Where Did I Go Wrong in Deriving Gravitational Potential Energy?

[Dead Boss]

Homework Statement

Derive the gravitational potential energy from Newton's law.

Homework Equations

$$\mathbf{F} = -G \frac{m_1m_2}{\left|\mathbf{r}\right|^3} \mathbf{r}$$

$$W = \int_A^B \mathbf{F} \cdot{} d\mathbf{r}$$

$$\Delta{}U = W$$

The Attempt at a Solution

I can use the scalar version of the law, but I want to do it with the vector form.
So, first some easy substitution:

$$W = \int_A^B \mathbf{F} \cdot{} d\mathbf{r} <br /> = -Gm_1m_2 \int_A^B \frac{\mathbf{r}\cdot{}d\mathbf{r}}{\left|\mathbf{r}\right|^3}<br /> = Gm_1m_2 \int_A^B \frac{r_xdr_x + r_ydr_y + r_zdr_z}{\left|\mathbf{r}\right|^3}$$

$$= -Gm_1m_2 \left\{<br /> \int_A^B \frac{r_x}{\left|\mathbf{r}\right|^3} dr_x +<br /> \int_A^B \frac{r_y}{\left|\mathbf{r}\right|^3} dr_y +<br /> \int_A^B \frac{r_z}{\left|\mathbf{r}\right|^3} dr_z<br /> \right\}$$

Now take one integrand, for example the x part,

$$\int_A^B \frac{r_x}{\sqrt{r_x^2 + r_y^2 + r_z^2}^3} dr_x =<br /> \int_A^B r_x \left( r_x^2 + r_y^2 + r_z^2 \right)^{-\frac{3}{2}} dr_x$$

and perform substitution:

$$t = r_x^2 + r_y^2 + r_z^2$$
$$dt = 2r_x dr_x$$
$$\frac{dt}{2} = r_x dr_x$$

That yields:

$$\frac{1}{2} \int_A^B t^{-\frac{3}{2}} dt_x =<br /> \left[ -t^{-\frac{1}{2}} \right]_A^B =<br /> \left[ -\frac{1}{\sqrt{r_x^2 + r_y^2 + r_z^2}} \right]_A^B =<br /> \left[ -\frac{1}{\left|\mathbf{r}\right|} \right]_A^B =<br /> -\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)$$

Doing the same for other two integrands and plunging them back:

$$W = -Gm_1m_2 \left\{<br /> -\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)<br /> -\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)<br /> -\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)<br /> \right\}$$

And finally

$$W = 3Gm_1m_2 \left(<br /> \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|}<br /> \right)$$

Now if we take B at infinity, the potential energy will be

$$U = - 3G\frac{m_1m_2}{\left|\mathbf{r}\right|}$$

which is three times as large as it should be. Don't tell me, let me guess. There's an error somewhere up there. *sigh*
Any help would be highly appreciated.

 

[hikaru1221]

This is where you were wrong:

$$<br /> dt = 2r_x dr_x<br />$$

t is a multi-variable function, therefore: $$dt = 2r_xdr_x+2r_ydr_y+2r_zdr_z$$

A suggestion: $$2\vec{r}d\vec{r} = d(\vec{r}^2) = d(r^2)$$

 

[Dead Boss]

Thanks, hikaru!
I knew I should have read that book about multivariable calculus.