This preview shows pages 1–2. Sign up to view the full content.
Phys 341: Homework #1 Solutions
1. (a) From dimensional analysis:
D
∼
(
GMT
2
)
1
/
3
= [
M

1
L
3
T

2
×
M
×
T
2
]
1
/
3
= [
L
]
(b) From dimensional analysis, the gravitational potential energy is
E
∼ 
GM
2
R
∼ 
GM
2
(
GMT
2
)
1
/
3
∼ 
G
2
/
3
M
5
/
3
T

2
/
3
Taking the derivative with respect to time (and dropping numerical factors of order
unity), we have
dE
dt
∼
G
2
/
3
M
5
/
3
T

2
/
3
dT
dt
(c) Now we assume that the change in energy is due entirely to gravitational radiation, so
we set
dE/dt
equal to the power we estimated. (With a minus sign, since radiating
power causes
E
to decrease.) Then solving for
dT/dt
we get:
dT
dt
∼ 
G

2
/
3
M

5
/
3
T
2
/
3
×
Mc
3
D
±
GM
c
2
D
²
n
∼ 
±
GM
c
3
T
²
(2
n

3)
/
3
where I used the result from part (a) to replace
D
, and then I collected all the terms.
Including the numerical prefactor then yields
dT
dt
∼ 
24
,
000
±
GM
c
3
T
²
(2
n

3)
/
3
(d) For
M
∼
M
±
and
T
= 27906
.
98 s, we have
GM
c
3
T
∼
[6
.
67
×
10

8
cm
3
g

1
s

2
][1
.
99
×
10
33
g]
[3
×
10
10
cm s

1
]
3
[27906
.
98 s]
∼
1
.
76
×
10

10
I then get the following estimates of
dT/dt
for di±erent values of
n
:
n
dT/dt
1

4
.
3
×
10
7
2

1
.
3
×
10
1
3

4
.
2
×
10

6
4

1
.
3
×
10

12
5

4
.
2
×
10

29
6

1
.
3
×
10

25
Clearly
n
= 4 is the exponent that gives something close to the observed value of
dT/dt
.
(Our estimate does not precisely equal the observed value mainly because the two bodies
are actually neutron stars that a slightly more massive than the Sun.)
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '08
 Keeton
 Physics, Energy, Potential Energy, Work

Click to edit the document details