PHYSICS 111 HOMEWORK SOLUTION #13
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:
mM F = G r2
=
6.67
?
10-11
?
2.10 ? 21 ? 10-3 (3.90 ? 10-2)2
= 1.93 ? 10-3 N
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 r2
and
the
free-fall force mg
:
mM mg = G r2
M g = G r2
=
6.67
?
10-11
?
6.68 ? 1019 (242 ? 103)2
= 0.0761 m/s2
3
b)
We can use the free-fall equation of motion under the above-calculated accel-
eration
h
=
1 2
gt2
to
get
the
time
in
flight:
2h t=
g
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 = vxt:
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:
vy2 - vy20 = 2gh
vy
=
2gh
= 2 ? 0.0761 ? 5000
= 27.59 m/s
and
v = 27.592 + 7.702 = 28.6 m/s
The direction of his impact is :
= arctan vy vx 27.59
= arctan 7.7
= 74.4
4
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 42
:
a3
=
GM 42
T
2
=
6.76
?
10-11 ? 1.989 42
?
1030
T
2
= 3.360 ? 1018T 2
= 3.360 ? 1018 ? (90.6 ? 365 ? 24 ? 3600)2
= 2.75833 ? 1037 m3
5
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