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Transcript
Chapter 7 Circular Motion and Gravitation
7.1 Calculate force of gravity using Newton's Law of Universal Gravitation.
24
5. What is the gravitational force between the Earth and the Sun? (Mass of Earth: 5.98 x 10
11
Mass of Sun: 1.99 x 1030,Radius of Earth's orbit around sun: 1.50 x 10 m)
a. 5.29 x 10.13 N
£b. 3.53x 10:;>
C.
c. 7.94
X
r.~ . G
r
•..• I
"". =- ,. H NO!'£.
J
L
ro
.
d. d. 5.29, 10"
,
![,. '7tlo~,"'..\(:.IJN'" )~""I'"'';
CO"
~'.J
(>'i.{/(l/'~.f
(p"
It'"
44
10
8
M,~ S,1J,<(()LVkj
~L
_~
kg
'"
?
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 two
masses
M. ,'1•..
a. Is always an attraction YGe
'j lJ ("l.
b. depends on how large the masses are
Y {.S
c. depends inversely on the square of the distances between the masses
C.J
[d. all of the aboYj
-r ./
Y
D
r..4G'" •••,;
7. Two balls near each other ha a.certaingravitational
triple the distance??
a. 9x bigger
j
{
b. 3x bigger
M M
C. 3x saller
to
_
. 9x smaller
j (3f) ~ -
F - G
.
.
force. What happens to the force if you
G
1\. 1'1,-
q r>-
D
8. Suppose the gravitational force between two massive spheres is 20 N. If the magnitude of each
mass doubles, then what is the force between the masses?
~
--4:-2e-~
F:.= / ~
j
v
y"'"
::.-
'LON
c. 40 N
r
/ 2.""2,,,...
d.5N
f-j,:;
l.t
e.2.5N
(1
1I(~
(~OAJ.[.A
1/
=-Yl'C"'f!")
.J
60?1
9. Suppose the gravitational force between two massive spheres is 40 N. If the distance between
the spheres is cut in half, the force between the masses is
a. 160 N
b. 20 N
c. 10 N
d. 5 N
e. 2.5 N
A
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
b. no
an e
J
decrease
disappear
f-:"C
~0
--------
C
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 radius
of 790,000 meters. Find the acceleration due to gravity (gravitational field strength) for dropped
objects on this planet.
a. 0
_
C( -
G !!J-
J
Ikg
.3200 N/k
c. 9.80 N/kg
d. 2.00 X 1015 N/kg
~
G ={. 67x/(;-tl ~
{t:
(/7
,0
l-j
. (,\~3 'I/o
~'7{)r::v•.
XIO -I'~'
r'
J
I-
r.-
t.
2~ Itt..
.J
7 Co
I ,O()Otll\
il,oa;",)
::3200""/s'/
B
7.6 Calculate the velocity&
period of an orbiting-satelhte.
the period of a satellite in orbit close
12. Compared to the period of a satellite in orbit far to the Ea
to the Earth is
ila
a. Lon er
"""r!
shorter
r ::2
e
not enough information
It.s
B
I?J
r;:;--
7f
(-r;;:;;
r '1' 71t
r+T.,
13. A satellite is placed in orbit around the Earth(mean radius 0
at a surface altitude of
6
3.00 x 10 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
(\~ ~ 5.1 e5,.Kr0 '-O/~
a. 772~
~20__
c. 7,907
d. 9.8 x10.6 m/s
.••.)
:=
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f. 38
X(06",
ftj
m/srz-;;;-
.
"l.
s. rB..('d~}-:
(-;-.
<J.\
c{J& r1-(
5,",-~( •• "
~
If;~'
~r-204<0
['t.S8Krd6~
B
raej,
(':
t
i 67)J {II",,..
-U ~ ~ ~:-
@;.
f)-II!!,,'"
7.7 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 8.4
units of distance from Jupiter, ~ow many days orbit is Zeus from the center of Jupiter?
~.
a. 2.6 days
b. 0.35 days
r{~.4rr'A = 3&
f'jI,
e
6.
<[email protected] Gif';)~ ~k
A
m
1
~. .U6
~.o days
days.
IB
C
~
~
'" ~
V&
f'!,
4;.2.iJ
fElJ
::.
f ~ (\"\2
= ~
Jg,'i.t (W1.
\e
fj.~"
C{
7.8 Conceptually
orbiting velocity
relate mass and distance separation to the gravitational
and period of orbit for planetary bodies.
force, field strength,
15. The only factor that affects the Earth's gravitational force of attraction on objects at or near the
Earth's surface is the mass of the Earth.
I, ,'"
a c...L.,Je.J
. Tr
. alse
~
7.9 Recognize fundamental
16. The sWe
)t'~f-a.II~
I
(at\. [.r.J.tI'Petf..e.
of ike. ab'
B
~
~~
applications of Earth satellites.
-~::J"
J~r
orbiting object around the Sun changes as it circles the Sun .
r.epU:ff2.
. True
.
~lI.I~
I.
A ,/~
/'":
A
Newton's
"/1..()rL.".'9
./
j..~
-
I
=
I
0'
LA-I).)
Fft5
Law of Gravitation
1) Consider the planet Mars:
[Earth mass; 5.98 x 1024kg = 9.3 x Mars mass;
Earth radius; 6380. km = 1.89 x Mars radius]
IYI
What is the accele~tion of gravity on Mars?
~.OXIO-II~ •. lUI
l.~
ref!::
/('
2
A nswer: 38. m/ s
(J'1..J)(IJ
1\
.r
3, tJ
,
?8
\
J( I ()l'
''I
4.~
b .~ tlo,
\
k
),')&' X Vli'Nl}.
Diff: 2
~
d' ~
OOQ 0\
~
'f.5 1'/'"
v " 2 s~
~,YoX (0
b1'\
o.Q
•U
I
C=-(,.67YIO"f!N>M"
"
1>1 3
=-
~7
G~
3) Newton never knew the numerical value of "his" universal gravitational constant G.
O~LV a fl-u- C-v cvdtJ.~
a. ~t el"'"' ~ e.vJ +0 1) Ji eel
Answer: TRUE
Diff: 1
J~Ll
cKjJe¥,'M.Mr
-
5) A 63. kg astronaut walks upon the surface of the planet Krypton which has 100. times the mass of
the 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) ~
Answer: 63.x 9.8\x 100/ 100
Oiff3
7) James
the Earth
A)
B)
C)
0)
E)
r ~b,b
\~
= 6.2NewtoMS"
S"
.
1/
re XI
.••..
Ol'
2~
9
\
7X1()-.~~\~J{ S, %XI() It}j)
/I)
,38 IO~M."1.
-
L b,2.tJ
f"
wei~s 180.pounds. T~s means t e Earm pulls Ja~8waFEI
it wit a 180. Ib force. Since
is 7.3 x 1022 times more massive than James, the Earth is attracted to James with a force of
O.lb.
7.2 x 1022 lb.
!-or\.'c,.
7.3 x 1024 lb.
180. lb.
WU'j a.J.'O(\
01'\
1.3 x 1025 lb.
fI) e-vJ
j,..J Law
l~a.f'~pJ~ .1~)
For
~
(1.,'1\.
tJtlMJ
\
o...J.
OfP~.f..e ~
lJct~~tu-UJOII
e~/'H,
)
Answer: D
Diff:1
9) The weight of an object on the
is what fraction of the weight of the same object on Earth?
22
(Mass of Moon: 7.35 x 10 kg . met of Moon:3476km)
~~rr,~ =
Answer: 116 1_ ,
6 7 XIO
Diff: 1
y".
( (),::
C
a
: G~
J"'«>t/~~ "~~0(-rJ
M'~'ktl\~:7,3S)(li~
= ~.,,"/~
-II •••• ,
'i1'
~87(,/Mf"
~ 1738000
1.61J"l.
~::
10) Consider a 2.4 kilo~ram feat er (Big'Bird)'on the planet Mars.~
3
[Earth mass = 6.0 x 10 4 kg = 9.3 x Mars mass; ~= (,'f.!X(I./ /i'j::: "'.
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.,<.{~
F> G
Answer: 9.1 Non Mars
24. Non Earth
Diff: 3
11:=
J'flt)
ea7J8
I'
t!,
' 1'0
/11:" 6 ]
1//)
~
Clct/. •
rs
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_----
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c= '.67 ,J(1()'"'7f~
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~
~~~
C~,~::lX(,,6_\1
7,()1J
-'J
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
11) The Uni
Answe. 0) N.m2(kg2
Diff:2
I'>t'I{:"-
kj'
J?,8h,'):- 2 s, ~2l=2.<{ IV)
::.M.~.{}. t.( Ij
~
'talional Constant is measured in which units?
13) A spherically symmetric planet has four times the Earth's mass and twice its radius. If a jar of
peanut butler weighs 12.N on the surface of the Earth, how much would it weigh on the surface of this
planet?
1"1, t>\ ,
Ih
P
~ p J.j
@ 1~;i>
C) 30 N
D)6N
E) 24 N
Answer: B
Diff: 2
G
12 tv
(''-<.::=
_'
~
Ul"ll.'
fLJ
G!",QI'I'J
(2.r-
J
- ~ G~~ ~{I)lI2 ~
~
J
l::.(,l.I'e••.• ~V"
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 Earth
B) when it gets above the atmosphere
_
C) when it is half-way there
A, '+\4
It Is never beyond the pull of Earth's griiYlfY;> .s
=)
(~
G
f- :-
C?)
11e.
Answer: D
Diff: 1
f\M)
"e
\
~ a. dill.\.
afPur tf.-.•..cJt~t...J
b \1 a•.•.J ~
v
j~M"~ ~'1
Moof\.J
b <.J-. t-
tJ
I
(1\ €.V<.
~ rt:lvd-",
Il.
-
17) By how manyJ'J~wtons does the weight of a 100. k1rperson change when he goes from sea level
toanaltitudeof5.0~?'
M,/Yl'l..
G,c,.b7X(O'"':::::,
f',= G.38xIO'",
~ 'S,97NO>~ ",=100h.
jl
r"T..
~
:J
F. = G
Answer: 1.54N
Diff: 2
f
=(U7jl()"':f:\C!061.~\{~.r7ttO'rt,
.':"
,)"
.f: '" i.n ';«O-jI~L)(\16
:¥ <-:
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_
7u
t'17/
27
N
_ ''f7g.2?
@
s8S"xlO' \
- 1 ~I
f
~7607~_.S;
19) A satellite encircles Mars at a distance above i s surfaceequal to 3 times the radius of Mars. The
acceleration of gravity at the satellite, as compared to the acceleration of gravity on the surface of
Mars, is
G' (nXI () -II!:::-'
A) one-ninth as much.
0:'~";;>
Co(
M
"1'
B) ze ro.
J \;=.
~
G
v---- .
C) the same.
D) ~ne-s~~eenth as much.
~ 1. ,:; (
8'0 :; (f(;z
I
-=-
rf
Answer: E
Diff: 2
20) What is the gravitational force on a 70. kg person on the Earth, due to the Moon? The mass of the
8
Moon is 7.36 x 1022kg and the distance to the moon is 3.82x 10 m.
C <:.
G ~~
or-.
:,,82..N
fI\, = 701(j
Answer: 0.0024N
Diff: 2
Wo., ~
()6",
'j
?
7.3' X(),.~
G ~ '.G 7,((;)"1/
16. ~
,,'l.
7 y'/()-I'Il;L
~
-:=
•••
~-
*l'
Z, 35" X(O-JrJ
22) An astronaut lands on a previous~ unknown planet of radius 2350. km and she measures the
acceleration 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?
Answer: (a) 0.26 x 1024 kg
(b) 125. Newtons
Diff:3
~G ~z
j
(
(3-,20
"'6- X.2,350,000 ••.)'
C' 7X(o"tJ.'/
'!l
G
M
~ -;;.
2, bS' X/O
I~'
0
-
I
:
"')t ~ ~37jlfrJ
"~P
~f
-'.---=
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.3~
3,Z.O.z;L _
r,8d"/s' -
2.S
~rJl
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. =
0(,1
t
-
1 ,.b f'J - M.~
j
Answer: (a) 5.0 x 10,11N
D~3
/"1IN
ML~2.00~
T ~~
f{\,'"
7
Z. Q:&...9
-:;
(" = 2..~•••..
G ~ ~. ~ 7XI 0-" I!::-'
~~
.5
,.-- elf'.,
r~=
t.
M..
y,
~d..l:.
i"~~l(z. ~",)
-r>:::>- ~.G7,.(10 -11.,::~ V:12.0
I~ Jo'-.O
1. )
7.6 Kepler's Laws and Earth Satellites
_----_
[s-. 0 X (\7/
f () -1/
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 its
period?
Answer: Remembering Kepler's 3rd law: T = 47. years at R=13. AU.
Oiff: 2
6) Kepler's discovery that T2/r3 = K applies
A) to orbits where K is a universal constant.
B) only to circular orbits.
C) to lunar motion provided K is.the same-for_planeta
0) to elliptical orbits where r is the average distance.
Answer: 0
Oiff: 2
7) The speed
A) 1.7
H) 7.6
C) 2.1
0) 2.5
E) 17.
for a "low" circular orbit about the Earth is about
4
x 10 km/s.
( .• ~,3e)(fO'",
i\\. ;.J,-,97X[OI~f{
km/s.
~
e:
~
km/s.
-U -: ~
x 104 km/s.
r
km/s.
r;
lI
::
'.b7'KIO'
....
=
= (, b7
-II AJ-'"
f,t()
,-~
,<~
YS.7YX/O'VIj
j
(b.:,8
Answer: B
Oiff: 2
G
f
Y(O',,)
3
7 110 X 10
M.
:::[7.?O IcM/sJ
9) An astronaut goes out for a "space-walk" at a distance above the Earth equal to the radius of the
Earth. Her acceleration will be
r ~2 (;,.3~ X,O ~ '\ f"\. ~ 5.9711 0
~!j
IDi~~
G" Go (,
A) g
•
O.25g
E)
.fig
&-'
j'
-
0,
7V{O"I~;/
aJ"';:"
'.
a -;
I'
J"'
Lj
G ~I
M
C?- r
r-
Answer: D
Dilf 2
10) If the Earth had four times its present mass, what would be its new period of revolution around the
Sun, compared to its present orbital period?
A) two times as much
B) one-fourth as much
C) four times as much
D ne-half as much
E) the same
Answer: E
Diff: 2
Answer: B
Diff 1
13) A planet is discovered to orbit around a star in the galaxy Andromeda, with the same orbital
diameter as the Earth around our sun. If that star has 4 times the mass of our sun, what will the
period of revolution of that new planet be, compared to the Earth's orbital period?
A) four times as much
B) one-half as much
C) one-fourth as much
D) twice as much
E)
.J2 times
as much
Answer: B
Diff:2
14) An Earth satellite is in circular orbit 230 km above the surface of the Earth. It is observed to have
a period of 89 min. From this information, estimate the mass of
r. b,300.xIO' t l.$O 'K({)~
,
2 the Earth. ,."" '<1l =?.
Answer: 6.0 x 1024 kg
Diff:2
Co
rTl_
I
l'
2.:rr { r'
Gr..
-l
"LfrrL
~
G "'.
fY\$ ~
i rr 1(3
C T~"
~161'jI06",
6. 67,((O"'AI.'
~l
.•.s. tf X/()?';:
y..
.n~o"
15) The following statements refer to man-made, artificial satellites in orbit around Earth. Which is an
accurate statement?
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.
re::J
<::g;;)
T~2l\[/~- ~-r-0-
Answer: C
Oiff:2
>?'."~d0 =
~
9
16) Europa, a moon of Jupiter, has an orbital d~'a eter of 1.34 x 10 m, and a period of 3.55 days.
fo,sS;
C ~tOjlfJ-II:;:..'" T.
~ L.
Wh";,them,,,ofJ"p;'e"
27
r=
Itj ~
Answer: 1.89 x 10 kg
(, ~ If X (()
O~2
.
rj1
I -::
-
'IT
rr ~:~n
~V'
p
G
f
{f\
M~
2.
c; If
",J
= ~17l'" ~,-,-I
':
4 If
~
J+
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3,()7XI().$
-~
7):/~8
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rn
M.n+
~:>
2
""iJ
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~.
(.3
1
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1.
t,loKtot •••..
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(,7/(()""N'"
;;Z ~Ol,(/~J
'
--I, B? XIC;:~r.
rri
18) Consider a small satellite movrng in a circular orbit (radius r) about a spherical planet (mass
The period does not depend upon
~
A) g at the satellite position.
'(
B) the lanet mass.
=:- 21f
G,~u.r
e satellite mass.
r
era IUS r.
E) the universal gravitational constant.
rr
Answer: C
Oiff: 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~ \
~
M41 ~
Answer: 1.7 x 107 m
Olff: 2
~
~.
'i" "2
p
hr
nXIO'~I~
.
"1 ~
iT
(,M.
rr
2
~~
C '"
r-r <
i.;,T~
J
;>
rr~ 2rr ~t;f'~~n'--"
'T' ':::t.(
(V\\"-
.-
C "'~.(,,7X(O-lt!!:.'
:;;2-1.600.5
4:
j
t
(7
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If
n
(If t
21, (~~
2.
-------
{J
r
1,68Y(07"'-----1
_'
Y(S.17f./r/Y;q)
1
21 ) Satellite A has twice the mass of satellite T, and rotates in the same orbit. Which of the following
is 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: D
Diff 2
23) A satellite is in a low circular orbit about the Earth (Le., it just skims the surface of the Earth). How
long does it take to make one revolution around the Earth?
if:: (',3£ Xlcfo 11\
7¥(d-tl~~
rr
Answer: 87 min
Diff" 2
~.2..1l
G' ,,{,
fli:..-
'rt\ ••.
$;,97 Xlo'YI'j
k"~<-
/"
.
;: L.1r
lJ. Si:>AlO '"
~
~
= tYi.r6 "- 8~1;)
'.'7,(IO-l!7;: S,r7)'ll~~'::S-{)7'fs ((~)
2
29) David lands on planet X and observ s objects fr.ee-falling with acceleration 4.9 m/s
is 3 times larger than the Earth, what is the planet MASS ex~ressed in Earth masses?
j::l.(,9~/{L
A) 9/4
B) 2/3
2li>
r .. 3(j.~kl/M)
g - (~
91
-
3r
Answer: E
Diff: 2
~
'X
1/
f'\ ~
{'
t.
=-
/.r/fld~
G "'\
l.lf.~)5~(/.?IXI07
{,. (, 7y( () '"II}:!.:.."
~t.
0
M
•• f"lN~r:
f'\~
z. 6f
2
,.(10
1
~,
£. It) L'f5
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M,
G
pla~
1...
Mtwi"'2.67 X I02.?~
••.r\)
line
r;"~.67i10-'I"'J'L
tj ~
1.. ;.
.
. st person to realize that the planets move in elliptical paths ar
30) W
A)
B
C)
0)
E)
ft
Kepler
opernicus
Brahe
Einstein
Galileo
Answer: A
Diff I
31) The net force on an object in orbit is zero.
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32) A satellite orbits the Earth once every 6.0 hours in a circle. What is the acceleration of the satellite
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Answer: 1.4 m/s2 toward center of Earth
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