CHAPTER
11
Gravity
1* ·
True or false:
(
a
) Kepler’s law of equal areas implies that gravity varies inversely with the square of the distance.
(
b
) The planet closest to the sun, on the average, has the shortest period of revolution about the sun.
(
a
) False
(
b
) True
2
·
If the mass of a satellite is doubled, the radius of its orbit can remain constant if the speed of the satellite
(
a
)
increases by a factor of 8.
(
b
) increases by a factor of 2.
(
c
) does not change.
(
d
) is reduced by a factor of 8.
(
e
) is
reduced by a factor of 2.
(
c
)
3
·
One night, Lucy picked up a strange message on her ham radio. “Help! We ran away from Earth to live in peace
and serenity, and we got disoriented. All we know is that we are orbiting the sun with a period of 5 years. Where are
we?” Lucy did some calculations and told the travelers their mean distance from the sun. What is it?
Use Equ. 112;
R
=
R
ES
(5/1)
2/3
;
R
ES
= 1.5
×
10
11
m
R
= (1.5
×
10
11
)(5)
2/3
m = 4.39
×
10
11
m
4
·
Halley’s comet has a period of about 76 y. What is its mean distance from the sun?
R
mean
= (1 AU)(76)
2/3
(see Problem 3)
R
mean
= 1.5
×
10
11
×
76
2/3
m = 26.9
×
10
11
m
5* ·
A comet has a period estimated to be about 4210 y. What is its mean distance from the sun? (4210 y was the esti
mated period of the comet Hale–Bopp, which was seen in the Northern Hemisphere in early 1997. Gravitational
interactions with the major planets that occurred during this apparition of the comet greatly changed its period, which is
now expected to be about 2380 y.)
R
mean
= (1 AU)(4210)
2/3
(see Problem 4)
R
mean
= 1.5
×
10
11
×
4210
2/3
m = 3.91
×
10
13
m
6
·
The radius of the earth’s orbit is 1.496
×
10
11
m and that of Uranus is 2.87
×
10
12
m. What is the period of
Uranus?
Use Equ. 112;
T
U
= (1 y)(
R
U
/1 AU)
3/2
T
U
= (2.87
×
10
12
/1.5
×
10
11
)
3/2
y = 83.7 y
7
·
The asteroid Hektor, discovered in 1907, is in a nearly circular orbit of radius 5.16 AU about the sun. Determine
the period of this asteroid.
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Gravity
T
H
= (1 y)(5.16/1)
3/2
(see Problem 6)
T
H
= 11.7 y
8
··
The asteroid Icarus, discovered in 1949, was so named because its highly eccentric elliptical orbit brings it close to
the sun at perihelion. The eccentricity
e
of an ellipse is defined by the relation
d
p
=
a
(1 
e
), where
d
p
is the perihelion
distance and
a
is the semimajor axis. Icarus has an eccentricity of 0.83. The period of Icarus is 1.1 years.
(
a
)
Determine the semimajor axis of the orbit of Icarus. (
b
) Find the perihelion and aphelion distances of the orbit of
Icarus.
(
a
) Use Kepler’s third law;
a
= (1 AU)(
T
I
)
2/3
(
b
)
d
p
=
a
(1 
e
);
d
a
+
d
p
= 2
a
;
d
a
= 2
a

d
p
a
= 1.5
×
10
11
×
1.1
2/3
m = 1.6
×
10
11
m
d
p
= (1.6
×
10
11
×
0.17) m = 2.72
×
10
10
m;
d
a
= 2.93
×
10
11
m
9* ·
Why don’t you feel the gravitational attraction of a large building when you walk near it?
The mass of the building is insignificant compared to the mass of the earth.
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 Spring '08
 staff
 Gravity, Mass, Potential Energy, General Relativity, Gravitational constant

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