Example 10-5 How High Does It Go?
You launch a rocket of mass 2.40 * 104 kg straight upward from Earth’s surface.
The rocket’s engines burn for a short time, giving the rocket an initial speed of
9.00 km>s (32,400 km>h or 20,100 mi>h), then shut off (Figure 10-9). To what
maximum height above Earth’s surface will this rocket rise? Earth’s mass is
5.97 * 1024 kg and its radius is 6370 km = 6.37 * 106 m.
Figure 10-9 ​A rocket launch What is the relationship between the speed imparted to the rocket
at launch and the maximum height that it reaches?
Set Up
We’ll ignore the force of air resistance on the rocket as it ascends
through the atmosphere (this force,
like kinetic friction, is nonconservative). Then Earth’s gravity is the
only force that does work on the
rocket, and total mechanical energy
is conserved.
With such a tremendous launch
speed, we expect the rocket to reach
a very high altitude. So the rocket
will not remain close to Earth’s surface, and we must use the more general form for gravitational potential
energy given by Equation 10-4.
Initially the rocket has speed
vi = 9.00 km>s = 9.00 * 103 m>s
and is a distance ri from Earth’s center equal to the radius of Earth. At its
maximum height the rocket is at rest
(vf = 0) and is a distance h above the
surface. Our goal is to determine h.
Solve
We first determine the total mechanical energy of the system using the
given values for the rocket’s initial
speed and position.
Total mechanical energy is conserved:
(6-23)
Ki + Ugrav, i = Kf + Ugrav, f
Gravitational potential energy:
GmEarth mrocket
Ugrav = r
Kinetic energy of rocket:
1
K = mrocket v 2
2
vf = 0
h=?
speed vi
(10-4)
ri = REarth
(6-8)
Earth
When the rocket is initially launched at the surface, its kinetic energy is
1
1
mrocket v 2i = 12.40 * 104 kg2 19.00 * 103 m>s2 2
2
2
= 9.72 * 1011 kg # m2 >s 2 = 9.72 * 1011 J
Ki =
(Recall that 1 J = 1 kg # m2 >s 2.)
The gravitational potential energy is
Ugrav, i = = -
GmEarth mrocket
GmEarth mrocket
= ri
R Earth
16.67 * 10-11 N # m2 >kg 2 2 15.97 * 1024 kg2 12.40 * 104 kg2
6.37 * 106 m
= 21.50 * 1012 N # m = 21.50 * 1012 J
(Recall that 1 J = 1 N # m)
The total mechanical energy is
Ei = Ki + Ugrav, i = (9.72 * 1011 J) + (21.50 * 1012 J) = 25.28 * 1011 J
When the rocket is at the high point
of its trajectory, it is momentarily
at rest and its kinetic energy is zero.
At this point the (conserved) total
mechanical energy is equal to the
gravitational potential energy. Use
this to solve for the rocket’s distance
rf from Earth’s center at its high
point.
At the high point of the trajectory, rocket speed is vf = 0 and its
kinetic energy is
1
K f = mrocket v 2f = 0
2
The total mechanical energy has the same value as when the
rocket was launched:
Ef = Kf + Ugrav, f = Ei = 25.28 * 1011 J
Since Kf = 0, this means that
Ugrav, f = -
GmEarth mrocket
= E i = -5.28 * 1011 J
rf
Solve for rf: rf = = -
GmEarth mrocket
Ei
16.67 * 10-11 N # m2 >kg 2 2 15.97 * 1024 kg2 12.40 * 104 kg2
1 -5.28 * 1011 J2
= 1.81 * 10 m = 1.81 * 104 km
7
Subtract Earth’s radius from rf to
find the rocket’s final height above
the surface.
The rocket’s final distance from Earth’s center, rf, equals the radius of Earth
(REarth) plus the rocket’s final height h above Earth’s surface:
rf = REarth + h
Solve for h:
h = rf 2 REarth = 1.81 * 104 km 2 6.37 * 103 km = 1.17 * 104 km
Reflect
The final height is nearly twice the
radius of Earth! This justifies our
decision to use Equation 10-4 for
gravitational potential energy. Had
we used the expression from Chapter 6, Ugrav = mrocket gy, we would
have gotten an incorrect answer because that expression assumes that
g has the same value at all heights.
In fact, g decreases substantially
in value at greater distances from
Earth, which means that the rocket
climbs much higher than the expression Ugrav = mrocket gy would predict.
From Example 10-3 (Section 10-2), the value
of g at a distance r from Earth’s center is
g =
g = 1.22 m/s2
GmEarth
r2
h ≈ 1.17 x 104 km
At the rocket’s maximum distance,
g =
=
g = 9.80 m/s2
GmEarth
r 2f
16.67 * 10-11 N # m2 >kg 2 2 15.97 * 1024 kg2
= 1.22 m>s 2
7
11.81 * 10 m2
Earth
2
which is much less than g = 9.80 m>s 2 at
Earth’s surface.
A common incorrect way to calculate h would be to
use Ugrav = mrocket gy, which is based on the (false!)
assumption that g = 9.80 m>s 2 at all altitudes. Then
conservation of total mechanical energy would have
told us that
1
1
mrocket v 2i + mrocket gyi = mrocket v 2f + mrocket gyf
2
2
With yi = 0 (rocket at surface), yf = h (rocket at maximum height), and vf = 0
(rocket momentarily at rest at maximum height), this becomes
1
m
v 2i + 0 = 0 + mrocket gh
2 rocket
19.00 * 103 m>s2 2
v 2i
h =
=
= 4.13 * 106 m = 4.13 * 103 km
2g
219.80 m>s 2 2
(Incorrect answer!)
The actual height that the rocket reaches is almost three times higher, 1.17 * 104
km, because in fact the value of g decreases with altitude.