224
5 Gravitation
The potential energy is unaltered and is therefore
U 1 = −
G Mm
r 1
= −
2G Mm
3a
Total energy E
1 = K + U 1 =
2G Mm
3a
−
2G Mm
3a
= 0
Case 2: Apse is nearer
It is sufficient to show that the total energy is positive.
r 2 = a(1 − ε) = a(1 − 0.5) = 0.5a
v
2
2 = G M
2
r 2
−
1
a
= G M
2
0.5a
−
1
a
=
3G M
a
New velocity v
2 = 2v 2 .
New kinetic energy K
2 =
1
2
m
v
2
2 =
1
2
m(2v 2 ) 2 =
6G Mm
a
Potential energy is unaltered and is given by
U 2 = −
G Mm
r 2
= −
G Mm
0.5a
= −
2G Mm
a
Total energy E 2 = K
2 + U 2 =
6G Mm
a
−
2G Mm
a
= +
4G Mm
a
,
which is a positive quantity.
5.42 The velocity of the particle in the orbit is given by
v
2
= G M
2
r
−
1
a
When the particle is at one extremity of the minor axis, r = a
v
2
= G M
2
a
−
1
a
=
G M
a
Let the new axes be 2a 1 and 2b 1 . By problem the force is increased by half,
but the velocity at r = a is unaltered.
v
2
= 1.5 G M
2
a
−
1
a 1
=
G M
a
∴ 2a 1 =
3a
2