- #1

Dead Boss

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## Homework Statement

Derive the gravitational potential energy from Newton's law.

## Homework Equations

[tex]

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

[/tex]

[tex]

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

[/tex]

[tex]

\Delta{}U = W

[/tex]

## 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:

[tex]

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

= -Gm_1m_2 \int_A^B \frac{\mathbf{r}\cdot{}d\mathbf{r}}{\left|\mathbf{r}\right|^3}

= Gm_1m_2 \int_A^B \frac{r_xdr_x + r_ydr_y + r_zdr_z}{\left|\mathbf{r}\right|^3}

[/tex]

[tex]

= -Gm_1m_2 \left\{

\int_A^B \frac{r_x}{\left|\mathbf{r}\right|^3} dr_x +

\int_A^B \frac{r_y}{\left|\mathbf{r}\right|^3} dr_y +

\int_A^B \frac{r_z}{\left|\mathbf{r}\right|^3} dr_z

\right\}

[/tex]

Now take one integrand, for example the x part,

[tex]

\int_A^B \frac{r_x}{\sqrt{r_x^2 + r_y^2 + r_z^2}^3} dr_x =

\int_A^B r_x \left( r_x^2 + r_y^2 + r_z^2 \right)^{-\frac{3}{2}} dr_x

[/tex]

and perform substitution:

[tex]

t = r_x^2 + r_y^2 + r_z^2

[/tex]

[tex]

dt = 2r_x dr_x

[/tex]

[tex]

\frac{dt}{2} = r_x dr_x

[/tex]

That yields:

[tex]

\frac{1}{2} \int_A^B t^{-\frac{3}{2}} dt_x =

\left[ -t^{-\frac{1}{2}} \right]_A^B =

\left[ -\frac{1}{\sqrt{r_x^2 + r_y^2 + r_z^2}} \right]_A^B =

\left[ -\frac{1}{\left|\mathbf{r}\right|} \right]_A^B =

-\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)

[/tex]

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

[tex]

W = -Gm_1m_2 \left\{

-\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)

-\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)

-\left( \frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|} \right)

\right\}

[/tex]

And finally

[tex]

W = 3Gm_1m_2 \left(

\frac{1}{\left|\mathbf{B}\right|} - \frac{1}{\left|\mathbf{A}\right|}

\right)

[/tex]

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

[tex]

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

[/tex]

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.

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