Download PHYSICS 111 HOMEWORK SOLUTION #13 May 1, 2013

PHYSICS 111 HOMEWORK
SOLUTION #13
May 1, 2013
0.1
In introductory physics laboratories, a typical Cavendish balance for measuring the gravitational constant G uses lead spheres with masses of 2.10 kg and 21.0 g whose centers are
separated by about 3.90 cm. Calculate the gravitational force between these spheres, treating
each as a particle located at the center of the sphere.
The gravitational force between the two masses is:
F
2
mM
r2
=
G
=
6.67 × 10−11 ×
=
1.93 × 10−3 N
2.10 × 21 × 10−3
(3.90 × 10−2 )2
0.2.
0.2
Miranda, a satellite of Uranus, is shown in part a of the figure below. It can be modeled as
a sphere of radius 242 km and mass 6.68 ×1019 kg.
• a) Find the free-fall acceleration on its surface.
• b) A cliff on Miranda is 5.00 km high. It appears on the limb at the 11 o’clock position
in part a of the figure above and is magnified in part b of the figure above. A devotee
of extreme sports runs horizontally off the top of the cliff at 7.70 m/s. For what time
interval is he in flight?
• c) How far from the base of the vertical cliff does he strike the icy surface of Miranda?
• d) What is his vector impact velocity?
a)
The free-fall acceleration on Miranda’s surface can be derived by equating the
gravitational force F = G mM
r 2 and the free-fall force mg :
mg
g
mM
r2
M
= G 2
r
= G
=
6.67 × 10−11 ×
=
0.0761 m/s2
6.68 × 1019
(242 × 103 )2
3
b)
We can use the free-fall equation of motion under the above-calculated acceleration ∆h = 12 gt2 to get the time in flight:
s
t =
2∆h
g
r
2 × 5000
0.0761
363 s
=
=
c)
How far from the base cliff will he strike can be evaluated by looking at the
horizontal component of the equations, x = vx t:
x
=
7.70 × 363
=
2791 m
d)
The horizontal component of his velocity is being constant throughout the
motion vx = 7.70m/s, we can evaluate the vertical component at the impact
by using the time-independent equation:
2
vy2 − vy0
=
vy
=
=
2g∆h
p
2g∆h
√
2 × 0.0761 × 5000
=
27.59 m/s
and
v
=
p
=
28.6 m/s
27.592 + 7.702
The direction of his impact is :
θ
4
vy
vx
27.59
= arctan
7.7
= 74.4◦
=
arctan
0.3.
0.3
A comet (see figure below) approaches the Sun to within 0.570 AU, and its orbital period
is 90.6 years. (AU is the symbol for astronomical unit, where 1 AU = 1.50 ×1011 m is the
mean EarthSun distance.) How far from the Sun will the comet travel before it starts its
return journey.
Kepler’s Law relates the square of the orbital period of a planet to the cube
of the semi-major axis (distance a in the figure), the proportionality constant
is GM
4π 2 :
a3
GM 2
T
4π 2
6.76 × 10−11 × 1.989 × 1030 2
=
T
4π 2
= 3.360 × 1018 T 2
=
=
3.360 × 1018 × (90.6 × 365 × 24 × 3600)2
=
2.75833 × 1037 m3
5
and
3.02145 × 1012 m
3.02145 × 1012
=
1.50 × 1011
= 20.143 AU
a
=
x
=
2a − 0.570
=
39.71 AU
0.4
Neutron stars are extremely dense objects formed from the remnants of supernova explosions.
Many rotate very rapidly. Suppose the mass of a certain spherical neutron star is twice the
mass of the Sun and its radius is 11.0 km. Determine the greatest possible angular speed it
can have so that the matter at the surface of the star on its equator is just held in orbit by
the gravitational force.
The matter at the surface of the neutron stars is subject to the gravitational
2
force G Mrn2m which is balanced by the centripetal mv
r force that keeps the star
in its orbit:
G
Mn m
r2
=
=
ω
=
=
=
6
mv 2
r
m(rω)2
r r
GMn
r3
r
6.76 × 10−11 ×
14122 rad/s
2 × 1.989 × 1030
110003
0.5.
0.5
How much work is done by the Moon’s gravitational field as a 995 kg meteor comes in from
outer space and impacts on the Moon’s surface?
The work done by the Moon’s gravitational field can be evaluated through
the change in potential energy:
W
−∆U
GM m GM m
= −(−
−
)
R
∞
GM m
= −(−
− 0)
R
6.67 × 10−11 × 7.36 × 1022 × 995
=
1.74 × 106
9
= 2.81 × 10 J
=
0.6
After the Sun exhausts its nuclear fuel, its ultimate fate will be to collapse to a white dwarf
state. In this state, it would have approximately the same mass as it has now, but its radius
would be equal to the radius of the Earth.
• a) Calculate the average density of the white dwarf.
• b) Calculate the surface free-fall acceleration.
• c) Calculate the gravitational potential energy associated with a 4.93-kg object at the
surface of the white dwarf.
a)
The white dwarf will have a mass almost the same mass as the sun Ms = 1.989×
1030 kg but will have a smaller radius comparable to earth’s r = 6.37 × 106 m.
7
The density is thus:
Ms
4πr 3
3
ρ =
1.989 × 1030
=
=
4π×(6.37×106 )3
3
9
1.84 × 10 kg/m3
b)
Following the same steps in problem 2-a), the free fall acceleration is :
g
= G
M
r2
1.989 × 1030
(6.37 × 106 )2
=
6.67 × 10−11 ×
=
3.27 × 106 m/s2
c)
Potential energy of the 4.93-kg object at the surface of the white dwarf is :
Ug
= −
GMs m
r
= −6.67 × 10−11 ×
1.989 × 1030 × 4.93
6.37 × 106
= −1.03 × 1014 J
0.7
• a) What is the minimum speed, relative to the Sun, necessary for a spacecraft to escape
the solar system if it starts at the Earth’s orbit?
• b) Voyager 1 achieved a maximum speed of 125,000 km/h on its way to photograph
Jupiter. Beyond what distance from the Sun is this speed sufficient to escape the solar
system?
The value of − GMr m is the potential energy that keeps the spacecraft in the
solar system and to overcome this barrier, the spacecraft will need a minimum
kinetic energy 12 mv 2 comparable to the first amount:
8
0.8.
1
mv 2
2
GM m
rr
2GM
=
r
r
2 × 6.67 × 10−11 × 1.989 × 1030
=
1.50 × 1011
= 42.1 km/s
=
v
b)
Again the same equation applies, knowing the acquired speed we can compute
the sufficient distance for the escape:
1
mv 2
2
=
r
=
=
=
GM m
r
2GM
v2
2 × 6.67 × 10−11 × 1.989 × 1030
6
2
( 125×10
3600 )
2.20 × 1011 m
0.8
A satellite of mass 190 kg is placed into Earth orbit at a height of 700 km above the surface.
• a) Assuming a circular orbit, how long does the satellite take to complete one orbit?
• b) What is the satellite’s speed?.
• c) Starting from the satellite on the Earth’s surface, what is the minimum energy input
necessary to place this satellite in orbit? Ignore air resistance but include the effect of
the planet’s daily rotation.
a)
In problem 3), using Kepler’s Law for elliptic motion, we were able to get the
semi-major axis distance in terms of the period of motion. Here we assume
a simpler circular motion and we will use the radius instead (we should also
9
include the satellite height above the surface ):
=
GM 2
T
4π 2
6.67 × 10−11 × 5.972 × 1024 2
T
4π 2
3.360 × 1018 T 2
=
5918.15 s
=
1.64 h
R3
=
(700 × 103 + 6.37 × 106 )3
=
T
b)
Knowing the period and the radius we can easily derive the speed:
v
=
=
=
2πR
T
2π(700 × 103 + 6.37 × 106 )
5918.15
7507 m/s
c)
The total energy of the {Earth+Satellite} system is to be conserved:
• Initially, on earth surface and by taking earth’s speed as ve =
2π×6.37×106
= 463.24 m/s
24×3600
Ei
=
1
Mm
mv 2 − G
2 e
r
Ef
=
1
Mm
mv 2 − G
2
R
• On orbit,
The change in energy is the minimum energy required to put this satellite on
orbit:
1
1
1
m(v 2 − ve2 ) + GM m( − )
2
r
R
1
= × 190 × (75072 − 463.242 ) + 6.67 × 10−11 × 5.972
2
1
1
× 1024 × 190 × (
−
)
6.37 × 106
6.37 × 106 + 700 × 103
= 6.52 × 109 J
∆E =
10
0.9.
0.9
Studies of the relationship of the Sun to our galaxythe Milky Way have revealed that the
Sun is located near the outer edge of the galactic disc, about 30 000 ly (1 ly = 9.46 ×1015
m) from the center. The Sun has an orbital speed of approximately 250 km/s around the
galactic center.
• a) What is the period of the Sun’s galactic motion?
• b)What is the order of magnitude of the mass of the Milky Way galaxy?
• c) Suppose the galaxy is made mostly of stars of which the Sun is typical. What is the
order of magnitude of the number of stars in the Milky Way?
a)
The period can be easily evaluated assuming a circular motion:
=
2πR
T
2πR
v
2π × 30000 × 9.46 × 1015
250 × 103
7.13267 × 1015 s
=
2.26 × 108 yrs
v
=
T
=
=
b)
Using the same procedure as in problem 8-a): the period , the radius and the
mass of the object giving rise to gravity are related with Kepler/Newton law:
R3
=
M
=
=
=
GM 2
T
4π 2
2 3
4π R
GT 2
4π 2 × (30000 × 9.46 × 1015 )3
6.67 × 10−11 × (7.13267 × 1015 )2
2.66 × 1041 kg
a 1041 order of magnitude
11
c)
Such a huge mass will roughly include a number of Suns about 1030 kg each:
41
11
an order of magnitude of 10
Suns.
1030 = 10
12