(This project involves solving a differential equation by
separation of variables.)

When an object is thrown upwards, we know by experience that it falls back to earth. How is it that we can launch rockets into space? We will investigate this question.

Let’s start with a mental experiment. When you throw a stone
upwards it eventually falls back to earth. Does it make a difference how hard
you throw the stone? Will it go higher if you throw it harder? Throwing the stone harder means a
larger ‘initial velocity’.

__ Physics background__:

Newton’s law of gravitation says
that the force acting between two objects (the earth and our rocket) is
proportional to their masses (*M* and *m*) and inversely proportional
to the distance between their centers of gravity *(d)*, that is,

*F = ( G M m ) / d ^{
2}*

*G* (a constant) is there to make
the units and values work out!
(*G *= 6.67 · 10^{-11} N-m^{2} /kg^{ 2} –
check these units).

*F = m *·* a *(mass times acceleration), so an
object on the earth’s surface is subject to a force of *F = m *·* g *from
the earth’s gravity, from which we conclude – if the two laws are consistent,
that *g = G M /R ^{2 }*=
(32 ft/s

*h* units (say miles) above the earth’s surface we can
calculate the force acting on it using the law of gravitation:

*F = (G M m) / (R+h) ^{2}=
(G M /R^{2}) m ( R^{2} / (R+h)^{ 2} ) = g m R^{2 }/
(R+h)^{2 }*(*)

*R ^{2}*).

__The differential equation__:

Let’s assume, unrealistically, that the rocket engines shut off after launch, and so the rocket gets its acceleration from the initial thrust of the engines.

As
mentioned, *F = m *·* a*, and *a = dv/dt*, so *F = m *·* dv/dt*
. Combine this with equation
(*) to find an expression for *dv/dt* (a differential equation). *F* and *dv/dt *are
negative since the rocket is slowing after launch, so don’t forget to take this
into account.

*dv/dt
*we are saying that *v* depends on *t*, ie, *v=v(t)*.But in
this case our differential equation shows that *v* also depends on *h*. Clearly *h* depends on *t *(*h(t)*)
and *dh/dt = v*, so we must think of *v* as a composition: *v=v(h(t)). *Thus, by the chain rule, *dv/dt =
dv/dh · dh/dt*. Use this
to rewrite the differential equation so that we have an equation for *dv/dh*.

__Your work__:

You will find the “escape velocity”, the initial velocity below which the rocket would fall back:

a) Verify the
value of *g* from part 1 (and remember that most calculators can convert
units).

b) Write the differential equation from part 2.

c) Solve the
differential equation by separation of variables (it’s ok to leave the solution
implicit: *v ^{2} =
···*).

d) When *h=0*,
*v=v _{0 }*(initial velocity). Use this initial value to find the constant from (c).

e) If the
rocket is going to “escape”, *h →∞*.
Use this to let one of the terms in your solution *→ 0*, and
simplify the solution accordingly.

f) Using the
data for the earth that you already have, calculate the “escape velocity”* v _{0}*
, which you can determine by making sure that

g) The moon
has a mass of 7.35 · 10^{22} kg and a radius of 1.738 · 10^{6}
m. Find the escape velocity
for the moon.

__Extra credit__
(use the library or the internet):

a) Find data for another planet and calculate its escape velocity.

b) Write a
paragraph about Isaac Newton using several sources and citing references.

© David Rutschman, 2002