5.3 Solutions
215
mv R = mv 0 d
(1)
or v =
v 0 d
R
(2)
Energy conservation requires
1
2
mv
2
0 =
1
2
mv
2
−
G Mm
R
(3)
or v
2
= v
2
0 +
2G M
R
(4)
Eliminating v between (2) and (4) the minimum value of v 0 is obtained.
v 0 =
2G M R
d 2 − R 2
5.27 According to Kepler’s third law
T
2
∝ r
3
(i) T 2
E
r 3
E
=
(365.3) 2
(1.5 × 10 11 ) 3 = 3.9539 × 10
−29 days
2
/m
3
T 2
v
r 2
v
=
(224.7) 2
(1.08 × 10 11 ) 3 = 4.0081 × 10
−29 days
2
/m
3
Thus Kepler’s third law is verified
(ii)
T = 2π
r 3
G M
(1)
where M is the mass of the parent body.
M =
4π 2
G
r 3
T 2
(2)
From (i) the mean value,
T 2
r 3
= 3.981 × 10 −29 days
2
/m 3 = 2.972 ×
10 −19 s 2 /m 3
M =
4π 2
6.67 × 10 −11 ×
1
2.972 × 10 −19 = 1.99 × 10
30 kg
5.28 At the perihelion (nearest point from the focus) the velocity (v p ) is maximum
and at the aphelion (farthest point) the velocity (v A ) is minimum. At both these