Velocity of a rocket which never returns to earth

[process91]

Homework Statement

With what speed should a rocket be fired upward so that it never returns to earth? (Neglect all forces except the Earth's gravitational attraction)

Homework Equations

The ONLY thing gone over so far involving rockets is the following formula:
Let the altitude of the rocket at time t be r(t), the mass be m(t) and the velocity of the exhaust matter, relative to the rocket be c(t). Then
m(t)r''(t)=-m'(t)c(t)-m(t)g

This is in a section focusing on differential equations.

The Attempt at a Solution

Solving for r'':
r''(t) = - \frac{m'(t)}{m(t)} c(t) - g

And now I don't really know where to go. I can't operate directly on this without making some assumptions about m and c, and based on the problems that we have worked on so far the procedure was similar to this:

Assume c(t) is constant, so c(t)=-c.

Let w be the initial weight of the rocket and fuel. Let k be the rate at which fuel is consumed. Then
m(t)=\frac{w-kt}{g} and m'(t) = -\frac{k}{g}.

Now we have
r''(t)=\frac{kc}{w-kt}-g

Integrating and using the initial condition r'(0)=0:
r'(t)=-c\ln(\frac{w-kt}{w})-gt

Integrating again and using the initial condition r(0)=0:
r(t)=\frac{c(w-kt)}{w}ln(\frac{w-kt}{w})-\frac{1}{2}gt^2+ct

It seems to me that, for this question, this model will not work. No matter what, gravity will pull back down. I really think I need to use a formula for gravity dependent on r(t) in order for this to work. Incidentally, the answer in the book is 6.96 mi/sec.

 

[PeterO]

Homework Statement

With what speed should a rocket be fired upward so that it never returns to earth? (Neglect all forces except the Earth's gravitational attraction)

Homework Equations

The ONLY thing gone over so far involving rockets is the following formula:
Let the altitude of the rocket at time t be r(t), the mass be m(t) and the velocity of the exhaust matter, relative to the rocket be c(t). Then
m(t)r''(t)=-m'(t)c(t)-m(t)g

This is in a section focusing on differential equations.

The Attempt at a Solution

Solving for r'':
r''(t) = - \frac{m'(t)}{m(t)} c(t) - g

And now I don't really know where to go. I can't operate directly on this without making some assumptions about m and c, and based on the problems that we have worked on so far the procedure was similar to this:

Assume c(t) is constant, so c(t)=-c.

Let w be the initial weight of the rocket and fuel. Let k be the rate at which fuel is consumed. Then
m(t)=\frac{w-kt}{g} and m'(t) = -\frac{k}{g}.

Now we have
r''(t)=\frac{kc}{w-kt}-g

Integrating and using the initial condition r'(0)=0:
r'(t)=-c\ln(\frac{w-kt}{w})-gt

Integrating again and using the initial condition r(0)=0:
r(t)=\frac{c(w-kt)}{w}ln(\frac{w-kt}{w})-\frac{1}{2}gt^2+ct

It seems to me that, for this question, this model will not work. No matter what, gravity will pull back down. I really think I need to use a formula for gravity dependent on r(t) in order for this to work. Incidentally, the answer in the book is 6.96 mi/sec.

Try searching the topic "escape velocity" and see what you find. You could even look up escape velocity in the Index of your textbook.

 

[process91]

I did some reading on escape velocity, and it does seem that is what I need. Just to be clear - this cannot be calculated from the equations I presented, correct?

 

[Curl]

This question doesn't make sense. It matters how long the thrusters are on. If the engine can provide 1m/s velocity forever then the rocket will never return.

 

[issacnewton]

here's the derivation of escape velocity

http://en.wikipedia.org/wiki/Escape_velocity#Deriving_escape_velocity_using_calculus

 

[process91]

Yes, I agree, the question has some ambiguity. This question is in a very well respected calculus book, Apostol's Calculus Vol I. Since the answer is equal to the escape velocity for earth, I assume some outside knowledge is required (literally the only equations talking about rockets in the book are the ones I presented).

 

[PeterO]

I did some reading on escape velocity, and it does seem that is what I need. Just to be clear - this cannot be calculated from the equations I presented, correct?

I don't think so.

Note: when they did the Moon trips in 1969,70, etc, they went very close to giving the craft escape velocity, which is why they had to be aimed carefully so the moon itself would swing them round and send them back to the Earth of anything went wrong - like with Apollo 13.