Solar sailing Mercury into Venus

1. Jun 25, 2012

Mou_Laurin

At Earth orbit, above the atmosphere, the solar power flux is roughly 1400 watts per square metre. This corresponds to a a pressure of 4.7 $\mu$N per square metre, or roughly one two thousandth of the weight of a paper clip at the Earth's surface. A perfectly reflecting material would feel double that force. If this pressure is inversely proportional to the square of earth's distance to the sun, which is equal to 1AU, you get that this force on Mercury is $$p=\frac{9.4}{0.387 098^2}\approx 68.3 \mu N/m^2$$
where 0.387098 is the length of Mercury's semi-major axis in AU.

If we build a perfectly reflecting sail that exactly covers the view of Mercury from the sun (it receives exactly the same amount of photons as Mercury does), its surface of incidence would be $$\pi (2.4397\cdot 10^6)^2=1.8699\cdot10^{13} m^2$$
where $2.4397\cdot10^6$ is the mean radius of Mercury. So the total force felt by the sail is $$F=68.3\cdot1.8699\cdot10^{13}=1.277\cdot10^{15}N$$.

Now the size of an orbit in this solar system depends only on the velocity of the orbiting object according to the equation $$v=\sqrt{\frac{GM}{r}}$$ which means that for one to push Mercury into a Venus orbit, all one must do is decelerate it to the right velocity. Pushing it parallel to its direction of travel, Mercury would climb the gravity well. In the counter-intuitive realm of orbital mechanics, the planet would slow down the whole time it is climbing the well. The optimal angle of incidence of the sail for this purpose is $35.3^\circ$. The mean orbital velocity of Mercury being $4.787\cdot10^4 m/s$ and that of Venus being $3.502\cdot10^4m/s$ the velocity difference is $Δ v=1.285\cdot10^4m/s$. And so, unsing Mercury's mass of $3.3022×10^{23} kg$ $$t=\frac{Δ v \cdot m}{F}=3.3225\cdot10^{12}s.$$
Which is roughly $105356$ years.

A lot of my numbers come from http://www.ugcs.caltech.edu/~diedrich/solarsails/intro/sailing.txt. The rest comes from Wikipedia.

2. Jun 26, 2012

jobigoud

Did you take into account the fact that as the device moves further away from the Sun towards the Venus orbit, its surface of incidence will decrease and it will not receive as many light than when it was at Mercury's distance ?

Also, the sail being at an angle, it will not receive the full F.

Last edited: Jun 26, 2012
3. Jun 26, 2012

To add to the above, surely since Mercury will no longer be receiving solar radiation directly, you'd need to take that into account and so the force will likely be halved. You're basically talking about changing Mercurys albedo, I suppose... Sail still has to push Mercurys mass so no great benefit from a high area to mass ratio obtained from a sail. Basically, setting Mercurys Albedo to 1ish would be the same idea perhaps..?

4. Jun 26, 2012

D H

Staff Emeritus
No. You are off by at least an order of magnitude.

As has already been noted, you did not account for the fact that tilting the solar sail reduces the force, and you did not account for the 1/r2 drop in solar flux.

You also did not account for gravity losses, aka gravity drag. Your Δv is that for a Hohmann transfer. Sans a gravity assist (and there is none between Mercury and Venus), a finite burn will always exceed the Δv for a Hohmann transfer. In this case, by more than a factor of two. You also didn't account for the fact that no solar sail is a perfect specular reflector. A specular reflector the size of Mercury? Come on.

What is the purpose of this thread?

5. Jun 26, 2012

phinds

My question as well. What's the point?

6. Jun 26, 2012

Chronos

A back door attempt at an 'original' theory would be my guess.

7. Jun 27, 2012

Mou_Laurin

@jobigoud: The distance has nothing to do with the surface of incidence, the angle does. It's not so hard to prevent the angle of incidence from changing, but it does mean that you will lose force, as you will have to use some of it to tilt the planet. Read the post carefully, I'm not assuming that the sail is circular, I'm assuming that the surface of incidence is circular.

@Deadstar: I'm not sure we're on the same page. You don't push a satellite perpendicular to its velocity when you want to change its orbit. You push it along said velocity. If you push it perpendicular, it won't budge a inch.

@D H: I think I've already answered the tilting bit. I indeed haven't taken into account the drop due to 1/r^2. I should have taken r to be the halfway point (short of doing the necessary calculus, which I will do). Also, the Δv would be half that for a Hoffmann transfer (right?), I did indeed use the correct Δv. Gravity drag is the only force that is doing work here, the Δv is the opposite direction as my force. But why am I off by at least an order of magnitude? With your arguments it seems that I would likely multiply the time by 3, maybe 4, but not ten. I'll grant you this, though, of all these people, you are the only one that had valid arguments amidst your post. I'd to hear more from you.

The point is to create a habitable planet that would be closer to Earth than Mars.

Anyway. You guys are funny. Taking this all seriously and all. And you are completely right, this could not be an original idea as you obviously have seen it elsewhere before. :P

8. Jun 27, 2012

D H

Staff Emeritus