208
5 Gravitation
Fig. 5.11
The potential at p from the entire rod is given by
V =
dV = −
G M
L
d+
L
2
d−
L
2
dx
x
= −
G M
L
ln
2d + L
2d − L
5.13 The linear speed of an object on the equator
v = ω R = (7.27 × 10
−5
)(6.4 × 10
6
) = 465.3 m/s
The orbital velocity of a surface satellite is
v 0 =
√ gr =
9.8 × 6.4 × 10 6 = 7920 m/s
When launched in the westerly direction the launching speed v 0 will be added
to v as the earth rotates from west to east, while in the easterly direction it will
be subtracted.
westerly launching speed
easterly launching speed
=
7920 + 465
7920 − 465
= 1.125
or 11%.
5.14 Consider an element of arc of length ds = R dθ , Fig. 5.12. The corresponding
mass element dm = λds = λR dθ .
The intensity at the origin where λ is the linear density (mass per unit
length) due to dm will be
GλR dθ
R 2 or
Gλ dθ
R
The x-component of intensity at the origin due to dm will be
dE x =
Gλ
R
dθ sin θ