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

[Dustinsfl]

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*}

 

[TSny]

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 $\ddot{\mathbf{r}} =\large \frac{\mathbf{F}_{\rm net}}{m}$.