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| 183_notes:grav_accel [2014/09/10 02:16] – [The Gravitational Force and the Momentum Principle] caballero | 183_notes:grav_accel [2021/02/05 00:02] (current) – [The Gravitational Force and the Momentum Principle] stumptyl | ||
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| + | Section 3.2 and 3.3 in Matter and Interactions (4th edition) | ||
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| ===== Gravitational Acceleration ===== | ===== Gravitational Acceleration ===== | ||
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| $$\vec{F}_{grav} = -G\dfrac{m_1 m_2}{|\vec{r}|^2}\hat{r}$$ | $$\vec{F}_{grav} = -G\dfrac{m_1 m_2}{|\vec{r}|^2}\hat{r}$$ | ||
| - | where object 1 has mass, $m_1$, and object 2 has mass, $m_2$. If the separation vector ($\vec{r}$) describes the relative position of object 2 with respect to object 1: | + | where object 1 has mass, $m_1$, and object 2 has mass, $m_2$. If the separation vector ($\vec{r}$) describes the relative position of object 2 with respect to object 1 (as shown in the figure to the right): |
| $$\vec{r} = \vec{r}_2 - \vec{r}_1$$ | $$\vec{r} = \vec{r}_2 - \vec{r}_1$$ | ||
| - | then the force expression above describes the force that object 1 exerts on object 2. That | + | then the force expression above describes the force that object 1 exerts on object 2. You can use [[183_notes: |
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| + | $$\vec{F}_{net, | ||
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| + | Using the formula for each force, we find: | ||
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| + | $$\dfrac{\Delta \vec{p}_2}{\Delta t} = m_2\dfrac{\Delta \vec{v}_2}{\Delta t} = m_2 \vec{a}_2 = -G\dfrac{m_1 m_2}{|\vec{r}|^2}\hat{r}$$ | ||
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| + | We then divide the mass of the object 2 out ($m_2$): | ||
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| + | $$\vec{a}_2 = -G\dfrac{m_1}{|\vec{r}|^2}\hat{r}$$ | ||
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| + | //__The resulting expression is the acceleration that object 2 experiences due to it's gravitational interaction with object 1__//. Notice that the acceleration of object 2 depends only on the mass of object 1 ($m_1$), and relative position of object 2 with respect to object 1 ($\vec{r}$). It also points towards object 1, which indicates that the object 2 is attracted (and will thus experience an acceleration along the line between object 1 and 2). | ||
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| + | So, __in general__: | ||
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| + | $$\vec{a} = -G\dfrac{m}{|\vec{r}|^2}\hat{r}$$ | ||
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| + | Sometimes, it's useful to think of this acceleration occurring in a single dimension (e.g., along the line that connects object 1 and object 2). Let's take that line to line in the $x$-direction. In that case, the expression for the magnitude of the acceleration in $x$-direction is given by: | ||
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| + | $$a_x = -G\dfrac{m}{x^2}$$ | ||
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| + | where the object with mass $m$ is the one that exerts the force on the mass in question (i.e., the object experiencing the acceleration) and $x$ is the distance between the objects. | ||
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| + | ==== The Local Gravitational Acceleration revisited ==== | ||
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| + | Earlier you read that the [[183_notes: | ||
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| + | For simplicity, let's take the downward vertical direction to be positive. Let's compute the acceleration due gravity at the surface of the Earth. Here the [[http:// | ||
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| + | $$a_y=G\dfrac{M_{Earth}}{R^2_{Earth}} = \left(6.67384 \times 10^{-11} \dfrac{m^3}{kg\: | ||
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| + | which is pretty close to the value we often use. In fact, the gravitational acceleration fluctuates a few percent over the surface of the Earth due to [[http:// | ||