(cyl327) lee HW06 gentle (56245)
This print-out should have 16 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
001 10.0 points
The innerspring mattress on your grandmothers bed is held up by 20 vertical springs,
each having a spring constant of 5000 N/m.
A 32 kg person jumps from a 1.99 m platform
onto the innersprings.
The acceleration of gravity is 9.8 m/s2 .
Assume: The springs were initially unstretched and that they stretch equally (typical old fashioned bed).
Determine the stretch of each of the springs.
1
then
(32 kg) (9.8 m/s2 )
20 (5000 N/m)
1.24911 108 N2
+
20(5000 N/m)
= 0.1149 m .
x=
002 10.0 points
A 9.2 kg block is dropped onto a spring of
spring constant 3723 N/m from a height of
700 cm.
9.2 kg
7m
Correct answer: 0.1149 m.
Explanation:
We have
Us
at end
+ Ug
end
= Ug
initial
.
Taking x = 0 at the initial level of the innersprings, this becomes (for n springs)
n
1
k x2 m g x = m g h ,
2
Find the speed of the block when the compression of the spring is 12 cm. The acceleration of gravity is 9.81 m/s2 .
Correct answer: 11.5701 m/s.
Explanation:
or
n
k x2 m g x m g h = 0 .
2
n
Let a = k , b = m g , and c = m g h. Using
2
the quadratic formula
b b2 4 a c
x=
2a
(and taking the positive solution to get the
lowest position of the person) gives
x=
mg +
m2 g 2 + 2 n k m g h
.
nk
Since
m2 g 2 + 2 n k m g h
= (32 kg)2 (9.8 m/s2 )2
+ 2 (20) (5000 N/m) (32 kg)
(9.8 m/s2 )(1.99 m)
= 1.24911 108 N2 ,
Let :
m = 9.2 kg ,
k = 3723 N/m ,
h = 700 cm = 7 m ,
x = 12 cm = 0.12 m , and
g = 9.81 m/s2 .
Applying conservation of energy,
U g + U s + K = 0
1
1
m g (h + x) + k x2 + m v 2 = 0 .
2
2
Since
k x2
v 2 = 2 g ( h + x)
m
2
= 2 (9.81 m/s ) (7 m + 0.12 m)
(3723 N/m) (0.12 m)2
9.2 kg
22
= 133.867 m /s
v=
133.867 m2 /s2 = 11.5701 m/s .
lee (cyl327) HW06 gentle (56245)
2
keywords:
Let :
003 10.0 points
The planet Mars has a mass of 6.1 1023 kg
and radius of 3.4 106 m.
What is the acceleration of an object
in free fall near the surface of Mars?
The value of the gravitational constant is
6.67259 1011 N m2 /kg2 .
h = 3.37 RE ,
r = 6.37 106 m ,
g = 9.8 m/s2 , and
G = 6.67259 1011 N m2 /kg2 .
Using Newtons law of gravity,
F =G
Correct answer: 3.521 m/s2 .
the gravitational acceleration on the surface
of the Earth is
Explanation:
Let :
M = 6.1 1023 kg ,
R = 3.4 106 m , and
G = 6.67259 1011 N m2 /kg2 .
Near the surface of Mars, the gravitation
force on an object of mass m is
Mm
F =G
,
R2
so the acceleration of an object in free fall is
a=
F
m
M
R2
= (6.67259 1011 N m2 /kg2 )
6.1 1023 kg
(3.4 106 m)2
=G
2
= 3.521 m/s .
004 10.0 points
When a falling meteor is at a distance 3.37
times the radius of the Earth above the
Earths surface, what is its free fall acceleration? The acceleration of gravity is
9.8 m/s2 , the universal gravitational constant
is 6.67259 1011 N m2 /kg2 , and the Earths
radius is 6.37 106 m.
Correct answer: 0.513172 m/s2 .
Explanation:
mM
,
r2
g=G
ME
2,
RE
where ME is the mass of the Earth and RE
is Earths radius. The free-fall acceleration of
the meteor is
ME
(3.37 RE + RE )2
ME
1
=
2 G R2
4.37
E
1
=
g
4.372
1
2
=
2 (9.8 m/s )
4.37
a=G
= 0.513172 m/s2 .
005 (part 1 of 3) 10.0 points
Given: G = 6.67259 1011 N m2/kg2
A 749 kg uniform solid sphere has a radius
of 0.549 m.
Find the magnitude of the gravitational
force exerted by the sphere on a 70.4 g particle located 0.861 m from the center of the
sphere.
Correct answer: 4.74616 109 N.
Explanation:
In this case the distance r1 from the particle
to the center of the sphere is greater than
the radius R of the sphere, so the particle is
outside the sphere. Thus, we can regard the
lee (cyl327) HW06 gentle (56245)
= 6.67259 1011 N m2 /kg2
70.4 g 749 kg
0.165 m
(0.549 m)3
= 3.50845 109 N .
sphere as a point mass and to apply Newtons
law of gravity in its simplest version:
F1 = G
mM
2
r1
= 6.67259 1011 N m2 /kg2
0.0704 kg 749 kg
(0.861 m)2
= 4.74616 109 N .
006 (part 2 of 3) 10.0 points
Find the magnitude of the gravitational force
exerted by the sphere on a 70.4 g particle
located at the surface of the sphere.
3
008 (part 1 of 2) 10.0 points
Given: G = 6.67259 1011 Nm2 /kg2
A spacecraft in the shape of a long cylinder
has a a mass with occupants of 1340 kg. It
has strayed in too close to a 1 km radius black
hole having a mass 137 times that of the Sun
(see the gure).
Black hole
l
Correct answer: 1.16736 10
8
N.
Explanation:
When the particle is on the surface of the
sphere, we can still consider it outside the
sphere and to calculate the gravitational force
on the particle using the law of gravity for
point masses:
mM
R2
= 6.67259 1011 N m2 /kg2
0.0704 kg 749 kg
(0.549 m)2
= 1.16736 108 N .
F2 = G
007 (part 3 of 3) 10.0 points
Find the magnitude of the gravitational force
exerted by the sphere on a 70.4 g particle
located 0.165 m from the center of the sphere.
Correct answer: 3.50845 109 N.
Explanation:
When the distance r3 from the particle to the
center of the sphere is smaller than the radius
R of the sphere, the particle is inside the
sphere and the gravitational force F3 exerted
on it increases linearly with the distance from
the particle to the center of the sphere:
F3 = G
mM
r3
R3
//
d
//
If the nose of the spacecraft points toward
the center of the black hole, and if the distance
between the nose of the spaceship and the
black holes center is 10.1 km, determine the
gravitational acceleration of the spaceship due
to the black hole.
Correct answer: 1.77059 1014 m/s2 .
Explanation:
We take the length of the spacecraft to be
small compared to the distance R between
the spacecraft and the black hole, we can
regard spacecraft the as a point mass and
apply Newtons law of gravity in its simplest
form:
GmM
F=
,
2
Ra
where m = 1340 kg is the mass of the spacecraft, M = 137 MSun = 2.72767 1032 kg
is the mass of the black hole and Ra is the
distance between the center of mass of the
spacecraft and the center of the black hole:
1
[R + (R + )]
2
1
= [10100 m + (10100 m + 77.5 m)]
2
= 10138.8 m .
Ra =
Thus, the total average acceleration of the
spaceship will be
a=
GM
F
=
2
m
Ra
lee (cyl327) HW06 gentle (56245)
(6.67259 1011 Nm2 /kg2 )
(10138.8 m)2
2.72767 1032 kg
= 1.77059 1014 m/s2 .
=
Correct answer: 75.1633 km.
Explanation:
Let :
009 (part 2 of 2) 10.0 points
What is the dierence in the acceleration felt
by the occupants in the nose of the 77.5 m
ship (those nearest the black hole) and those
in the rear of the ship (those farthest from the
black hole)?
Correct answer: 2.70693 1012 m/s2 .
Explanation:
If Rf = R and Rb = R + are respectively the
distances from the front and from the back
of the spacecraft to the black hole, then the
dierence in the gravitational accelerations in
the nose and in the rear of the spacecraft will
be
a = G M
T = 2.95 ms = 0.00295 s ,
G = 6.67259 1011 N m2 /kg2 ,
M = 14.5 MS , and
MS = 1.991 1030 kg .
The gravitational acceleration is supplied
by the centripetal acceleration:
ac = ag
v2
GM
=2
r
r
2
2r
G (14.5 MS )
=
T
r
r=
3
=
1
1
2
2
Rf
Rb
3
(14.5 MS ) T 2
G
4 2
20(1.991 1030 kg)(0.00295 s)2
4 2
1
1
R 2 (R + )2
(R + )2 R 2
= GM
R 2 (R + )2
= GM
= (2.72767 1032 kg)(1.48727 1010
4
3
6.67259 1011 N m2 /kg2
= 75163.3 m
= 75.1633 km .
1
)
m2
(6.67259 1011 Nm2 /kg2 )
= 2.70693 1012 m/s2 .
010 10.0 points
X-ray pulses from Cygnus X-1, a celestial xray source, have been recorded during highaltitude rocket ights. The signals can be
interpreted as originating when a blob of ionized matter orbits a black hole with a period
of 2.95 ms.
If the blob were in a circular orbit about
a black hole whose mass is 14.5 times the
mass of the Sun, what is the orbit radius?
The value of the gravitational constant is
6.67259 1011 N m2 /kg2 and mass of the
Sun is 1.991 1030 kg.
011 10.0 points
A satellite moves in a circular orbit around
the Earth at a speed of 6.2 km/s.
Determine the satellites altitude above
the surface of the Earth.
Assume the
Earth is a homogeneous sphere of radius
6370 km and mass 5.98 1024 kg. The
value of the universal gravitational constant
is 6.67259 1011 N m2 /kg2 .
Correct answer: 4010.36 km.
Explanation:
Let :
v = 6.2 km/s ,
Re = 6370 km ,
Me = 5.98 1024 kg , and
G = 6.67259 1011 N m2 /kg2 .
lee (cyl327) HW06 gentle (56245)
The gravitational force provides the centripetal acceleration, so
G m Me
m v2
=
r2
r
G Me
r=
v2
= (6.67259 1011 N m2 /kg2 )
5.98 1024 kg 1 km
(6.2 km/s)2
1000 m
= 10380.4 km ,
3
and the height of the satellite above the
Earths surface is
h = r Re
= 10380.4 km 6370 km = 4010.36 km .
012 10.0 points
An apple on the surface of the earth falls 4.9 m
in its rst second of fall from rest. The moon
is about 60 times farther away from the center
of the earth than is the apple.
How far does the moon fall toward the earth
each second?
Correct answer: 0.00136111 m.
Explanation:
Mearth
1
2.
2
r
r
By uniformly accelerated motion in a
straight line,
1
1
s = a t2 a 2 .
2
r
For the same time interval,
2
sm =
1
60
2
=
1
60
=
ra
60 ra
2
=
1
60
Explanation:
Basic Concepts: Solution:
According to Keplers third law
T2 =
4 2 3
r
GM
where r is the radius of the satellites orbit.
Thus, solving for r
1
GM T2 3
r=
4 2
(6.67 1011 N m2 /kg2 )
=
4 (3.1415926)2
= 1.81879 108 m .
a=G
2
Correct answer: 1.11979 108 m.
(2.4 1027 kg) (38520 s)2
By universal gravitation,
ra
rm
013 10.0 points
A synchronous satellite is put in orbit about
a planet to always remain above the same
point on the planets equator. The planet
rotates once every 10.7 h, has a mass of 2.4
1027 kg and a radius of 6.99 107 m.
Given: G = 6.67 1011 N m2 /kg2 .
Calculate how far above the planets surface
the satellite must be.
1
3
Let : d1 = 4.9 m and
rm = 0.00136111 m .
sm
=
sa
5
2
sa
(4.9 m) = 0.00136111 m .
Now, the altitude h of the satellite (measured
from the surface of Jupiter) is
h=rR
= (1.81879 108 m) (6.99 107 m)
= 1.11979 108 m .
014 10.0 points
Two satellites A and B with the same mass
orbit the Earth in concentric orbits. The
distance of satellite B from the earths center
is four times that of satellite A.
lee (cyl327) HW06 gentle (56245)
What is the ratio of the tangential speed of
satellite B to that of satellite A?
vB
1
1.
=
vA
4
vB
2.
=2
vA
1
vB
=
3.
vA
16
vB
4.
=4
vA
vB
1
5.
= correct
vA
2
vB
= 64
6.
vA
vB
1
7.
=
vA
64
vB
8.
= 16
vA
Explanation:
Let :
Ta =
=
3/2
Te
422 million km
150 million km
3/2
(1 year)
016 (part 2 of 2) 10.0 points
What is the orbital velocity of the asteroid?
Assume there are 365 days in one year.
1
.
R
Thus the ratio is
1
1
=
.
4
2
015 (part 1 of 2) 10.0 points
The period of the earth around the sun is
1 year and its distance is 150 million km from
the sun. An asteroid in a circular orbit around
the sun is at a distance 422 million km from
the sun.
What is the period of the asteroids orbit?
Correct answer: 4.7188 year.
Explanation:
ra
re
= 4.7188 year .
mA = mB , and
RB = 4 RA .
RA
=
RB
and
2
Te
T2
=a
3
3
re
ra
v2
GM m
=
R
R2
GM
v=
R
vB
=
vA
Let : Te = 1 year ,
re = 150 million km ,
ra = 422 million km .
From Keplers laws,
Let M be the mas of the Earth. Using
Newtons second law, the tangential speed for
either satellite is obtained from
m
6
Correct answer: 17817.8 m/s.
Explanation:
2 ra
Ta
2 (4.22 1011 m) 1 y
=
4.7188 year
365 d
1d 1h
24 h 3600 s
= 17817.8 m/s .
va =