220
5 Gravitation
Using (1) in (7) we find
v
=
√
2v
(8)
If θ is the angle between v and the radius vector R angular momentum
conservation gives
M C v R = M C v
R sin θ = M C
√
2v R sin θ
or sin θ =
1
√
2
θ = 45
◦
5.35 At both perigee and apogee the velocity of the satellite is perpendicular to the
radius vector. In order to show that the angular momentum is conserved we
must show that
mv p r p = mv A r A
or v p r p = v A r A
where m is the mass of the satellite.
v p r p = 10.25 × 6570 = 67342.5
v A r A = 1.594 × 42250 = 67346.5
The data are therefore consistent with the conservation of angular momentum.
5.36 (a) v 0 =
G M
2
r
−
1
a
(1)
G M = (6.67 × 10
−11
)(6 × 10
24
) = 4 × 10
14
r = R = 6.4 × 10
6
a = 8 × 10
7 m
v 0 = 1.095 × 10
4 m/s = 10.095 km/s
(b) ε =
1 +
2E J 2
G 2 M 2 m 3
(2)
J = m Rv 0 sin 45
◦
=
m Rv 0
√
2
(3)
E = −
G Mm
2a
(4)
Combining (1), (2), (3) and (4)
5 Gravitation
Using (1) in (7) we find
v
=
√
2v
(8)
If θ is the angle between v and the radius vector R angular momentum
conservation gives
M C v R = M C v
R sin θ = M C
√
2v R sin θ
or sin θ =
1
√
2
θ = 45
◦
5.35 At both perigee and apogee the velocity of the satellite is perpendicular to the
radius vector. In order to show that the angular momentum is conserved we
must show that
mv p r p = mv A r A
or v p r p = v A r A
where m is the mass of the satellite.
v p r p = 10.25 × 6570 = 67342.5
v A r A = 1.594 × 42250 = 67346.5
The data are therefore consistent with the conservation of angular momentum.
5.36 (a) v 0 =
G M
2
r
−
1
a
(1)
G M = (6.67 × 10
−11
)(6 × 10
24
) = 4 × 10
14
r = R = 6.4 × 10
6
a = 8 × 10
7 m
v 0 = 1.095 × 10
4 m/s = 10.095 km/s
(b) ε =
1 +
2E J 2
G 2 M 2 m 3
(2)
J = m Rv 0 sin 45
◦
=
m Rv 0
√
2
(3)
E = −
G Mm
2a
(4)
Combining (1), (2), (3) and (4)