232
5 Gravitation
Fig. 5.21
1
p 2 =
1
r 2 +
1
r 4
dr
dθ
2
(1)
r = 2a cos θ
(2)
dr
d θ
= −2a sin θ
(3)
∴
dr
dθ
2
= 4a
2
(1 − cos
2
θ) = 4a
2
− r
2
(4)
∴
1
r 4
dr
dθ
2
=
4a 2
r 4 −
1
r 2
(5)
Using (5) in (1) and simplifying
p =
r 2
2a
(6)
d p
dr
=
r
a
(7)
f = −
h 2
p 3
d p
dr
= −
h 2 a 4
r 5
5.54 Initially the earth’s orbit is circular and its kinetic energy would be equal to
the modulus of potential energy
1
2
mv
2
0 =
1
2
mG M
r
(1)
Suddenly, sun’s mass becomes half and the earth is placed with a new quantity of potential energy, its instantaneous value of kinetic energy remaining
unaltered.
5 Gravitation
Fig. 5.21
1
p 2 =
1
r 2 +
1
r 4
dr
dθ
2
(1)
r = 2a cos θ
(2)
dr
d θ
= −2a sin θ
(3)
∴
dr
dθ
2
= 4a
2
(1 − cos
2
θ) = 4a
2
− r
2
(4)
∴
1
r 4
dr
dθ
2
=
4a 2
r 4 −
1
r 2
(5)
Using (5) in (1) and simplifying
p =
r 2
2a
(6)
d p
dr
=
r
a
(7)
f = −
h 2
p 3
d p
dr
= −
h 2 a 4
r 5
5.54 Initially the earth’s orbit is circular and its kinetic energy would be equal to
the modulus of potential energy
1
2
mv
2
0 =
1
2
mG M
r
(1)
Suddenly, sun’s mass becomes half and the earth is placed with a new quantity of potential energy, its instantaneous value of kinetic energy remaining
unaltered.