Differential equations and escape velocities

[BrettJimison]

Homework Statement

Good day all! I'm stumped on a question:

If I fire a bullet straight up what will be the initial velocity such that the bullet doesn't come back down?
I need to model a differential equation (it will be first order) some how!
Also, Gravity is not constant, but rather, the acceleration due to gravity dv/dt is -k/r^2 where k is a positive constant and r is the distance to the center of the Earth (4000 mi)

Homework Equations

Also, dv/dt=(dv/dr)v

The Attempt at a Solution

My teacher said something about at the point where the initial velocity is great enough to over come gravity, the root in the de will become complex. That's all I know. Any help would be appreciated!
So far I know:

m(dv/dt)=-mg and g=-k/r^2
I can find a de for v(r) easily since the eqn is seperable, but I'm not sure what to do with it...
Also, the problem gives g at the surface of Earth as -32 ft/s^2, and r in miles, so unfortunately we aren't using metric here.

Thanks!

 

[BrettJimison]

Also, I believe we are ignoring air resistance of the bullet since it is not mentioned in the statement.

 

[Ray Vickson]

Homework Statement

Good day all! I'm stumped on a question:

If I fire a bullet straight up what will be the initial velocity such that the bullet doesn't come back down?
I need to model a differential equation (it will be first order) some how!
Also, Gravity is not constant, but rather, the acceleration due to gravity dv/dt is -k/r^2 where k is a positive constant and r is the distance to the center of the Earth (4000 mi)

Homework Equations

Also, dv/dt=(dv/dr)v

The Attempt at a Solution

My teacher said something about at the point where the initial velocity is great enough to over come gravity, the root in the de will become complex. That's all I know. Any help would be appreciated!
So far I know:

m(dv/dt)=-mg and g=-k/r^2
I can find a de for v(r) easily since the eqn is seperable, but I'm not sure what to do with it...
Also, the problem gives g at the surface of Earth as -32 ft/s^2, and r in miles, so unfortunately we aren't using metric here.

Thanks!

Solving the DE is doing it the hard way; using conservation of energy is doing it the easy way.

 

[BrettJimison]

Solving the DE is doing it the hard way; using conservation of energy is doing it the easy way.

Yes, my first instinct is to use energy. Unfortunately, this is for a de class and it needs to be solved using a de.

 

[Dick]

Yes, my first instinct is to use energy. Unfortunately, this is for a de class and it needs to be solved using a de.

You could use de's to prove conservation of energy holds first. If dv/dt=-k/r^2 then (dv/dr)(dr/dt)=-k/r^2. dr/dt=v. So v*dv/dr=-k/r^2. Integrate it.

 

[BrettJimison]

You could use de's to prove conservation of energy holds first. If dv/dt=-k/r^2 then (dv/dr)(dr/dt)=-k/r^2. dr/dt=v. So v*dv/dr=-k/r^2. Integrate it.

Hello dick, thanks for the response. I solved the de using a similar method you mentioned above. I checked my solution using a work integral and it was .001 off of the "accepted value". It involved solving the de and finding the critical values and setting a part of the velocity function derived greater than the critical value (so I wouldn't get complex numbers). I'm sure it could have been done many ways. Also, I checked my answer by a simple energy equation as Ray eluded to.