Escape velocity and tangential impulse

[christianwos]

Here is the problem.

A rocket of mass m is in a circular orbit around the Earth at a distance R from the center.
(a) What tangential impulse, mΔv, must be given to the body so that it just escapes to infinity?

My attempt:
I set the problem in terms of energy, E=1/2m((dr/dt)^2+r^2(dø/dt)^2) V(r) and using conservation of angular momentum I get

dr/dt=[2mMG/r-(l/mr)^2]^1/2,
where l=mr^2dø/dt.

Am I missing something?

Thank you.

 

[Janus]

Do you know how to find the orbital velocity for the object?
Do you know how to find the escape velocity for the object?
What's the difference?

 

[BvU]

Hello Chris, welcome to PF.
Just to help me understand what you are doing: when you

set the problem in terms of energy, E=1/2m((dr/dt)^2+r^2(dø/dt)^2) V(r)

what is it you are going to solve ? V = 0 ?

Second, is there a way to justify

using conservation of angular momentum

when you whack this rocket with a big mΔv ?

 

[christianwos]

I am just setting the energy equation up. I generally find it easier to work with energy than with forces. Regarding the angular momentum, I meant that L is conserved until the impulse is delivered.

 

[christianwos]

Do you know how to find the orbital velocity for the object?
Do you know how to find the escape velocity for the object?
What's the difference?

The orbital velocity is jusr rdø/dt, while the escape velocity, for a stationary object shot vertically in a gravitational field, would be v=(MG/r)^1/2, where M is the mass of the planet or whatever the object is stationary on.

 

[Janus]

The orbital velocity is jusr rdø/dt, while the escape velocity, for a stationary object shot vertically in a gravitational field, would be v=(MG/r)^1/2, where M is the mass of the planet or whatever the object is stationary on.

First off, your escape velocity equation is incorrect (and it doesn't have to be the vertical direction)

Secondly, what is orbital velocity in the terms of M, G and r?

 

[christianwos]

The orbital velocity should be v=(MG/r)^1/2 which I get from mv^2/r=mMG/r^2, right?

 

[Janus]

The orbital velocity should be v=(MG/r)^1/2 which I get from mv^2/r=mMG/r^2, right?

Correct. Now note that this is the answer you gave for escape velocity.

So what is the correct expression for escape velocity?

 

[christianwos]

So the energy is E=-mMG/2R and the escape velocity should be twice the orbital?

 

[Janus]

So the energy is E=-mMG/2R and the escape velocity should be twice the orbital?

You're dancing around the answer.

Your energy equation is correct, but in it's present form doesn't help much as it does not have velocity in it.

The other way to express the energy is as the sum of kinetic and potential energies.

Once you have this, all you need to know is that the total energy of an object at escape velocity is 0.

With this information, you can solve for escape velocity.

 

[christianwos]

Now I get v=(2MG/R)1/2.

 

[Janus]

Now I get v=(2MG/R)1/2.

Good! Now compare this to the expression you got for the orbital velocity and you have your required delta v.

 

[christianwos]

Got it. Thank you for your infinite patience.

 

[physicsnmathstudent0]

Good! Now compare this to the expression you got for the orbital velocity and you have your required delta v.

I'm still a little lost in this last step, how can I get delta v by comparing the orbital velocity and the escape velocity? Is V_esc = V_orb + delta v? I think I just answered myself but I still will like to have a little help. Thank you!

 

[jbriggs444]

I'm still a little lost in this last step, how can I get delta v by comparing the orbital velocity and the escape velocity? Is V_esc = V_orb + delta v? I think I just answered myself but I still will like to have a little help.

Yes, it is that simple. As long as it is a tangential velocity and a tangential impulse. Since this is an eight year old thread relating to someone else's homework exercise, further discussion belongs in a fresh thread.