5.3 Solutions
205
(a) θ will be maximum when sin 2λ is maximum, i.e. 2λ = 90 ◦ or λ = 45 ◦ .
(b) At the poles λ = 90 ◦ and so θ = 0 ◦ .
(c) At the equator λ = 0 ◦ and so θ = 0 ◦ .
5.6 Consider a spherical shell of radius r and thickness dr concentric with the
sphere of radius R. If ρ is the density, then
ρ =
3M
4π R 3
(1)
The mass of the shell = 4πr 2 drρ.
The mass of the sphere of radius r which is equal to 4πr 3 /3 may be considered
to be concentrated at the centre.
The gravitational potential energy between the spherical shell and the sphere of
radius r is
dU = −
G(4πr 2 drρ)
4π
3
r 3 ρ
r
= −
16π 2 Gρ 2 r 4 dr
3
(2)
The total gravitational energy of the earth
U =
dU = −
16π 2 Gρ 2
3
R
0
r
4 dr = −
16π 2 Gρ 2 R 5
15
= −
3
5
G M 2
R
(3)
where we have used (1).
U = −
6.67 × 10 −11 × 0.6 × (6 × 10 24 ) 2
6.4 × 10 6
= 2.25 × 10
32 J
5.7 If g 0 is the gravity at the earth’s surface, g h at height h and g d at depth d, then
g h = g 0
R 2
(R + h) 2
(1)
g d = g 0
1 −
d
R
(2)
By problem, g d = g h at h = d,
(3)
From (1) and (2) we get
d
2
+ d R − R
2
= 0