Environmental Engineering Reference
In-Depth Information
Example 14.2.1 The rate at which a clock ticks in the Earth's gravitational field
can be approximated by the formula
1
t ( 0 ),
(h)
c 2
t (h)
=
+
where t (h) is the time interval between two ticks of a clock at height h as deter-
mined by an observer at height h and t ( 0 ) is the time interval between the same
two events measured by an observer on the ground (h
=
0 ). (h) is the Newtonian
gravitational potential, defined such that ( 0 )
=
0 .
(i) Show that this expression gives Eq. (14.47) in the case that h
R where R is
the radius of the Earth.
(ii) Now consider a clock on a GPS satellite which orbits the Earth at a speed
of 3.9 km/s at an altitude of 20 . 2
10 3 km. Use the result quoted above to
determine by how much the GPS clock speeds up every day compared to an
identical clock located on the Earth's surface due to the fact that it is in a
weaker gravitational field. Now compute the amount by which the clock slows
down as a result of time dilation. Which effect wins? [You may neglect the
rotation of the Earth.]
×
Solution 14.2.1 (i) The Newtonian potential at a height h above the Earth's sur-
face is just
GM
R
GM
R
(h)
=−
h +
+
1
GM
R
R
=
R
+
h
GM
R
h
=
R ,
h
+
where M is the mass of the Earth and R is its radius. In the limit h
R this
reduces to
GM
R 2
(h)
h,
GM/R 2 and hence we have Eq. (14.47).
(ii) The gravitational speeding of the clock is determined, relative to an observer
on the Earth's surface, by the factor
=
which is equal to gh once we identify g
(h)
c 2
gh
c 2
R
1
+
=
1
+
h
+
R
10 6
9 . 81
×
20 . 2
×
1
10 10
=
1
+
20 . 2 / 6 . 4 =
1
+
5 . 3
×
9
×
10 16
1
+
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