

4πr2M

R.

(c) (�) The gravitational force on a particle ofmass m and distance r from the center of theplanet is

F = −Gmr2M(r)r = −GmM

rRr, r < R,

where r is the unit vector along a radial axisextending from the planet. The negative signindicates that the direction of the force istoward the center. If r > R, then

F = −GmMr2

.

13-5 Gravitational Potential Energy

1. The gravitational potential energy U(r) of asystem of two particles, with masses M and m andseparated by a distance r, is the negative of thework that would be done by the gravitational forceof either particle acting on the other if theseparation between the particles was changed frominfinite (very large) to r.




(a) (�) A conventional choice of the referencepoint is r →∞ and the boundary value forthe gravitational potential energy is

U(r →∞)→ 0.

(b) (�) Then the gravitational potential energy is

U(r) = −∫

Fg · dr = −GMm

r.

(c) (�) If a system contains more than twoparticles, its total gravitational potentialenergy U is the sum of the terms representingthe potential energies of all the pairs. As anexample, for three particles, of masses m1,m2, and m3,

U = −(Gm1m2

r12+Gm1m3

r13+Gm2m3

r23

),

where rij is the distance between the twoparticles of masses mi and mj .

(d) (�) An object will escape the gravitationalpull of an astronomical body of mass M andradius R (that is, it will reach an infinitedistance). If the object’s speed near thebody’s surface is at least equal to the escapespeed, given by

vescape =

√2GM

R.

2. Two particles, each of mass m, are a distance dapart.

(a) (�) The potential energy of this system is

U1+2 = −Gm2

d.

(b) (�) The work required the two particles fromfar away to this situation is

W = −Gm2

d.

(c) (�) If there is another particle of mass 2m isat rest at the midpoint, then the potentialenergy of the 1 + 2 + 3 system is

U1+2+3 = −Gm2

d− 2

G(2m)m

d

2

= −9Gm2

d.

(d) (�) To bring a third particle, with mass 2m,from far away to a resting point midwaybetween the two particles, an external agentmust do work equal to

U1+2+3 − U1+2 = −8Gm2

d.

3. There are four particles of mass m at

r1 = +ia+ ja,

r2 = −ia+ ja,

r3 = −ia− ja,

r4 = +ia− ja.

(a) (�) The work required to bring the fourparticles from a→∞ to a = d1 is

W = −Gm2

(1

2d1× 4 +

1

2√

2d1× 2

)= −Gm

2

d1

(2 +

1√2

).

(b) (�) The work required to bring the fourparticles from a = d1 to a = d2 is

W = Gm2

(1

d1− 1

d2

)(2 +

1√2

).

13-6 Planets and Satellites: Kepler’s Laws

1. The motion of satellites, both natural andartificial, is governed by Kepler’s laws.




(a) (�) The Law of Orbits: All planets move inelliptical orbits with the Sun at one focus.

(b) (�) The Law of Areas: A line joining anyplanet to the Sun sweeps out equal areas inequal time intervals. (This statement isequivalent to conservation of angularmomentum.)

(c) (�) The Law of Periods: The square of theperiod T of any planet is proportional to thecube of the semimajor axis a of its orbit. Forcircular orbits with radius r,

T 2 =

(4π2

GM

)r3,

where M is the mass of the attracting body,the Sun in the case of the solar system. Forelliptical planetary orbits, the semimajor axisa is substituted for r.

2. The Law of Orbits

(a) (�) The semimajor axis a is half the width ofthe ellipse.

(b) (�) The eccentricity e is defined by

e =f

a,

where f is the absolute value of the xcoordinate of a focus when the center is at theorigin.

(c) (�) An eccentricity of zero corresponds to acircle.

(d) (�) The eccentricity of Earth’s orbit is only0.0167. Thus the orbit is close to a circle.

3. The Law of Areas

(a) (�) This law is closely related to theconservation of angular momentum.

(b) (�) The area of the shaded wedge in thefigure closely approximates the area swept outin time ∆t by a line connecting the Sun andthe planet, which are separated by distance r.

Sun

The area ∆A of the wedge is approximately

∆A ≈ 1

2r2∆θ.

(c) (�) The instantaneous rate at which area isbeing swept out is then

dA

dt=

1

2r2dθ

dt=

1

2r2ω.

(d) (�)

Sun

The angular momentum of the planet is

L = rp⊥ = r(mv⊥) = r(mrω) = mr2ω.

(e) (�) Therefore,

dA

dt=

L

2m.




4. The Law of PeriodsConsider the circular orbit with radius r. Theradius of a circle is equivalent to the semimajoraxis of an ellipse.

(a) (�) The centripetal force for the uniformcircular motion is the gravitational force.

(b) (�) Applying Newton’s second law F = ma tothe orbiting planet,

GMm

r2= mrω2.

(c) (�) The angular velocity ω can be replacedwith

ω =2π

T,

where T is the period of the orbital motion.

(d) (�) We obtain Kepler’s third law

T 2 =

(4π2

GM

)r3,

where the quantity in parentheses is aconstant that depends only on the mass M ofthe central body about which the planetorbits.

13-7 Satellites: Orbits and Energy

1. Consider a planet or satellite with mass m thatmoves in a circular orbit with radius r.

(a) (�) Its potential energy U is given by

U = −GMm

r.

(b) (�) Its kinetic energy K is given by

K =GMm

2r.

(c) (�) The total mechanical energy is given by

E = −GMm

2r.

(d) (�) For an elliptical orbit of semimajor axis a,the total mechanical energy is given by

E = −GMm

2a.

2. (�) Assume that Earth is in circular orbit aroundthe Sun with kinetic energy K and potentialenergy U , taken to be zero for infinite separation.Then, the relationship between K and U is

K = −1

2U.

3. As is shown in the figure, two satellites, A and B,both of equal mass m, move in the same circularorbit of radius r around the Earth but in oppositesenses of rotation and therefore on a collisioncourse. We ignore the gravitational force betweenthe two satellites.

Earth

(a) The total mechanical energy EA of thesatellite A and Earth before the collision is

KA =GMm

2r,

UA = −GMm

r,

EA = −GMm

2r,

where M is the Earth mass. Because A and Bare identical,

KA +KB =GMm

r,

UA + UB = −2GMm

r,

EA + EB = −GMm

r.

(b) (�) The total linear momentum of the systemis vanishing just before the collision.




(c) (�) The total angular momentum of thesystem is 0 before the collision.

(d) (�) The total linear momentum of the systemis 0 right after the collision.

(e) (�) The total angular momentum of thesystem is 0 after the collision.

(f) (�) If the collision is completely inelastic sothat the wreckage remains as one piece oftangled material of mass 2m, then the totalkinetic energy is 0. Thus the total potentialenergy just before the collision is the totalmechanical energy E′ right after the collision.

E′ = UA + UB = −2GMm

r.

(g) (�) Just after the collision, the wreckage fallsdirectly toward Earth’s center.

13-8 Einstein and Gravitation

1. (�) Einstein pointed out that gravitation andacceleration are equivalent. This principle ofequivalence led him to a theory of gravitation (thegeneral theory of relativity) that explainsgravitational effects in terms of a curvature ofspace.

2. Principle of Equivalence:

(a) (�) This is the fundamental postulate ofEinstein’s general theory of relativity.

(b) (�) Gravitation and acceleration areequivalent.

(c) (�) Einstein’s principle of equivalence statesthat gravitational mass and inertial mass arethe same.

3. Curvature of Space:

(a) (�) Einstein showed that gravitation is due toa curvature of space that is caused by themasses.

(b) (�) Space and time are entangled, so thecurvature of which Einstein spoke is really acurvature of spacetime, the combined fourdimensions of our universe.

(c) (�) When light passes near Earth, the path ofthe light bends slightly because of thecurvature of space there, an effect calledgravitational lensing .

(d) (�) When light passes a more massivestructure, like a galaxy or a black hole havinglarge mass, its path can be bent more.

(e) (�) In some situations, the quasars we seeblend together to form a giant luminous arc,which is called an Einstein ring .

(f) (�) Although our theories about gravitationhave been enormously successful in describingeverything from falling apples to planetaryand stellar motions, we still do not fullyunderstand it on either the cosmologicalscale or the quantum physics scale.


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