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

[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 someone 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}

 

[vela]

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

The quantity

\frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R})

is not positive; it's negative. Because r>R, i.e. the particle is above the planet's surface, the first exponential will be smaller than the second exponential, right?

so we let

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

The original inequality actually gives you

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

The exponential isn't in the denominator. So now you have

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

 

[cloud360]

The quantity

\frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R})

is not positive; it's negative. Because r>R, i.e. the particle is above the planet's surface, the first exponential will be smaller than the second exponential, right?

The original inequality actually gives you

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

The exponential isn't in the denominator. So now you have

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

Now do we find the limit?

 

[Mark44]

The quantity

\frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R})

is not positive; it's negative. Because r>R, i.e. the particle is above the planet's surface, the first exponential will be smaller than the second exponential, right?

The original inequality actually gives you

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

The exponential isn't in the denominator. So now you have

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

Now do we find the limit?

No.
First, simplify the expression at the right in the inequality above.
Next, remind yourself what question you're trying to answer (part d of this problem).
Does the simplified inequality suggest an answer to this question?

 

[cloud360]

No.
First, simplify the expression at the right in the inequality above.
Next, remind yourself what question you're trying to answer (part d of this problem).
Does the simplified inequality suggest an answer to this question?

Following your help, i have attached my solution.Please can you kindly tell me if it is correct or not?
[PLAIN]http://img862.imageshack.us/img862/4393/2010b5d.gif

 

[Mark44]

You're going through a lot of motions that you don't need.
You're given that
V \geq \sqrt{2g/\gamma}e^{-\gamma R/2}
\Rightarrow V^2 \geq 2g/\gamma e^{-\gamma R}


In your first inequality, you can say this
v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R})
\geq 2g/\gamma e^{-\gamma R} + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R})
= 2g/\gamma e^{-\gamma r}
\text{So } v^2 \geq 2g/\gamma e^{-\gamma r}

The line where you have
r' >= sqrt(2g/gamma * e^{- gamma * r) is incorrect.

If y² >= A, with A a positive number,
then y >= sqrt(A) or y <= -sqrt(A)
It is incorrect to say that y >= +/-sqrt(A).

What you should have is
v \geq \sqrt{2g/\gamma e^{-\gamma r}}

You don't need to concern yourself with negative values of v, since if v < 0, the particle is moving back toward the planet. We're interested only in the case where the particle is moving in the positive direction; i.e., v > 0, and the particle is moving away from the planet.

Most of what you have in the last seven or eight lines should go away, as it is either unnecessary or incorrect.

Since you have shown that v (= r') > 0, it's not zero, so the particle never stops, and continues moving away from the planet. That's all you need to say.

 

[cloud360]

You're going through a lot of motions that you don't need.
You're given that
V \geq \sqrt{2g/\gamma}e^{-\gamma R/2}
\Rightarrow V^2 \geq 2g/\gamma e^{-\gamma R}


In your first inequality, you can say this
v^2 = V^2 + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R})
\geq 2g/\gamma e^{-\gamma R} + \frac{2g}{\gamma}(e^{-\gamma r} - e^{-\gamma R})
= 2g/\gamma e^{-\gamma r}
\text{So } v^2 \geq 2g/\gamma e^{-\gamma r}

The line where you have
r' >= sqrt(2g/gamma * e^{- gamma * r) is incorrect.

If y² >= A, with A a positive number,
then y >= sqrt(A) or y <= -sqrt(A)
It is incorrect to say that y >= +/-sqrt(A).

What you should have is
v \geq \sqrt{2g/\gamma e^{-\gamma r}}

You don't need to concern yourself with negative values of v, since if v < 0, the particle is moving back toward the planet. We're interested only in the case where the particle is moving in the positive direction; i.e., v > 0, and the particle is moving away from the planet.

Most of what you have in the last seven or eight lines should go away, as it is either unnecessary or incorrect.

Since you have shown that v (= r') > 0, it's not zero, so the particle never stops, and continues moving away from the planet. That's all you need to say.

thanks a lot for your help :)