?Use Newton’s second law F = ma to prove this equation

[cloud360]

?Use Newton’s second law F = ma to prove this equation

Homework Statement

[PLAIN]http://img834.imageshack.us/img834/338/2010b6.gif

Homework Equations

Can somone show me the proof i don't no anything about this, any websites which tell u how to do questions like these. Is there any formulas i need to know, or is this a simple re arranging question

(i know the formulae for velocity and acceleration v=r', a=v'')

The Attempt at a Solution

(a) do i equate the total force to ma i.e

-mge^-γrȓ=ma

divide both sides by m

-ge^-γrȓ=a

then i don't know what to do

(b) i don't know where to start with b

 

[cloud360]

is it possible to drop the unit vector for the first question, if so, how?

 

[Pengwuino]

a) Yes, by definition, $$\Sigma F = ma$$. And as cloud said, what about the problem implies you can drop the radial unit vector (you can by the way)?

b) Look at the problem in a way that's more obvious to what you're looking for. Does anything stand out if you write the equation as

$${{dV} \over {dt}} = -ge^{{-\gamma r}}$$

You are not given any time information, so you suspect you'll need to integrate the RHS with respect to 'r'. What happens to both sides of the equation when you multiply by $$dr$$? Remember, $${{dr} \over {dt}} = V$$. Keep in mind that I'm being a little sloppy and the V in your solution really is something more like $$V_{initial}$$ whereas my V, or I probably should have put $$\dot r$$ is the integration variable.

 

[AlexChandler]

is it possible to drop the unit vector for the first question, if so, how?

You can either drop the unit vector by erasing it, or you can draw a vector over the a. Either is okay.
You already have basically reached the answer to part a.
you have written -ge-γrȓ=a
Actually this equation does not make sense unless a is a vector. You cannot equate a vector and a scalar.

This would be correctly written in either of the following ways:

$$\vec a = -ge^{{-\gamma r}} \hat r$$

or

$$a = \dot{v} = \ddot{r} = -ge^{{-\gamma r}}$$

Also, this may give some insight into the second part.

You see that you could rewrite the given formula as

$$\dot{r}^2 - V^2 = \frac{2g}{ \gamma } ( e^{- \gamma r} - e^{- \gamma R} )$$

Which you can also rewrite as:$$\frac{1}{2} m ( \dot{r}^2 - V^2 ) = \frac{1}{2} m \frac{2g}{ \gamma } ( e^{- \gamma r} - e^{- \gamma R} )$$

Now the left side can be interpreted as $$\Delta K.E.$$

What does this say about the right hand side?

 

[cloud360]

a) Yes, by definition, $$\Sigma F = ma$$. And as cloud said, what about the problem implies you can drop the radial unit vector (you can by the way)?

b) Look at the problem in a way that's more obvious to what you're looking for. Does anything stand out if you write the equation as

$${{dV} \over {dt}} = -ge^{{-\gamma r}}$$

You are not given any time information, so you suspect you'll need to integrate the RHS with respect to 'r'. What happens to both sides of the equation when you multiply by $$dr$$? Remember, $${{dr} \over {dt}} = V$$. Keep in mind that I'm being a little sloppy and the V in your solution really is something more like $$V_{initial}$$ whereas my V, or I probably should have put $$\dot r$$ is the integration variable.

are you saying initially its respect to t, then it becomes with respect to r.

i.e (dr/dt)*dt=dr?

 

[cloud360]

You can either drop the unit vector by erasing it, or you can draw a vector over the a. Either is okay.
You already have basically reached the answer to part a.
you have written -ge-γrȓ=a
Actually this equation does not make sense unless a is a vector. You cannot equate a vector and a scalar.

This would be correctly written in either of the following ways:

$$\vec a = -ge^{{-\gamma r}} \hat r$$

or

$$a = \dot{v} = \ddot{r} = -ge^{{-\gamma r}}$$

Also, this may give some insight into the second part.

You see that you could rewrite the given formula as

$$\dot{r}^2 - V^2 = \frac{2g}{ \gamma } ( e^{- \gamma r} - e^{- \gamma R} )$$

Which you can also rewrite as:


$$\frac{1}{2} m ( \dot{r}^2 - V^2 ) = \frac{1}{2} m \frac{2g}{ \gamma } ( e^{- \gamma r} - e^{- \gamma R} )$$

Now the left side can be interpreted as $$\Delta K.E.$$

What does this say about the right hand side?

can you please kindly expand on this a little bit, i am not following how you got from


$$\dot{r}^2 - V^2 = \frac{2g}{ \gamma } ( e^{- \gamma r} - e^{- \gamma R} )$$

to


$$\frac{1}{2} m ( \dot{r}^2 - V^2 ) = \frac{1}{2} m \frac{2g}{ \gamma } ( e^{- \gamma r} - e^{- \gamma R} )$$


I know how to derive the energy equation T+V=E (if conserved) or T+V=c (if F is a force field). don't know if this is related to b?

 

[Pengwuino]

are you saying initially its respect to t, then it becomes with respect to r.

i.e (dr/dt)*dt=dr?

Yes. You could solve it as a function of time, but the problem you want is in terms of radii. So you want to get your integrations in the form of an integral over the radial distance. What you initially have, $$F = m\ddot x$$, is derivatives with respect to time, which is what you don't want.

 

[HallsofIvy]

Why are you doing all that? You are given that F= ma and that $F= mge^{-\gamma r}$. From that it follows immediately that $F= ma= -mge^{-\gamma r}$ and so $a= -ge^{-\gamma r}$. For an object falling directly toward the planet, which has only the "r" component, $a= \ddot{r}= -ge^{-\gamma r}$.

For b, first use the chain rule: da/dt= dv/dr dr/dt= v dv/dr so the equation becomes
$v dv/dr= -ge^{-\gamma r}$ and integrate:
$$\int v dv= -g \int e^{-\gamma r} dr$$

I see no reason to bring "energy" into this.

 

[AlexChandler]

For b, first use the chain rule: da/dt= dv/dr dr/dt= v dv/dr so the equation becomes
$v dv/dr= -ge^{-\gamma r}$ and integrate:
$$\int v dv= -g \int e^{-\gamma r} dr$$

I see no reason to bring "energy" into this.

Yes this way is better. Do it this way

 

[AlexChandler]

can you please kindly expand on this a little bit, i am not following how you got from


$$\dot{r}^2 - V^2 = \frac{2g}{ \gamma } ( e^{- \gamma r} - e^{- \gamma R} )$$

to


$$\frac{1}{2} m ( \dot{r}^2 - V^2 ) = \frac{1}{2} m \frac{2g}{ \gamma } ( e^{- \gamma r} - e^{- \gamma R} )$$

I multiplied both sides by m/2 to get this. I was referring to the work energy theorem. $$W = \int \vec F \cdot d \vec r = \Delta K.E.$$ But it is really unnecessary to do it this way. You should do it the way outlined above.

 

[AlexChandler]

My point was that you could consider the change in K.E. then set it equal to the work done which is the integral of the force that was given and this immediately gives you the equation for b.

 

[cloud360]

Why are you doing all that? You are given that F= ma and that $F= mge^{-\gamma r}$. From that it follows immediately that $F= ma= -mge^{-\gamma r}$ and so $a= -ge^{-\gamma r}$. For an object falling directly toward the planet, which has only the "r" component, $a= \ddot{r}= -ge^{-\gamma r}$.

For b, first use the chain rule: da/dt= dv/dr dr/dt= v dv/dr so the equation becomes
$v dv/dr= -ge^{-\gamma r}$ and integrate:
$$\int v dv= -g \int e^{-\gamma r} dr$$

I see no reason to bring "energy" into this.

thank you so much, i am very grateful for you help !

but i don't understand why "da/dt= dv/dr dr/dt= v dv/dr "

please can you kindly explain this part

i also don't understand how you changed the integral on RHS from being in terms of dt, to being in terms of dr !

i know 0.5d/dt(v^2)=a i.e 0.5d/dt(r'^2)=a

 

[AlexChandler]

thank you so much, i am very grateful for you help !

but i don't understand why "da/dt= dv/dr dr/dt= v dv/dr "

please can you kindly explain this part

It is just the chain rule in calculus.

In Leibniz notation the chain rule is just

$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$

You can basically think of it the same way as the following:

$$\frac{a}{b} = \frac{a}{c} \frac {c}{b}$$

where a,b,c are just numbers

After you use then chain rule, you can recognize dr/dt as the rate of change of position with time, or just velocity v

But actually you've written something incorrectly. It should be

$$a = \frac{dv}{dt} = \frac{dv}{dr} \frac{dr}{dt} = v \frac{dv}{dr}$$

 

[cloud360]

Should i also assume the initial velocity is V i.e v(0)=V?

or

At t=0 velocity =R
At t=0, |v|=speed=Vi got this, but can't get any further

[PLAIN]http://img856.imageshack.us/img856/1953/12754659.gif

 

[cloud360]

It is just the chain rule in calculus.

In Leibniz notation the chain rule is just

$$\frac{dy}{dx} = \frac{dy}{du} \frac{du}{dx}$$

You can basically think of it the same way as the following:

$$\frac{a}{b} = \frac{a}{c} \frac {c}{b}$$

where a,b,c are just numbers

After you use then chain rule, you can recognize dr/dt as the rate of change of position with time, or just velocity v

But actually you've written something incorrectly. It should be

$$a = \frac{dv}{dt} = \frac{dv}{dr} \frac{dr}{dt} = v \frac{dv}{dr}$$

dr/dt*dv/dr=dv/dt=a', where i think of v as r'

then how is dv/dt=da/dt

also how is a=v*(dv/dr)

 

[AlexChandler]

dr/dt*dv/dr=dv/dt=a', where i think of v as r'

then how is dv/dt=da/dt

also how is a=v*(dv/dr)

where did you get this from??

$$\frac{dv}{dt}$$

is surely not the same as

$$\frac{da}{dt}$$

this is just not true... (unless they are both zero) which they are not

 

[AlexChandler]

If you are having trouble with the chain rule, I do recommend looking it up in a calculus book. However, let's try a different approach. The work energy theorem should be quite straightforward in this case.

We know the gravitational force as

$$mge^{{-\gamma r}}$$

The work energy theorem is simply

$$\Delta K.E. = \int \vec F \cdot d \vec r$$

then since the dot product must give a negative sign, (negative work done) we must have

$$\frac{1}{2} m (v^2 - V^2) = - \int_R^r mge^{{-\gamma r}} dr$$

or

$$\frac{1}{2} m ( v^2 - V^2 ) = \frac{ m g }{ \gamma } ( e^{ - \gamma r} - e^{- \gamma R } )$$

From here, some algebra gets you to the answer. This approach may be better if you have an intuition for physics, but not much experience with math.

 

[cloud360]

where did you get this from??

$$\frac{dv}{dt}$$

is surely not the same as

$$\frac{da}{dt}$$

this is just not true... (unless they are both zero) which they are not

doesnt the dr cancel out here?

dr/dt*dv/dr, so were left with dv/dt

 

[cloud360]

[tex] \frac{1}{2} m (v^2 - V^2)

i don't understand why it is v^2-V^2

why is it now just 0.5mv^2, i understand that 0.5mv^2=integral of a

 

[AlexChandler]

i don't understand why it is v^2-V^2

why is it now just 0.5mv^2, i understand that 0.5mv^2=integral of a

You're going to have to try a bit harder, we can't do all of the work here. It is a bounded integral, that's where you get the difference of the squares of the velocities.

And yes the dr's cancel out. But you have said that:

$$\frac{da}{dt} = \frac{dv}{dt}$$

This is not true. Where did you get this from??

 

[cloud360]

You're going to have to try a bit harder, we can't do all of the work here. It is a bounded integral, that's where you get the difference of the squares of the velocities.

And yes the dr's cancel out. But you have said that:

$$\frac{da}{dt} = \frac{dv}{dt}$$

This is not true. Where did you get this from??

sorry for not clearing that up.

but halls of ivy said:

da/dt= dv/dr dr/dt= v dv/dr

but now i think he forget to put a comma and meant to say

da/dt= dv/dr , dr/dt= v dv/dr

your opinion?

 

[AlexChandler]

sorry for not clearing that up.

but halls of ivy said:

da/dt= dv/dr dr/dt= v dv/dr

but now i think he forget to put a comma and meant to say

da/dt= dv/dr , dr/dt= v dv/dr

your opinion?

Ah yes, he has made a mistake. I didn't see that before. But think about it... how could that be true? He did not forget to put a comma, and what you have typed (da/dt= dv/dr , dr/dt= v dv/dr ) is completely wrong.
All that you need is

$$a = \frac{dv}{dr} \cdot \frac{dr}{dt} = v \frac{dv}{dr}$$

Either use this, or simply use the work energy theorem as I have shown.

 

[cloud360]

ok i get the correct answer.but how did u know the integral limits are r and R?

 

[AlexChandler]

ok i get the correct answer.


but how did u know the integral limits are r and R?

The work done in displacing an object from some location R to another location r is defined as

$$\int_R^r \vec F \cdot d \vec r$$

 

[cloud360]

so is this just a formulae i have to remember. i saw it in my notes, but it is not r and R, instead my teacher put it as t1 and t2, different times !

also thanks for your help. but one final question.

it says something about t=0. does this mean there is also an initial condition?

 

[AlexChandler]

so is this just a formulae i have to remember. also thans for your help. but one final question.

it says something about t=0. does this mean there is also an initial condition?

Yes it just means that

$$\dot r (t=0) = V$$

but you have already used this. And you really don't have to worry about time when using the work energy theorem.

 

[cloud360]

thanks so much, i understand now, so

$$\int_R^r \vec F \cdot d \vec r$$=

(integral t1 to t2) F.r' dt

 

[AlexChandler]

thanks so much, i understand now, so

$$\int_R^r \vec F \cdot d \vec r$$=

(integral t1 to t2) Fr' dt

yes you can do this, but you don't have to. F is already in terms of r. So you should just integrate with dr. But yes you can say that:

$$W = \int_{t1}^{t2} F \frac{dr}{dt} dt$$

but it is really unnecessary.

 

[HallsofIvy]

dr/dt*dv/dr=dv/dt=a', where i think of v as r'

The "a' " here must be a typo. dr/dt*dv/dr= dv/dt= a, not the derivative of a.

then how is dv/dt=da/dt

also how is a=v*(dv/dr)

a= dv/dt. By the chain rule, dv/dt= (dv/dr)(dr/dt) and dr/dt= v.