Morby, Grant – Homework 28 – Due: Apr 14 2006, noon – Inst: Drummond
1
This
printout
should
have
13
questions.
Multiplechoice questions may continue on
the next column or page – fnd all choices
beFore answering.
The due time is Central
time.
001
(part 1 oF 1) 10 points
Consider the orbit oF a typical comet around
the sun, which is marked at fve di±erent po
sitions,
X
,
S
,
Z
,
P
, and
U
.
Sun
a
b
P
S
U
X
Z
Using Kepler’s second law oF planetary mo
tion, rank those positions in order oF thier rel
ative speeds, with the position For the Fastest
speed frst.
1.
X
Z
P
S
U
2.
Z
P
S
X
U
3.
U
X
S
P
Z
4.
Z
X
U
S
P
5.
U
S
P
Z
X
6.
P
S
U
X
Z
correct
Explanation:
Note :
(
f,
0) =
‡
p
a
2

b
2
,
0
·
e
=
√
a
2

b
2
a
.
The closer the comet is the the Sun, the
Faster it is traveling, thereFore (
P
,
S
,
U
,
X
,
Z
) is correct.
002
(part 1 oF 1) 10 points
A “geosynchronous” Earth satellite can re
main directly overhead in which oF the Follow
ing cities?
1.
Sidney
2.
Singapore
correct
3.
San ²rancisco
4.
London
5.
Moscow
Explanation:
An Earth satellite is geosynchronous when
its period and the direction oF motion are the
same as Earth’s spin motion. This occurs only
as the satellite is overhead in a point which
lies on the Earth’s equator.
003
(part 1 oF 1) 10 points
The radius oF the satellite’s orbit around the
center oF the Earth is
r
=
9
2
R
.
Hint:
You may fnd it useFul to take into
account that the gravitational Force is a con
servative Force.
Hint:
The universal gravitational Force law
is
~
F
=
G
M m
r
2
ˆ
r .
Earth
Satellite
R
9
2
Caution:
Neglect the rotational kinetic en
ergy due to the Earth’s rotation.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentMorby, Grant – Homework 28 – Due: Apr 14 2006, noon – Inst: Drummond
2
Find the energy required to launch a satel
lite from Earth into the circular orbit at the
speci±ed radius
r
=
9
2
R
.
1.
E
=
10
9
GM m
R
2.
E
=
7
8
GM m
R
3.
E
=
2
3
GM m
R
4.
E
=
7
9
GM m
R
5.
E
=
8
9
GM m
R
correct
6.
E
=
7
6
GM m
R
7.
None of these
8.
E
=
11
9
GM m
R
9.
E
=
9
8
GM m
R
10.
E
=
3
2
GM m
R
Explanation:
Basic Concepts:
Force of gravity between
two masses
m
1
and
m
2
at a distance
r
F
g
=
G
m
1
m
2
r
2
and
U
g
=
G
m
1
m
2
r
,
where
G
is the gravitational constant.
Centripetal acceleration for circular motion
is
a
r
=
v
2
r
,
for the radial acceleration, where
v
is the
tangential speed.
Solution:
Since the gravitational force sup
plies the centripetal force,
ma
r
=
m
v
2
r
=
G
mM
r
2
so that
mv
2
=
G
mM
r
=
G
mM
9
2
R
.
Then the kinetic energy of the satellite in
orbit is found to be
K
r
=
1
2
mv
2
=
G
mM
2
µ
9
2
R
¶
.
Conservation of energy gives us
U
R
+ Δ
E
=
U
r
+
K
r
Δ
E
=
U
r

U
R
+
K
r
=

GM m
9
2
R
+
GM m
R
+
GM m
9
R
=
GM m
R
µ

2
9
+
9
9
+
1
9
¶
=
8
9
GM m
R
.
004
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '09
 KLEINMAN
 Physics, Work

Click to edit the document details