Chapter 7 Circular Motion and Gravitation

7.1 Calculate force of gravity using Newton's Law of Universal Gravitation.

5. What is the gravitational force between the Earth and the Sun? (Mass of Earth: 5.98 x 1024 kgMass of Sun: 1.99 x 1030,Radius of Earth's orbit around sun: 1.50 x 1011m)

a. 5.29 x 10.13 N r. .G ~L M,~ S,1J,<(()LVkj "". =- ,. H NO!'£. •..• I ro (p"£b. 3.53x 10:;> ~ r L J . It'" '"C. c. 7.94 X 1044

d. d. 5.29, 10" , ![,. '7tlo~,"'..\(:.IJN'" )~""I'"''; _~ ?8 CO" ~'.J (>'i.{/(l/'~.f C~,<;"2X/~1

7.2 Calculate force of gravity at a given distance given the force of gravity at another distance(making use of the inverse square relationship).

6. According to Newton's Universal Law of Gravitation, the gravitational force between twomasses -r ./ M. ,'1 •..

a. Is always an attraction YGe 'j lJ ("l.b. depends on how large the masses are Y {.Sc. depends inversely on the square of the distances between the masses Y C.J[d. all of the aboYj

D

7. Two balls near each other ha4a.certaingravitational force. What happens to the force if youtriple the distance?? r.. G'" •••,; .

a. 9x bigger j {b. 3x bigger M MC. 3x saller F - G . to _ G 1\. 1'1,-. 9x smaller j - (3f) ~ - q r>-

D

8. Suppose the gravitational force between two massive spheres is 20 N. If the magnitude of eachmass doubles, then what is the force between the masses? F:.= / ~

~ j v y"'" ::.- 'LON--4:-2e-~c. 40 N r / 2.""2,,,... 1/d.5N f-j,:; l.t (1 =-Yl'C"'f!")e.2.5N 1I(~ .J

(~OAJ.[.- 60?1A

9. Suppose the gravitational force between two massive spheres is 40 N. If the distance betweenthe spheres is cut in half, the force between the masses is

a. 160 Nb. 20 Nc. 10 Nd. 5 Ne. 2.5 N

A

~0--------

10. If the radius of the Earth increased (or you went on a mountain top), with no change in mass,your weight would technically

a. Increase f-:"Cb. no an e J

decreasedisappear

C

7.6

7.3 Calculate the gravitational field strength surrounding a planetary body.

11. Lola the explorer stands atop a planet. The planet's mass is 3.0 x 1025 kg and has a radiusof 790,000 meters. Find the acceleration due to gravity (gravitational field strength) for dropped

objects on this planet. _ G !!J- G ={. 67x/(;-tl ~C( - {t: l-j t.

a. 0 Ikg J . (,\~3 'I/o 2~ Itt...3200 N/k .J

c. 9.80 N/kg (/7 -I'~'~'7{)r::vr. 7Co~ ,0 XIO r' •. - I ,O()Otll\

d. 2.00 X 1015 N/kg J I- il,oa;",)

B ::3200""/s'/Calculate the velocity& period of an orbiting-satelhte.

the period of a satellite in orbit closer;:;--r ::2 7f (-r;;:;;It.s r '1' 71tI?J r+T.,B

12. Compared to the period of a satellite in orbit far to the Eato the Earth is ila

a. Lon er """r!shorter

enot enough information

13. A satellite is placed in orbit around the Earth(mean radius 0 at a surface altitude of3.00 x 106 meters. The satellite's ma~ is 750 kg and the mass of the Earth is 5.98 x 1024 kg. Find the

speed of the Earth satellite. (' -=Cb.3:lJX to 'I\. T 3.00;r. Ii .••.) := f. 38 X(06",a. 772~ (\~ ~5.1 e5,.Kr0 '-O/~ G~~.(,7XI f)-II!!,,'" @;.~20__ ftj "l. (': raej, <J.\ c{J& r1-(c. 7,907 m/srz-;;;- i 67)J {II",,.. t 5,",-~( ••" If;~'d. 9.8 x10.6 m/s -U ~ ~ ~:- . (-;-. s.rB..('d~}-: ~

B ['t.S8Krd6~ ~r-204<07.7 Calculate either the orbiting period or orbiting radius of a satellite using Kepler's 3,0 law.14. Gaia is 4.2 units from the center of Jupiter and orbits Jupiter once every 3 days. Zeus orbits 8.4units of distance from Jupiter, ~ow many days orbit is Zeus from the center of Jupiter? ~. Jg,'i.t (W1.

a. 2.6 days r{~.4rr'A = 3& f'jI, e 6.<tu.@J1. Gif';)~ ~k 4;.2.iJb. 0.35 days A '" ~

~. ~.o days 1 m V fElJ.U6 days. I B & ::. fj.~"C ~ f ~ (\"\2 C{

~ = ~ \ef'!,

7.8 Conceptually relate mass and distance separation to the gravitational force, field strength,orbiting velocity and period of orbit for planetary bodies.

15. The only factor that affects the Earth's gravitational force of attraction on objects at or near theEarth's surface is the mass of the Earth.

~

. Tr I, ,'"c...L.,Je.J )t'~f-a.II~ (at\. [.r.J.tI'Petf..e. "/1..()rL.".'9. alse a I -~::J"

B ~ ~~ of ike. ab'7.9 Recognize fundamental applications of Earth satellites. J~r

16. The sWe orbiting object around the Sun changes as it circles the Sun .. True r.epU:ff2. I.. ~lI.I~ /'": A ,/~

A ./ j..~- I Fft5 LA-I).)

Newton's Law of Gravitation

Diff: 2

Answer: TRUEDiff: 1

1) Consider the planet Mars: IYI ~ ?8 J(I ()l'[Earth mass; 5.98 x 1024kg = 9.3 x Mars mass; d' ~ , ''I = I 'f.5 v " 2 s~4.~ 0' 1'/'"Earth radius; 6380. km = 1.89 x Mars radius]What is the accele~tion of gravity on Mars? ref!:: b .~ tlo, OOQ 0\ ~ ~,YoX (0 b1'\

~.OXIO-II~ •. lUI l.~ \ o.Q/(' (J'1..J)(IJ k •U I

A 38 /2 1\ C=-(,.67YIO"f!N>M"nswer: . m s ),')&' X Vli'Nl}. \ " ~7.r 3, tJ1>1 3 =- G ~

3) Newton never knew the numerical value of "his" universal gravitational constant G.

O~LV a fl-u- C-v cvdtJ.~ J~Ll cKjJe¥,'M.Mr

a. ~t el"'"' ~ e.vJ+0 1) Ji eel-

fI) e-vJ !-or\.'c,. j,..J Law

ForWU'j a.J.'O(\ l~a.f'~pJ~01'\ .1~)~ (1.,'1\. tJtlMJ \ o...J. OfP~.f..e ~ lJct~~tu-UJOII

e~/'H,)

5) A 63. kg astronaut walks upon the surface of the planet Krypton which has 100. times the mass ofthe Earth and 100. times the diameter of the Earth. What does she weigh on Krypton?

f:::G ~ '\=-~31~ 'f"":;6,38.X'/~''''J(OP 't.38,A'IOt .••..,.) (' fill."~.?~)(I()"-'f~ \/4) ~ S" re Ol'

Answer: 63.x 9.8\x 100/ 100 = 6.2NewtoMS" . XI 9Oiff3 \~ 1/ 2~ \r~b,b 7X1()-.~~\~J{ S, %XI() It}j) L b,2.tJ

/I) ,38 IO~M."1.7) James wei~s 180.pounds. T~s means t e Earm pulls Ja~8waFEI it wit a 180. Ib force. Sincethe Earth is 7.3 x 1022 times more massive than James, the Earth is attracted to James with a force of

A) O.lb.B) 7.2 x 1022 lb.C) 7.3 x 1024 lb.0) 180. lb.E) 1.3 x 1025 lb.

f"-

Answer: DDiff:1

Choose from the following list:(a) J(b) m/s(c) rad(s(d) N m2/kg2(e) rad/s2

(f) radians(g) seconds(h) s2/m3

9) The weight of an object on the is what fraction of the weight of the same object on Earth?(Mass of Moon: 7.35 x 1022 kg . met of Moon:3476km) a : G ~

~~rr,~ = C M'~'ktl\~:7,3S)(li~ J"'«>t/~~ "~~0(-rJ1//)Answer: 116 1_ , -II••••, = ~.,,"/~ kj' I' J'flt) ~Diff: 1 y". 6 7 XIO 'i1' e 7J8 Clct/. •

( (),:: ~87(,/Mf" ~ 1738000 ~:: 1.61J"l. 11:= a t!, ~10) Consider a 2.4 kilo~ram feat er (Big'Bird)'on the planet Mars.~ ' 1'0 rs[Earth mass = 6.0 x 10 4 kg = 9.3 x Mars mass; ~= (,'f.!X(I./3/i'j::: "'. /11:" 6 ]Earth radius = 6380. km = 1.89 x Mars radius 1 {'If':: 3!' _----

(a) What does the feather weigh on Mars? • 8 XIO '" '" (' •..(b) What would the same feather weigh on Earth? M,-~ 2.,<.{~ c= '.67 ,J(1()'"'7f~

Answer: 9.1 Non Mars F> G I'>t'I{:"- >~,('fA(I)'"t:;.\f.l.I.(~X~,I.(JWl3~)!Iiil-24. Non Earth ~ ~~~ C~,~::lX(,,6_\1 7,()1J

Diff: 3 -'J

~ ::.M.~.{}. t.(IjJ?,8h,'):- 2 s,~2l=2.<{IV)

11) The Uni 'talional Constant is measured in which units?Answe. 0) N.m2(kg2

Diff:2

Answer: BDiff: 2

13) A spherically symmetric planet has four times the Earth's mass and twice its radius. If a jar ofpeanut butler weighs 12.N on the surface of the Earth, how much would it weigh on the surface of thisplanet? G 1"1, t>\ , Ih P ~ p J.j J@ 1~;i> (''-<.::= 12 tv _' ~ Ul"ll.' l::.(,l.I'e••.• ~V"

C) 30 N fLJD)6N G!",QI'I'JE) 24 N (2.r-J

- ~ G~~ ~{I)lI2 ~~

16) A spaceship is traveling to the moon. At what point is it beyond the pull of Earth's gravity?A) when it is closer to the moon than it is to EarthB) when it gets above the atmosphere _C) when it is half-way there f- :- G A, '+\4C?) It Is never beyond the pull of Earth's griiYlfY;> .s =) (~

Answer: 1.54NDiff: 2

11e. Moof\.J j~M"~ ~'1 afPur tf.-.•..cJt~t-Answer: D f\M) "e b t- t ...J bDiff:1 \ <.J-. I J (1\ €.V<. v Il. \1 a•.•.J ~

~ a.dill.\. ~ rt:lvd-", -17) By how manyJ'J~wtons does the weight of a 100. k1rperson change when he goes from sea leveltoanaltitudeof5.0~?' F. = G M,/Yl'l.. G,c,.b7X(O'"':::::, f',= G.38xIO'", ~ 'S,97NO>~ ",=100h.

jl r"T.. ~ :Jf =(U7jl()"':f:\C!061.~\{~.r7ttO'rt, . C'Jx 27.':" ,)" 3~{O'M)' 1 fU. N

.f: '" i.n ';«O-jI~L)(\16 '(~.n (Jrv _ t'17/ 7u _ ''f7g.2? @:¥ <-: '1' s8S"xlO' \ - 1 ~ I f ~7607~_.S; ~rJl19) A satellite encircles Mars at a distance above i s surfaceequal to 3 times the radius of Mars. Theacceleration of gravity at the satellite, as compared to the acceleration of gravity on the surface ofMars, is G' (nXI () -II!:::-'

A) one-ninth as much. 0:'~";;> Co( G M "1'B) ze ro. v---- . J \;=. ~

C) the same. I

D) ~ne-s~~eenth as much. ~ 1. ,:; ( 8'0 :; (f(;z -=- rf

Answer: EDiff: 2

3,Z.O.z;L _r,8d"/s' -

"')t ~ ~37jlfrJ

"~P ~f-'.---=~" .3~

Answer: 0.0024NDiff: 2

20) What is the gravitational force on a 70. kg person on the Earth, due to the Moon? The mass of theMoon is 7.36 x 1022kg and the distance to the moon is 3.82x 108 m. C <:. G ~~

or-. :,,82..N ()6", 'j ,,'l.fI\, = 701(j ? 16. ~ 7y'/()-I'Il;LWo., ~ 7.3' X(),.~ ~ *l'G ~ '.G 7,((;)"1/ •••~- -:= Z, 35" X(O-JrJ

22) An astronaut lands on a previous~ unknown planet of radius 2350. km and she measures theacceleration of gravity to be 3.20 mls . 0,' .3.20 ~/\'- ,....2..:'50 eXiO '"

(a) What is the mass of this new planeN 4",' 7no."N",/,~~(b) What does the astronaut weigh on this new planet if she weighs 384. N on Earth?

j ~G~z (3-,20 "'6- X.2,350,000 ••.)' I 2.S

( C' 7X(o"tJ.'/ : 2, bS' X/OAnswer: (a) 0.26 x 1024 kg 0 I~'

(b) 125. Newtons '!l MDiff:3 - ~ -;;.

G

23) A mass which weighs 19.6 newtons on Earth is 2.3 meters away from a 2.00 kg mass.(a) Determine their mutual gravitational attraction to each other.

r. 10(,1 - t /"1IN ML~2.00~

j = ,.b f'J - M.~ f{\,'" T ~~ 7 -:; Z. Q:&...9 (" = 2..~•••..Answer: (a) 5.0 x 10,11N .5 G ~ ~. ~ 7XI 0-" I!::-'D~3 ~~,.-- elf'., M.. t. y,r~= -r>:::>- ~.G7,.(10 -11.,::~ V:12.0 I~ Jo'-.O ~d..l:.

i"~~l(z. ~",)1. )7.6 Kepler's Laws and Earth Satellites _----_[s-.0 X f () -1/ (\7/3) As you know, the Earth has an orbital period of one year at an orbital radius of one AU. If a new"minor planet" were to be found in a circular orbit with radius 13. AU, what would be itsperiod?

Answer: Remembering Kepler's 3rd law: T = 47. years at R=13. AU.Oiff: 2

Answer: 0Oiff: 2

6) Kepler's discovery that T2/r3 = K appliesA) to orbits where K is a universal constant.B) only to circular orbits.C) to lunar motion provided K is.the same-for_planeta0) to elliptical orbits where r is the average distance.

7) The speed for a "low" circular orbit about the Earth is aboutA) 1.7 x 104 km/s. ( .• ~,3e)(fO'", i\\. ;.J,-,97X[OI~f{H) 7.6 km/s. ~ e: ~C) 2.1 km/s. -U -: ~0) 2.5 x 104 km/s. rE) 17. km/s.

-II AJ-'"G = (, b7 f,t() ,-~,<~

Answer: BOiff: 2

:: '.b7'KIO'lIr; YS.7YX/O'VIj

.... j f(b.:,8 Y(O',,)

= 7 110 X 103M. :::[7.?O IcM/sJ9) An astronaut goes out for a "space-walk" at a distance above the Earth equal to the radius of theEarth. Her acceleration will be r ~2 (;,.3~ X,O ~ '\ f"\. ~ 5.9711 0~!j

A) g G" Go (, 7V{O"I~;/IDi~~ &-' -0 '. a - G ~• O.25g j' , J"';:" I

E) .fig a -; I' MJ"' Lj C?- rr-

Answer: DDilf 2

10) If the Earth had four times its present mass, what would be its new period of revolution around theSun, compared to its present orbital period?

A) two times as muchB) one-fourth as muchC) four times as muchD ne-half as muchE) the same

Answer: EDiff: 2

Answer: BDiff 1

13) A planet is discovered to orbit around a star in the galaxy Andromeda, with the same orbitaldiameter as the Earth around our sun. If that star has 4 times the mass of our sun, what will theperiod of revolution of that new planet be, compared to the Earth's orbital period?

A) four times as muchB) one-half as muchC) one-fourth as muchD) twice as muchE) .J2 times as much

Answer: BDiff:2

14) An Earth satellite is in circular orbit 230 km above the surface of the Earth. It is observed to havea period of 89 min. From this information, estimate the mass of the Earth. "" =? r. b,300.xIO' l.$O ,2 ,. '<1l . t 'K({)~

Answer: 6.0 x 1024 kg Co ~161'jI06",Diff:2 rTl_ 2.:rr { r' 6. 67,((O"'AI.'

I -l Gr.. fY\$ ~ i rr 1(3 ~l .n~o"y..l' "LfrrL ~ C T~" .•.s. tf X/()?';:

G "'.

15) The following statements refer to man-made, artificial satellites in orbit around Earth. Which is anaccurate statement? re::J

A) The velocity required to keep a satellite in a given orbit depends on the mass of the ~e.B) The peri~volution of a sate lite oving about the Earth is independent of the size of

the orbit it travels.~ v.::C) A satellite in a large diameter circular orbit will always have a longer period of revolution

abo rth than will.a atellite. in a smaller circular orbit.0) It is possible ~ satellite traveling at either a high speed or at a IOWspeedi

circular orbit.E) Only circular are possible for artificial satellites. <::g;;)

T~2l\[/~- ~-r-0-Answer: COiff:2

16) Europa, a moon of Jupiter, has an orbital d~'a eter of 1.34 x 109 m, and a period of 3.55 days.

Wh";,them,,,ofJ"p;'e" T. fo,sS; >?'."~d0 ~ ..f"C ~tOjlfJ-II:;:..'" ~ L. ""iJ ~V' -~ = 3,()7XI().$Itj~ r = (, ~ If X (() p -::: ~ 7):/~8 ..1

Answer: 1.89 x 1027kg rn 2. I ': ",J c; If 1. t,loKtot •••..\O~2 f = ~ ~, J. ~:> M.n+ 17l'" -,-- ~. (,7/(()""N'" _'

rj1 2 - ;;Z ~Ol,(/~JI -:: 'IT G M~ {f\ ~ 4 If 1 (.3 _ 'rr ~:~n "1 /~ J+ G rri --I,B?XIC;:~r.

18) Consider a small satellite movrng in a circular orbit (radius r) about a spherical planet (massThe period does not depend upon ~

A) g at the satellite position. '(B) the lanet mass. rr =:- 21f G,~u.r

e satellite mass. rera IUS r.

E) the universal gravitational constant.

Answer: COiff: 2

20) A satellite orbits the Earth once every 6.0 hours in a circle. What is the radius of the orbit if it is a

2000. kg satellite? rp -=f,O~rV~O(v\I'A V~ \ :;;2-1.600.5 C "'~.(,,7X(O-lt!!:.'~ ~ hr p (V\\"- J (If t

Answer: 1.7 x 107 m M41 ~ ~. nXIO'~I~ ;> j 4: (7 X{U'''''-:1JJ{21, (~~ 1Y(S.17f./r/Y;q)Olff: 2 'i" "2 If n 2.. rr2 -------rr~2rr ~t;f'~~n'--" {J t 1,68Y(07"'-----1

'T' ':::t.( "1 ~ ~~ C '" r-r < .- riT (,M. i.;,T~

21 ) Satellite A has twice the mass of satellite T, and rotates in the same orbit. Which of the followingis true?

A) The speed of T is twice the speed of A.B) The speed of T is one-fourth the speed of A.C) The speed of T is half the speed of A.D) The speed of T is equal to the speed of A.

Answer: DDiff 2

30) W ftA) KeplerB opernicusC) Brahe0) EinsteinE) Galileo

Answer: EDiff: 2

23) A satellite is in a low circular orbit about the Earth (Le., it just skims the surface of the Earth). Howlong does it take to make one revolution around the Earth? if:: (',3£ Xlcfo 11\ G' ,,{, 7¥(d-tl~~

rr ~.2..1l fli:..- 'rt\ ••.$;,97 Xlo'YI'j k"~<-

Answer: 87 min /"Diff" 2 lJ. Si:>AlO '" ~ ~

. ;: L.1r '.'7,(IO-l!7;: S,r7)'ll~~'::S-{)7'fs ((~) = tYi.r6 "-8~1;)29) David lands on planet X and observ s objects fr.ee-falling with acceleration 4.9 m/s2

. line pla~is 3 times larger than the Earth, what is the planet MASS ex~ressed in Earth masses?

A) 9/4 j::l.(,9~/{L r .. 3(j.~kl/M) =- /.r/fld~ r;"~.67i10-'I"'J'LB) 2/3 1/ 'X 1... ~t.2li> g - (~ f'\ ~ l.lf.~)5~(/.?IXI07 0 M 1

91 - {' t. {,. (, 7y( () '"II}:!.:.." •• f"lN~r:tj ~ f'\~

3 r 1.. ;. G "'\ Mtwi"'2.67 X I02.?~ z. 6f ,.(102 ~,

~ ••.r\) £. It) L'f5G M, ::: If.sTlr.,

. st person to realize that the planets move in elliptical paths ar

Answer: ADiff I

31) The net force on an object in orbit is zero.

Answer: ~ L or-Diff:1~ \c...=~

'(\e.,e..J.j 0 f!o.".. C- ~

q Q""JQ..32) A satellite orbits the Earth once every 6.0 hours in a circle. What is the acceleration of the satellite(direction too)? /r',,? rT1 1/ ~/O ' ••vLrlJJd_ d 2.('( ('

"I. , I "'l!?'()~ ~~ = 21 &CXJ-s LJ:: t '0" -

Answer: 1.4 m/s2 toward center of Earth G _ I I -., ' , WI' ( rrDiff: 2 t:" - ~." 7 'flf)~... _ _ -v 1 ("2.-Tlff. ":x{~'.,IY

~::%Z.~ ~,,~ 0-t.S-,97X!t!1.1{~1 1--:-/-,L(-I-"(.-s~-1C/C.- -r"'\..: 'Z.I"""'J;)

0t.L l",ohl<Of", r •.J.t..~) ~---- /.&S,.rro..,