Gravitational potential energy

In summary, hikaru attempted to solve the gravitational potential energy equation with the vector form of the law, but there was an error in the substitution.
  • #1
Dead Boss
150
1

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.
 
Last edited:
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  • #2
This is where you were wrong:
[tex]

dt = 2r_x dr_x

[/tex]

t is a multi-variable function, therefore: [tex]dt = 2r_xdr_x+2r_ydr_y+2r_zdr_z[/tex]

A suggestion: [tex]2\vec{r}d\vec{r} = d(\vec{r}^2) = d(r^2)[/tex]
 
  • #3
Thanks, hikaru!
I knew I should have read that book about multivariable calculus.
 

1. What is gravitational potential energy?

Gravitational potential energy is the energy an object possesses due to its position in a gravitational field. It is the potential for an object to do work as a result of its position in relation to other objects affected by gravity.

2. How is gravitational potential energy calculated?

Gravitational potential energy is calculated using the equation U = mgh, where U is the potential energy, m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object relative to a chosen reference point.

3. Does gravitational potential energy depend on the mass of the object?

Yes, gravitational potential energy is directly proportional to the mass of the object. This means that as the mass of the object increases, so does its potential energy.

4. How does the height of an object affect its gravitational potential energy?

The higher an object is positioned in a gravitational field, the greater its potential energy will be. This is because the higher the object is, the more work it can potentially do when it falls towards the ground due to gravity.

5. Can gravitational potential energy be negative?

Yes, gravitational potential energy can be negative. This occurs when an object is positioned below the reference point, such as when an object is underground or below the surface of a body of water. In this case, the potential energy is considered to be "negative" because work would need to be done to move the object to a higher position where it would have more potential energy.

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