MHB Why is there a Mass Term in the Denominator of the Spacecraft's Motion Equation?

AI Thread Summary
The discussion focuses on the motion equation of a spacecraft with a solar sail, highlighting the role of mass in the denominator of the acceleration equation. The force from solar radiation pressure is expressed as F = (2SA/c)(r/r^3), while the gravitational force is Fg = - (mμ/r^3)r. The net force combines these two forces, leading to the equation of motion that incorporates mass, ensuring dimensional consistency. It emphasizes the distinction between force and acceleration, clarifying that the term involving mass in the denominator is essential for accurate representation. Understanding these dynamics is crucial for analyzing spacecraft motion under solar radiation pressure.
Dustinsfl
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How does the solution have a m in the denominator next to c?

A perfectly reflecting solar sail is deployed and controlled so that its normal vector is always aligned with the radius vector from the sun.
The force exerted on the spacecraft due to solar radiation pressure is
$$
\mathbf{F} = \frac{2 S A}{c}\frac{\mathbf{r}}{r^3}
$$
where $S$ is the solar intensity, $A$ is the sail area and $c$ is the speed of light.

Show that the equation of motion for the spacecraft is
$$
\ddot{\mathbf{r}} + \left(\mu - \frac{2 S A}{cm}\right)\frac{\mathbf{r}}{r^3} = 0.
$$

The equation of motion for a spacecraft in the absence of a solar sail is
$$
\ddot{\mathbf{r}} = -\frac{\mu}{r^3}\mathbf{r}.
$$
Summing the forces, we have
\begin{alignat*}{3}
\ddot{\mathbf{r}} & = & \frac{2 S A}{c}\frac{\mathbf{r}}{r^3} - \frac{\mu}{r^3}\mathbf{r}\\
\ddot{\mathbf{r}} - \frac{2 S A}{c}\frac{\mathbf{r}}{r^3} + \frac{\mu}{r^3}\mathbf{r} & = & 0\\
\ddot{\mathbf{r}} + \left(\mu - \frac{2 S A}{c}\right)\frac{\mathbf{r}}{r^3} & = & 0
\end{alignat*}
 
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Dustinsfl said:
Summing the forces, we have
\begin{alignat*}{3}
\ddot{\mathbf{r}} & = & \frac{2 S A}{c}\frac{\mathbf{r}}{r^3} - \frac{\mu}{r^3}\mathbf{r}\\
\end{alignat*}
It's important to distinguish between force and acceleration. The left side of the equation above is acceleration. But the first term on the right side is a force while the second term on the right is acceleration. So the equation is not dimensionally consistent.

The gravitational force is ## \mathbf{F}_{\rm g} = - \frac{m\mu}{r^3}\mathbf{r}##.

So, the total force is $$ \mathbf{F}_{\rm net} = \mathbf{F}_{\rm sail} +\mathbf{F}_{ \rm g} = \frac{2 S A}{c}\frac{\mathbf{r}}{r^3} - \frac{m\mu}{r^3}\mathbf{r} $$

Now use ##\ddot{\mathbf{r}} =\large \frac{\mathbf{F}_{\rm net}}{m}##.
 
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