216
5 Gravitation
points the velocity is perpendicular to the radius vector. Since the angular
momentum is constant
mv A r A = mv p r p
or r A =
v p r p
v A
(1)
where r A = r max and r p = r min
The eccentricity
ε =
r max − r min
r max + r min
=
r A − r p
r A + r p
=
v p − v A
v p + v A
(2)
where we have used (1)
ε =
30.0 − 29.2
30.0 + 29.2
= 0.0135
A small value of eccentricity indicates that the orbit is very nearly circular.
5.29 (a) For circular orbit,
T =
2πa
v
T = 14.4 days = 1.244 × 10
6 s
2a =
vT
π
=
2.2 × 10 5 × 1.244 × 10 6
3.1416
= 8.7 × 10
10 m
(b) Since the velocity of each component is the same, the masses of the components are identical.
v
2
=
G(M + m)
a
=
2G M
a
(∵ m = M)
∴ M =
av 2
2G
=
(4.35 × 10 10 )(2.2 × 10 5 ) 2
2 × 6.67 × 10 −11
= 1.58 × 10
31 kg
5.30 At the surface, the component of velocity of the satellite perpendicular to the
radius R is
v 0 sin 30
◦
=
v 0
2
(Fig. 5.16)
Therefore, the angular momentum at the surface =
mv 0 R
2