Su (ycs73) – HW04 – Tsoi – (58020)
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001
(part 1 of 2) 10.0 points
Two satellites
A
and
B
orbit the Earth in
the same plane.
Their masses are
m
and
6
m
, respectively, and their radii
r
and 3
r
,
respectively.
3
r
r
6
m
B
m
A
What is the ratio of the orbital speeds?
1.
v
B
v
A
=
√
3
2.
v
B
v
A
= 3
3.
v
B
v
A
=
1
√
2
4.
v
B
v
A
=
1
3
5.
v
B
v
A
=
1
√
3
correct
6.
v
B
v
A
=
1
9
7.
v
B
v
A
= 2
8.
v
B
v
A
=
1
2
9.
v
B
v
A
=
√
2
10.
v
B
v
A
= 9
Explanation:
The force of gravity is responsible for hold
ing a satellite in its orbit, so the orbital cen
tripetal force is equal to the force of gravity:
F
r
=
m
v
2
r
=
G
M
E
m
r
2
,
where
M
E
is the mass of the Earth,
m
is the
mass of the satellite, and
r
is the radius of
the orbit (from the Earth’s center). Thus the
tangential speed
v
of an orbit at distance
r
is
v
=
radicalbigg
G M
r
∝
radicalbigg
1
r
.
Since
r
A
r
B
=
1
3
, the ratio
v
B
v
A
=
radicalbigg
r
A
r
B
=
1
√
3
.
002
(part 2 of 2) 10.0 points
Let 10
R
E
be the distance of the satellite
A
from the center of the Earth, where
R
E
is the
radius of the Earth.
What is the gravitational acceleration due
to the Earth at satellite
A
?
g
is the gravita
tional acceleration at the surface of the Earth.
1.
g
A
=
g
√
11
2.
g
A
=
g
100
correct
3.
g
A
=
g
81
4.
g
A
=
g
11
5.
g
A
=
g
3
6.
g
A
=
g
√
10
7.
g
A
=
g
9
8.
g
A
=
g
121
9.
g
A
=
g
10
10.
g
A
=
g
Explanation:
g
A
=
G M
E
r
2
A
=
G M
E
(10
R
E
)
2
=
1
100
g
.
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Su (ycs73) – HW04 – Tsoi – (58020)
2
003
(part 1 of 2) 10.0 points
A small metal ball is suspended from the ceil
ing by a thread of negligible mass.
The ball
is then set in motion in a horizontal circle so
that the thread’s trajectory describes a cone.
The acceleration of gravity is 9
.
8 m
/
s
2
.
v
r
g
ℓ
m
θ
What is the speed of the ball when it is in
circular motion? Answer in terms of
g
,
ℓ
and
θ
.
1.
v
=
g ℓ
tan
θ
2.
v
=
radicalbig
g ℓ
sin
θ
3.
v
=
g
ℓ
tan
θ
4.
v
=
radicalbig
g ℓ
cos
θ
5.
v
=
√
ℓ
tan
θ
6.
v
=
radicalbig
g ℓ
cos
θ
7.
v
=
radicalbig
g ℓ
sin
θ
8.
v
=
√
g ℓ
sin
θ
9.
v
=
radicalbig
g ℓ
tan
θ
10.
v
=
radicalbig
g ℓ
tan
θ
sin
θ
correct
Explanation:
Use the free body diagram below.
T
m g
θ
The tension on the string can be decom
posed into a vertical component which bal
ances the weight of the ball and a horizontal
component which causes the centripetal ac
celeration,
a
centrip
that keeps the ball on its
horizontal circular path at radius
r
=
ℓ
sin
θ
.
If
T
is the magnitude of the tension in the
string, then
T
vertical
=
T
cos
θ
=
m g
(1)
and
T
horiz
=
m a
centrip
or
T
sin
θ
=
m v
2
ball
ℓ
sin
θ
.
(2)
Solving (1) for
T
yields
T
=
m g
cos
θ
(3)
and substituting (3) into (2) gives
m g
tan
θ
=
m v
2
ball
ℓ
sin
θ
.
Solving for
v
yields
v
=
radicalbig
g ℓ
tan
θ
sin
θ .
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 Spring '09
 TSOI
 Force, Friction, Mass, Heat, Correct Answer

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