# Homework Help: Equilibrium distance for solar sail

1. Oct 12, 2015

### cyberdiver

This is not actually a homework assignment, but something I decided to try in my own time. I wanted to find the radius from a star at which a solar sail would be held at equilibrium (radiation pressure = gravity), given mass per unit area and stellar luminosity at a reference radius.

So I attempted the following:
Pressure of radiation and pressure of gravity are equal and opposing:
$$p_{radiation}=-p_{gravity}$$
Subsititute radiation pressure equation and gravity equation (rho_A is areal density, E_F is energy flux):
$$2 \cdot \frac{E_F}{c} = -{\rho}_A \cdot g$$
Substitute inverse square law equation into E_F, areal density equation into rho_A, and the law of universal gravitation equation:
$$2 \cdot \frac{E_{F0} \cdot (\frac{r_0}{r})^2}{c} = -\frac{m}{A} \cdot \frac{G \cdot M}{r^2}$$
Attempt to make r^2 the subject:
$$2 \cdot E_{F0} \cdot \frac{r_0^2}{r^2} \cdot A \cdot r^2 = -m \cdot G \cdot M$$

The problem here is that r^2 and r^2 will cancel out, making the equation useless. How else could I solve this problem? Does it require calculus?

2. Oct 12, 2015

### haruspex

You are assuming there is a particular distance where a balance is struck. Your equations are telling you something else.

3. Oct 12, 2015

### cyberdiver

Hold on. Is it because radiation pressure and gravity both follow the inverse square law, so the equilibrium is at all distances?

4. Oct 12, 2015

### haruspex

Yes.

5. Oct 12, 2015

### cyberdiver

But if the solar sail is heavy enough, it would start falling toward the star, wouldn't it?

6. Oct 12, 2015

### haruspex

whether the sail drifts away from the star or falls towards it will depend on its mass to area ratio and the star's mass to flux ratio. What it won't depend on is the starting position.

7. Oct 12, 2015

### cyberdiver

So the sail can only reach a state of equilibrium if its areal density is just right, otherwise it would drift?

8. Oct 12, 2015

### haruspex

Yes.

9. Oct 12, 2015

### cyberdiver

I understand now. Thank you very much!