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Mou_Laurin
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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 [itex]\mu[/itex]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 [tex] p=\frac{9.4}{0.387 098^2}\approx 68.3 \mu N/m^2[/tex]
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 [tex]\pi (2.4397\cdot 10^6)^2=1.8699\cdot10^{13} m^2[/tex]
where [itex]2.4397\cdot10^6[/itex] is the mean radius of Mercury. So the total force felt by the sail is [tex]F=68.3\cdot1.8699\cdot10^{13}=1.277\cdot10^{15}N[/tex].
Now the size of an orbit in this solar system depends only on the velocity of the orbiting object according to the equation [tex]v=\sqrt{\frac{GM}{r}}[/tex] 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 [itex]35.3^\circ[/itex]. The mean orbital velocity of Mercury being [itex]4.787\cdot10^4 m/s[/itex] and that of Venus being [itex]3.502\cdot10^4m/s[/itex] the velocity difference is [itex]Δ v=1.285\cdot10^4m/s[/itex]. And so, unsing Mercury's mass of [itex]3.3022×10^{23} kg[/itex] [tex]t=\frac{Δ v \cdot m}{F}=3.3225\cdot10^{12}s.[/tex]
Which is roughly [itex]105356[/itex] years.
A lot of my numbers come from http://www.ugcs.caltech.edu/~diedrich/solarsails/intro/sailing.txt. The rest comes from Wikipedia.
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 [tex]\pi (2.4397\cdot 10^6)^2=1.8699\cdot10^{13} m^2[/tex]
where [itex]2.4397\cdot10^6[/itex] is the mean radius of Mercury. So the total force felt by the sail is [tex]F=68.3\cdot1.8699\cdot10^{13}=1.277\cdot10^{15}N[/tex].
Now the size of an orbit in this solar system depends only on the velocity of the orbiting object according to the equation [tex]v=\sqrt{\frac{GM}{r}}[/tex] 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 [itex]35.3^\circ[/itex]. The mean orbital velocity of Mercury being [itex]4.787\cdot10^4 m/s[/itex] and that of Venus being [itex]3.502\cdot10^4m/s[/itex] the velocity difference is [itex]Δ v=1.285\cdot10^4m/s[/itex]. And so, unsing Mercury's mass of [itex]3.3022×10^{23} kg[/itex] [tex]t=\frac{Δ v \cdot m}{F}=3.3225\cdot10^{12}s.[/tex]
Which is roughly [itex]105356[/itex] years.
A lot of my numbers come from http://www.ugcs.caltech.edu/~diedrich/solarsails/intro/sailing.txt. The rest comes from Wikipedia.