Preparation problems

9-32. What vertical launch speed is necessary to get a rocket to 1100 km altitude?

Solution:

Here we note the similarity between the problem of throwing a rock in the air with an initial speed and determining its maximum height. To solve that problem, and this one, we use the principle of the conservation of energy. We will neglect any frictional or drag forces in the problem. Thus we can write

Einitial
=
Efinal
Ki + Ui
=
Kf + Uf
We know that the final kinetic energy will be zero, as the rocket will have stopped at its final altitude. Additionally, we recall that we can write the gravitational potential energy of any object as
U = - GMm
r
where M is the mass of the Earth, m is the mass of the rocket, and r is the distance from the object to the center of the Earth. Recalling that our final distance is RE + h, where RE is the radius of the Earth and h is the final altitude of the rocket, we have
Ki + Ui
=
0 + Uf
Ki
=
Uf - Ui
1
2
 mvi2
=
é
ê
ë 		- GMm
(RE+h)
 ù
ú
û 		- é
ê
ë 		- GMm
RE
 ù
ú
û
vi2
=
2GM
RE
 - 2GM
(RE+h)
vi
=
  æ
Ö
2GM
RE
 - 2GM
(RE+h)
 
Plugging in our known values, we get an answer of
vi = 4290  m/s

9-39. By what factor must the speed of an object in circular orbit be increased to reach escape speed from its orbital altitude?

Solution:

We need to find a relationship between the altitude of a circular orbit and its period. We can start by considering the relationship between the gravitational force and the radial force keeping the satellite in uniform circular motion

Fg
=
m ar
GMm
r2
=
m v2
r
v2
=
GM
r
vorb
=
  æ
Ö
GM
r
 
Now we need to find an expression for the escape speed of an object. We start by recalling
Eescape = K + U = 0
This gives us
0
=
1
2
 m vesc2 - GMm
r
1
2
 m vesc2
=
GMm
r
vesc2
=
2GM
r
vesc
=
  æ
Ö
2GM
r
 

We can quickly see now that

vesc
vorb
 = Ö2 = 1.41

9-42. Determine the escape speed from (a) Saturn's moon Iapetus, with mass 1.9 ×1021 kg and radius 7.3×105 m, and (b) a neutron star, with the Sun's mass crammed into a sphere 6.0 km in radius.

Solution:

We need to find an expression for the escape speed of an object. We start by recalling

Eescape = K + U = 0
This gives us
0
=
1
2
 m vesc2 - GMm
r
1
2
 m vesc2
=
GMm
r
vesc2
=
2GM
r
vesc
=
  æ
Ö
2GM
r
 

(a) Having found the expression for escape speed, we simply plug in our numbers to find

vesc = 590  m/s

(b) Again, having found the expression for escape speed, we simply plug in our numbers to find

vesc = 2.1 ×108  m/s  !



Follow-up problems

9-13. Find the speed of a satellite in geosynchronous orbit?

Solution:

We need to find a relationship between the radius of a circular orbit and its velocity. We can start by considering the relationship between the gravitational force and the radial force keeping the satellite in uniform circular motion

Fg
=
m ar
GMm
r2
=
m v2
r
v2
=
GM
r
v
=
  æ
Ö
GM
r
 
We can look up the radius for a geosynchronous orbit (it is given on page 212 of Wolfson), which is r = 4.22 ×107 m. Plugging this into our equation, we get an answer of
v = 3070  m/s

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