Derive Formula for Velocity of Object Approaching Earth

[directory_NA]

Homework Statement

A particle of mass m is dropped from a height h, which is not necessarily small compared with the radius of the earth. Show that if air resistance is neglected, the speed of the particle when it reaches the surface of the Earth is given by $\sqrt {2gh}$ $\sqrt {\frac {R_E} {R_E + h}}$ .
($R_E$ = radius of Earth)

Homework Equations

$KE = \frac 1 2$mv²
$PE = -G \frac {Mm} {r}$
$PE= mgh$ (unsure if this one is relevant to the question)

The Attempt at a Solution

I started with the basic rule of conservation of energy and wrote out my equation:
$\frac {1} {2} mv^2_i - \frac {GMm} {R_E + h} = \frac {1} {2} mv^2_f - \frac {GMm} {R_E}$
Since it was dropped, thus, I went ahead and presumed $v_i = 0$ and also canceled out all the "m"'s. Giving the equation:
$- \frac {GM} {R_E + h} = \frac {1} {2} v^2_f - \frac {GM} {R_E}$
If I solve for $v_f$, I would get the following:
$\sqrt {2GM[R_E - (R_E + h)]}$
... which consequently looks nothing like the desired equation.

I'm not sure if I'm on the right track or something is going wrong within my derivation so I would like some advice regarding how to approach this problem and what would lead me on the right track.

 

[PeroK]

Your approach is correct, but your algebra went astray!

 

[directory_NA]

Your approach is correct, but your algebra went astray!

May I ask where the problem arose? I'm having trouble finding the root of the error.

 

[directory_NA]

Your approach is correct, but your algebra went astray!

Ah, now looking back I realize that the
$R_E - (R_E + h)$
should be instead,
$\frac {1} {R_E} - \frac {1} {R_E + h }$

Yet that doesn't quite account for why $R_E$ would be in the numerator and there is a $gh$ instead of $GM$

 

[PeroK]

May I ask where the problem arose? I'm having trouble finding the root of the error.

You'll need to show your work. But, I could ask what happened to the denominators?

 

[PeroK]

Ah, now looking back I realize that the
$R_E - (R_E + h)$
should be instead,
$\frac {1} {R_E} - \frac {1} {R_E + h }$

Yet that doesn't quite account for why $R_E$ would be in the numerator and there is a $gh$ instead of $GM$

What is $g$?

 

[directory_NA]

What is $g$?

$\frac {GM} {r^2}$, if the m was taken out from $mg = \frac {GMm} {r^2}$

 

[PeroK]

$\frac {GM} {r^2}$, if the m was taken out from $mg = \frac {GMm} {r^2}$

Does that give you the book answer?

 

[directory_NA]

Does that give you the book answer?

Sorry, I'm not quite sure what you mean.
I tried substituting $\frac {GM} {r^2}$ into the desired equation in the question, but it didn't seem to connect with what I had.

 

[PeroK]

Sorry, I'm not quite sure what you mean.
I tried substituting $\frac {GM} {r^2}$ into the desired equation in the question, but it didn't seem to connect with what I had.

What's $r$?

 

[directory_NA]

What's $r$?

$R_E$

 

[PeroK]

$R_E$

That must help!

 

[directory_NA]

That must help!

Let me take a look at it for a second...

 

[directory_NA]

That must help!

I managed to get
$v = \sqrt {2gR_E} \sqrt {(1-\frac {R_E} {R_E+h})}$
However, now there seems to be an extra $R_E$ and an extra $(1-)$

 

[PeroK]

I managed to get
$v = \sqrt {2gR_E} \sqrt {(1-\frac {R_E} {R_E+h})}$
However, now there seems to be an extra $R_E$ and an extra $(1-)$

Well, your first approach is to keep going with the algebra. I think you have to be more positive about the idea that one expression might be the same as another, just simplified or reorganised.

Or, if you get stuck like this, then you can always plug in the numbers and see whether your answer is the same numerically. Take suitable values for $g, R_E$ and $h$ and see whether your answer is numerically the same as the book answer.

 

[directory_NA]

Well, your first approach is to keep going with the algebra. I think you have to be more positive about the idea that one expression might be the same as another, just simplified or reorganised.

Or, if you get stuck like this, then you can always plug in the numbers and see whether your answer is the same numerically. Take suitable values for $g, R_E$ and $h$ and see whether your answer is numerically the same as the book answer.

So the implication is that algebraically my equation is the same as the one in the question, but it is just organized in a different manner?
And the only way to get to the question equation would be to fiddle with the algebra a bit more?
Is it possible to give any hints regarding how to approach the equation given in the question?

 

[PeroK]

So the implication is that algebraically my equation is the same as the one in the question, but it is just organized in a different manner?
And the only way to get to the question equation would be to fiddle with the algebra a bit more?
Is it possible to give any hints regarding how to approach the equation given in the question?

Well, it seems to me that there is only really one thing you could do with:

$1 - \frac{R}{R+h}$

Hint: common denominator.

 

[directory_NA]

Well, it seems to me that there is only really one thing you could do with:

$1 - \frac{R}{R+h}$

Hint: common denominator.

I see it now, that was sly.
Thanks so much!