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:
F
2
mM
r2
=
G
=
6.67 ¡Á 10?11 ¡Á
=
1.93 ¡Á 10?3 N
2.10 ¡Á 21 ¡Á 10?3
(3.90 ¡Á 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 ¡Á1019 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 ¡Á 10?11 ¡Á
=
0.0761 m/s2
6.68 ¡Á 1019
(242 ¡Á 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 ¡Á 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 ¡Á 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
¡Ì
2 ¡Á 0.0761 ¡Á 5000
=
27.59 m/s
and
v
=
p
=
28.6 m/s
27.592 + 7.702
The direction of his impact is :
¦È
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 ¡Á1011 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
4¦Ð 2 :
a3
GM 2
T
4¦Ð 2
6.76 ¡Á 10?11 ¡Á 1.989 ¡Á 1030 2
=
T
4¦Ð 2
= 3.360 ¡Á 1018 T 2
=
=
3.360 ¡Á 1018 ¡Á (90.6 ¡Á 365 ¡Á 24 ¡Á 3600)2
=
2.75833 ¡Á 1037 m3
5
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