5.3 Solutions
219
L
= L
∴ v
=
L
r =
L
r
=
3.9375 × 10 15
0.39 × 1.5 × 10 11 = 6.73 × 10
4 m/s = 67.3 km/s
Total energy per unit mass
E =
1
2
v
2
−
G M
r
=
1
2
(3×10
4
)
2
−
6.67 × 10 −11 × 2 × 10 30
1.75 × 10 11
= −3.12×10
8 J
a negative quantity. Therefore the orbit is bound.
5.34 (a) The centripetal force is provided by the gravitational force.
G M E M S
R 2
=
M E v 2
R
or G M S = v
2 R
(1)
(b) Total energy of the comet when it is closest to the sun
E =
1
2
M C (2v)
2
−
G M C M S
R/2
(2)
Using (1) in (2) we find E = 0.
(c) At the distance of the closest approach, the comet’s velocity is perpendicular to the radius vector. Therefore the angular momentum
L = M C (2v)
R
2
= M C v R
(3)
Let v t be the comet’s velocity which is tangential to the earth’s orbit at P.
Then the angular momentum at P will be
L
= M c v t R
(4)
Angular momentum conservation gives
M C v t R = M C v R
(5)
or v t = v
(6)
(d) The total energy of the comet at P is
E
=
1
2
M C (v
)
2
−
G M S M C
R
= 0
( 7 )
where v is the comet’s velocity at P, because E = E = 0, by energy
conservation.