?Do you have to integrate velocity (v) to get the position function r(t)?

[cloud360]

?Do you have to integrate velocity (v) to get the position function r(t)?

Homework Statement

[PLAIN]http://img94.imageshack.us/img94/238/2010b5e.png

Homework Equations

The Attempt at a Solution

attempted solution to first question below, don't know what to do for second:

[PLAIN]http://img838.imageshack.us/img838/2129/2010b5qesol.png

i would be really grateful if you can help me out, i got an exam within a week

 

[Mark44]

It looks to me like R is a constant, the radius of the planet. Little r is the distance from the center of the planet to some point.

I think your equation for (r'(t))² is incorrect - V² shouldn't have that radical in it.

 

[HallsofIvy]

"Small r" is the variable, "R" is a constant.

However, it appears that what you are supposed to substitute for V is
$$\sqrt{\frac{2g}{\gamma}}e^{-\gamma r/2}$$
and you are to substitute that into $V^2$.

If that, you get
$$V^2= \frac{2g}{\gamma}e^{-\gamma R}$$
NOT the
$$\sqrt{\frac{2g}{\gamma}e^{R/2}}$$
that you have.

$$\dot{r}(\gamma)^2= V^2+ \frac{2g}{\gamma}\left(e^{-\gamma r}- e^{-\gamma R}\right)$$
should be
$$\frac{2t}{\gamma}e^{-\gamma R}+ \frac{2g}{\gamma}e^{-\gamma r}- \frac{2g}{\gamma}e^{-\gamma R}$$
$$= \frac{2g}{\gamma}e^{-\gamma r}$$
which is much simpler.

 

[cloud360]

"Small r" is the variable, "R" is a constant.

However, it appears that what you are supposed to substitute for V is
$$\sqrt{\frac{2g}{\gamma}}e^{-\gamma r/2}$$
and you are to substitute that into $V^2$.

If that, you get
$$V^2= \frac{2g}{\gamma}e^{-\gamma R}$$
NOT the
$$\sqrt{\frac{2g}{\gamma}e^{R/2}}$$
that you have.

$$\dot{r}(\gamma)^2= V^2+ \frac{2g}{\gamma}\left(e^{-\gamma r}- e^{-\gamma R}\right)$$
should be
$$\frac{2t}{\gamma}e^{-\gamma R}+ \frac{2g}{\gamma}e^{-\gamma r}- \frac{2g}{\gamma}e^{-\gamma R}$$
$$= \frac{2g}{\gamma}e^{-\gamma r}$$
which is much simpler.

thanks for helping me again :)

i will try this

 

[cloud360]

is my solution to C correct?
[PLAIN]http://img163.imageshack.us/img163/4480/b5sol.gif

but i don't know answer to D, is it something to do with unbounded growth i.e V(x) and E don't interesect (E=T+V, were T=kinetic, V=potential)

 

[cloud360]

anyone?

 

[vela]

Your final answer for part c is correct, but your work needs to be cleaned up a lot.

What were parts a and b of the problem? I think knowing the complete problem will provide better context for answering part d.

 

[cloud360]

Your final answer for part c is correct, but your work needs to be cleaned up a lot.

What were parts a and b of the problem? I think knowing the complete problem will provide better context for answering part d.

this is part a

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

 

[vela]

Hint: If the particle returns to the planet, that means it reached some maximum distance and then turned around. What do you know about its velocity when it reaches the maximum distance?

 

[cloud360]

Hint: If the particle returns to the planet, that means it reached some maximum distance and then turned around. What do you know about its velocity when it reaches the maximum distance?

E<V(x)

where E is a straight line on the graph V vs x, where V represents kinetic energy.

is that correct?

 

[vela]

No, it's not. Try again.

 

[cloud360]

No, it's not. Try again.

V(x) does not have any maximum or minimum points?

also, isn't there 2 solutions for C, one for negative square and positive square root? so is my solution for C complete, can you please kindly confirm i got one and only solution?

 

[vela]

If the particle turns around, it reaches a maximum distance from the planet. Mathematically, what does attaining a maximum mean? Think back to basic calculus.

 

[cloud360]

If the particle turns around, it reaches a maximum distance from the planet. Mathematically, what does attaining a maximum mean? Think back to basic calculus.

there is no limit ?

it diverges

 

[Mark44]

If the particle turns around, it reaches a maximum distance from the planet. Mathematically, what does attaining a maximum mean? Think back to basic calculus.

there is no limit ?

it diverges

Neither of these follows from what vela is suggesting. When you were first studying calculus, how did you find the maximum value of a function?

 

[cloud360]

Neither of these follows from what vela is suggesting. When you were first studying calculus, how did you find the maximum value of a function?

find solutions to r''=0, sub in the dependant variable an see if >0 or less than 0 or equal to zero, this indicates whether it is maximum or minimum

if there is a maximum it will return to planet, does that mean if there is a minimum, it will not return but go away from planet (this is kind of weird if true)?

because i am using the logic that if the particle turns around from a maximum point, it will return to the planet, if it turns around from a minimum point, it will go away from planet?

 

[vela]

Nope. The particle is initially launched from the planet's surface. If the particle returns to the planet, it must have turned around somewhere. That somewhere is when it reaches the maximum distance from the planet. If you can show there is no maximum distance, then you can conclude that the particle didn't turn around. The easiest way to do this is to try find the maximum distance and show there is no solution.

 

[cloud360]

sorry

 

[Mark44]

find solutions to r''=0, sub in the dependant variable an see if >0 or less than 0 or equal to zero, this indicates whether it is maximum or minimum

No. At the maximum distance, r' = 0.

if there is a maximum it will return to planet, does that mean if there is a minimum, it will not return but go away from planet (this is kind of weird if true)?

You don't need to be concerned about the minimum distance, which is 0 (i.e., the particle is on the surface of the planet).

because i am using the logic that if the particle turns around from a maximum point, it will return to the planet, if it turns around from a minimum point, it will go away from planet?

 

[cloud360]

No. At the maximum distance, r' = 0.You don't need to be concerned about the minimum distance, which is 0 (i.e., the particle is on the surface of the planet).

this is my solution to d, is it correct?
[PLAIN]http://img713.imageshack.us/img713/2870/solnew.gif

also, when solving c, we have r'^2 , if we square root both sides, don't we also get a negative soluton, do we ignore that negative solution because gamme >0

i.e r'= - sqrt(V^2 +(2g/gamma)*e^(gamma*r)-e^(gamma*R))...do we ignore this, as it says gamma>0
r'= + sqrt(V^2 +(2g/gamma)*e^(gamma*r)-e^(gamma*R))

 

[vela]

No, it's not correct. You've argued that if V=sqrt(2g/gamma)e^(-gamma R/2), the particle won't return to the planet. You haven't answered the actual question asked, which concerns other values of V as well.

 

[Mark44]

There are some very simple ideas at play here, but you need to understand what derivatives mean.

In the following, t* is some arbitrary time.
If v(t*) = 0, the particle is at rest.
If v(t*) > 0, the particle is moving away from its starting point (r is increasing).
If v(t*) < 0, the particle is moving toward its starting point (r is decreasing).

 

[cloud360]

There are some very simple ideas at play here, but you need to understand what derivatives mean.

In the following, t* is some arbitrary time.
If v(t*) = 0, the particle is at rest.
If v(t*) > 0, the particle is moving away from its starting point (r is increasing).
If v(t*) < 0, the particle is moving toward its starting point (r is decreasing).

If V’’(x)<0 , point x unstable (as subbing gives exponential function which is unbounded as t-->infinity)

do i need to show the second derivative is negative?

 

[Mark44]

I think you're just guessing now.

Given:
$$v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R})$$

(Above I'm using v for x'.)

$$\text{If} V \geq \sqrt{2g/\gamma}e^{-\gamma R/2}$$
why doesn't the particle return to the planet?

If you replace V² in the first equation by something that is less than or equal to it, what inequality do you get?

For the particle to stop and go in the other direction, back toward the planet, what must be true about v?

 

[cloud360]

I think you're just guessing now.

Given:
$$v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R})$$

(Above I'm using v for x'.)

$$\text{If} V \geq \sqrt{2g/\gamma}e^{-\gamma R/2}$$
why doesn't the particle return to the planet?

If you replace V² in the first equation by something that is less than or equal to it, what inequality do you get?

For the particle to stop and go in the other direction, back toward the planet, what must be true about v?

do i need to go the the extreme point of when v'=0, and test both sides, if both sides are same sign, then it proves it won't return to the planet. what i mean is do i sub a value greater that it (the value of V they gave) , then less than it and show v' is the positive in both cases?

like testing for a point of inflection?

i will admit, i am guessing because i really am lost, honestly all the things i have said has been wrong so far, even the limit test was not good enough :(

 

[Mark44]

$$V \geq \sqrt{2g/\gamma}e^{-\gamma R/2}$$
$$\Rightarrow V^2 \geq 2g/\gamma e^{-\gamma R}$$
Right?

Then
$$v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R}) \geq ??$$

If you can show that v² is >= something, then that gives you some crucial information that can be used to answer the question you're trying to answer. You remember what that is, right?

 

[cloud360]

$$V \geq \sqrt{2g/\gamma}e^{-\gamma R/2}$$
$$\Rightarrow V^2 \geq 2g/\gamma e^{-\gamma R}$$
Right?

Then
$$v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R}) \geq ??$$

If you can show that v² is >= something, then that gives you some crucial information that can be used to answer the question you're trying to answer. You remember what that is, right?

since we are tane the square o something, then v² is >=0

so

$$v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R}) \geq ??$$>=0

 

[Mark44]

since we are tane the square o something, then v² is >=0

v² >= 0 for any real number v, so that is not news.

so

$$v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R}) \geq ??$$>=0

You didn't use the inequality that I provided. Don't you think that might be important?

 

[cloud360]

s

$$v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R}) \geq ??$$>=0

v^2>V^2, ?as the bracket 2g/gamma(stuff inside bracket), part is positive?

 

[Mark44]

Let me say it again. Do you think that the first inequality here might be important?
$$V^2 \geq 2g/\gamma e^{-\gamma R}$$

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

 

[cloud360]

Let me say it again. Do you think that the first inequality here might be important?
$$V^2 \geq 2g/\gamma e^{-\gamma R}$$

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

The limit of this first inequality is infinity, as R goes to infinity, R represents a position function?

$$V^2 \geq 2g/\gamma e^{-\gamma R}$$

this means that the velocity v^2 is also infinity? so it won't return to the planet?

 

[cloud360]

anyone? please can you help.

how do i solve this, can somone tell me steps

 

[Mark44]

Let me say it again. Do you think that the first inequality here might be important?
$$V^2 \geq 2g/\gamma e^{-\gamma R}$$

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

From the inequalilty, substitute for V² in the equation.

$$v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R}) \geq \text{something}$$

 

[cloud360]

so we let

V^2=2g/(gamma*e^(-gamma*R))

to get

v^2=2g/(gamma*e^(-gamma*R)) + (2*g/gamma)(e^(-gamma*r)-e^(-gamma*R))

can you please tell me what that something is, so i can go to next step

 

[Mark44]

The "something" is what you get if you replace V² by an expression that is greater than or equal to V².

$$V^2 \geq \frac{2g}{\gamma} e^{-\gamma R}$$


$$v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R}) \geq \text{something}$$