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where
T
E
= 365.25 days is Earth’s orbital period and
r
E
= 1.50
×
10
11
m is its mean
distance from the Sun. In this case, it is perfectly legitimate to take logarithms and obtain
o
21
log
log
log
33
EE
M
rT
aT
M
§·
=+
¨¸
©¹
© ¹
(written to make each term positive) which is the way we plot the data (log (
r
E
/
a
) on the
vertical axis and log (
T
E
/T
) on the horizontal axis).
(b) When we perform a leastsquares fit to the data, we obtain
log (
r
E
/
a
) = 0.666 log (
T
E
/
T
) + 1.01,
which confirms the expectation of slope = 2/3 based on the above equation.
(c) And the 1.01 intercept corresponds to the term 1/3 log (
M
o
/
M
) which implies
3.03
oo
3
10
.
1.07
10
MM
M
M
=
¡
=
×
Plugging in
M
o
= 1.99
×
10
30
kg (see Appendix C), we obtain
M
= 1.86
×
10
27
kg for
Jupiter’s mass. This is reasonably consistent with the value 1.90
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics

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