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Unformatted text preview: MA 36600 LECTURE NOTES: MONDAY, JANUARY 26 Modeling with First Order Equations (cont’d)
Escape Velocity. Say that we have a rocket of mass m which blasts oﬀ from the surface of the Earth at
an initial velocity v0 . Gravity pulls back on the rocket no matter how high the rocket goes, so is it possible
that the rocket can ever leave the Earth and go to another planet? The escape velocity ve of a rocket is that
velocity such that the pull of gravity is too weak to force the rocket to return. That is, if the rocket has an
initial velocity v0 > ve then it will never return!
We derive a formula for the escape velocity ve . Recall Newton’s Law of Gravity and Newton’s Second
Law of Motion:
d2 x
mM
m 2 = −G 2
dt
r
where we deﬁne the variables
m = mass of the rocket
M = mass of the Earth
r = distance between the center of masses of the rocket and the Earth
R = radius of the Earth = 6.38 × 106 meters = 3963.11 miles x = r − R = height of the rocket from the surface of the Earth G = universal gravitational constant
GM
g=
= 9.81 meters/sec2 = 6.10 × 10−3 miles/sec2
R2
The diﬀerential equation above reduces to the equation
m d2 x
m g R2
=−
2
dt2
(x + R) which can also be written as d2 x
=−
dt2 g .
x 2
R
(Compare with Computer Lab #2.) We wish to solve this diﬀerential equation. Unfortunately, we do not
yet know how to solve second order diﬀerential equations which are nonlinear, so we perform a trick. We
will write a diﬀerential equation which involves x and v instead of x and t.
To this end, we note the following using the Chain Rule:
1+ d2 x
dv
dv dx
dv
=
=
·
=v
.
dt2
dt
dx dt
dx
Hence we ﬁnd the initial value problem
mv dv
m g R2
=−
2,
dx
(x + R) x(0) = 0, v (0) = v0 . The lefthand side of this diﬀerential equation is the exact derivative of a function:
dv
d1
2
mv
=
mv .
dx
dx 2 Upon integrating both sides we ﬁnd that can solve this diﬀerential equation implicitly:
1
m g R2
m v2 =
+C
2
x+R
for some constant C . This is an integral curve for the diﬀerential equation. This constant C can be found
from the initial condition:
1
1
1
m g R2
2
2
C = m v0 − m g R
=⇒
m v 2 − m v0 =
− m g R.
2
2
2
x+R
1 ...
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This note was uploaded on 11/30/2011 for the course MATH 366 taught by Professor Edraygoins during the Spring '09 term at Purdue UniversityWest Lafayette.
 Spring '09
 EdrayGoins
 Equations

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