Edit: Although this question has been marked as a duplicate of this one about solar sail thrust calculation, they are different because I ask about a solar sail that has an orbital trajectory and will have a low trust transfer, going from one circular orbit to another.

Notice: An answer to this question may be similar to this one, but the energy calculation there uses the mass of propellant !

This would be the starting point of the journey.


Image: NASA/JPL/University of Arizona

Could it be advantageous to saw a rectangular block of 5 x 7 x 8 meters out of Phobos and bring it to a low orbit around Mars and change it into a space station ?

By creating a space of 3 x 5 x 6 meters within this block and a construction to prevent it from falling apart, with a mass density of 1.876 g/cm$^3$ the mass of the block would be about 3.6 x 10$^5$ kg.
Consequently this block from Phobos would have 2 meter thick walls and would give radiation protection to life within it.

For almost circular orbits the orbital speed can be calculated from v$_0$$^2$ x r = GM, with GM being 4.282 x 10$^1$$^3$ for Mars. Thus a space station 100 km above Mars will have an orbital speed of about 3.5 km/sec, while that of Phobos is about 2.1 km/sec.

For a low thrust transfer with a solar sail, going from one circular orbit to another simply requires the same delta-v as the difference between the two speeds, so in this case the delta-v to get the block from Phobos to a 100 km above Mars will be about 1400 m/sec.

But how does one calculate the timespan and energy needed to get the block station in that low orbit around Mars with a solar sail with an area of 100 x 100 m for instance, assuming the solar sail is always facing the direction of travel and for half of the orbit it is in the shadow of Mars ?

  • $\begingroup$ To give a ratio here, you're asking about a 10,000 ton cube of Phobos. The maximum payload of falcon heavy is 64 metric tons. To move this cube from Earth to LEO would take 156.25 falcon heavy launches. To move it from Earth to trans-mars injection would be 595.2 falcon launches. To move it from Phobos to Mars LEO is 1.4 delta-v and from Earth to Earth LEO is 9.3 delta-v. Approximations would say the maneuver you list to be 15% of the effort of Earth to LEO. Meaning you'd need ~24 fully fueled falcon heavy's at Phobos to do it. $\endgroup$ Aug 16, 2018 at 16:48
  • $\begingroup$ Hi, Conelisinspace. I think you've attracted a downvote because people take your question as proposing moving Phobos with a solar sail, even though your actual question is just about moving an arbitrary cube between two orbits using a solar sail, correct? Can you explain why space.stackexchange.com/questions/4587/… doesn't answer your question? $\endgroup$
    – Bear
    Aug 16, 2018 at 17:25
  • $\begingroup$ @Bear I linked that in my first comment too :P $\endgroup$ Aug 16, 2018 at 17:40
  • $\begingroup$ Note, turning a stone into a satellite requires a... whole civilization. $\endgroup$
    – peterh
    Aug 16, 2018 at 18:19
  • 1
    $\begingroup$ If you want to try the calculation yourself, I'd probably try something more simple... Like calculating the time it would take for a solar sail to go from 100km to 200km altitude over mars given certain coefficients, maybe assuming that the tilt will always be in the correct direction to maximize thrust and that the distance to the sun will remain constant. $\endgroup$ Aug 16, 2018 at 18:31

2 Answers 2


Let's make a quick simplified estimate.

According to Wikipedia, the solar sail will exert a force of $8.17 \mu N / m^2$, when the Sun rays are perpendicular to the sail. So for a 100x100m sail this will be $0.0817 N$.

However, the $8.17 \mu N / m^2$ figure is for a sail at Earth distance from the Sun. Our sail will be at Mars distance, so there will be less solar radiation pressure and the force will decrease by a factor proportional to $R_{Mars} / R_{Earth}$ where $R_{Mars}$ and $R_{Earth}$ is the Sun radiance at Mars ($561 W/m^2$) and Earth distance ($1361 W / m^2$) respectively. So this will be $561 / 1361 \approx 0.43$ and our force will reduce to about $0.0352 N$.

Now let's assume that the solar sail will be in the shadow for half of the orbit and produce no force (this is not true, it will be in the shadow for less than that). A quarter of the orbit will be in the sunlight and moving towards the Sun, so it will decelerate our Phobos block. For another quarter, it will be in the sunlight and moving away from the Sun, so the Sun would accelerate our block - to avoid this, we will keep the sail parallel to the Sun rays, so that again it won't produce any force. During the quarter orbit where it will decelerate our Phobos block, we will keep the sail perpendicular to our direction of travel, and the force produced will be proportional to $sin(\theta)$ where $\theta$ is the angle formed by the sail and the Sun rays. The average of this factor during this quarter orbit can be calculated by integrating $sin(\theta)$ from 0 to 90 degrees and taking its average. $$ \alpha _ {1/4 orbit} = \frac{\int_{0}^{90} cos(\theta) d\theta}{90} \approx 0.636 $$ For the whole orbit, it will be $0.636 / 4 = 0.159$. This means that on average, during the whole orbit, we will output only $0.159$ of the force compared to as if we were always with the sail perpendicular to the Sun rays.

All in all, our average force will be $0.0352 \cdot 0.159 \approx 0.0056 N$.

According to the comments to the question, our Phobos block will weight 360 tonnes (360.000 Kg). So our acceleration will be: $a = F / m = 0.0056 / 360000 \approx = 0.0000000156 m/s$

So to reach our $\Delta v$ of $1400 m/s$ it will take us $1400 / a \approx 90000000000s !$

That is about 2853 years!

Some notes on the estimate:

  • Given the very small force, in this estimation we have assumed as if the orbit will stay circular (it will actually decrease very slowly in a spiral).
  • We have also exaggerated Mars' shadow
  • We have ignored the sail mass (I hoped it will be negligible compared to the 360 tons of the block of Phobos)
  • We have assumed no aerobraking (which could be a thing, see SF comment).

Removing totally Mars's shadow would mean we will use half orbit instead of a quarter, so the time will be halved ("only" 1426 years).

Below some Python code with the calculations:

import scipy.integrate as integrate 
import math

deltaV = 1400 #m/s

sailArea = 100*100 #m2
# Sail force per square meter from https://en.wikipedia.org/wiki/Solar_sail, assuming Earth distance
sailForce = 8.17e-6 # N/m2 

#Integrate from 0 to PI/2.0, which is the same as from 0 to 90 degrees
avgEfficiencyDuringQuarterOrbit = integrate.quad(lambda x: math.sin(x), 0, math.pi/2.0)[0] / (math.pi/2.0)
avgEfficiencyDuringWholeOrbit = avgEfficiencyDuringQuarterOrbit / 4

# Radiances from https://nssdc.gsfc.nasa.gov/planetary/factsheet/marsfact.html
radianceOnMars = 586.2 # W/m2
radianceOnEarth = 1361 # W/m2 
radianceReduction = radianceOnMars/radianceOnEarth

avgForce = avgEfficiencyDuringWholeOrbit * radianceReduction * sailForce * sailArea # Newton
F = avgForce #N Force
# Mass is 360 tons of Phobos (from the comments to the question)
M = 360*1000 # Kg 

a = F / M

time = deltaV / a #s needed to accelerate to that speed

timeInDays = time / (60*60*24)
timeInYears = timeInDays / 365
print("It will take {} days, or {} years.".format(timeInDays, timeInYears))
  • $\begingroup$ Would the decresasing spiral and distance to mars be the biggest error in this calculation? Would a solar sail with constant low-level thrust maintain a concentric orbit or, as it spirals inwards, begin to gain eccentricity? The question asked deltaV but would the final orbit be a low eccentricity? $\endgroup$ Sep 16, 2018 at 14:59
  • $\begingroup$ @MagicOctopusUrn Because Mars orbits the Sun, the part of the orbit of the sail around Mars where the force on the sail works has moved a bit along that orbit with every single orbit, so after a martian year that force has worked evenly at all the parts of the orbit of the sail around Mars . $\endgroup$
    – Cornelis
    Sep 16, 2018 at 16:00
  • 1
    $\begingroup$ @uhoh Yes, I didn't realize while typing the question that I was using the wrong function (the "quote" button instead of the one for code/pre-formatted text). Thanks for the tips :) $\endgroup$
    – BlueCoder
    Sep 16, 2018 at 16:15
  • $\begingroup$ Is your calculation of the average of the factor during the quarter of the orbit when the force is at it's maximum right ? Should it not vary between 0.71 (half of the square root of 2) and 1 ? $\endgroup$
    – Cornelis
    Sep 16, 2018 at 16:15
  • 1
    $\begingroup$ @MagicOctopusUrn The link in question to en.wikipedia.org/wiki/… then leads to ocw.mit.edu/courses/aeronautics-and-astronautics/… which says that in going from one circular orbit to another with continuous low-thrust it is possible to assume that the orbit will stay circular. $\endgroup$
    – BlueCoder
    Sep 16, 2018 at 16:24

Could it be advantageous to saw a rectangular block of 5 x 7 x 8 meters out of Phobos and bring it to a low orbit around Mars and change it into a space station?

Orbital mechanics aside, a block of Phobos would not probably not make a good space station.

We often think about asteroids and small moons as being solid hunks of rock, but an object as small as Phobos which does not have enough gravity to pull itself into a sphere would not have gone through a molten stage to turn itself into solid rock. And it does not have enough gravity nor depth to produce the pressure necessary to produce rock, nor the seismic activity and weathering effects to reveal it on the surface.

Instead, Phobos is likely a pile of rubble weakly held together by its gravity.

The recent thinking, however, is that the interior of Phobos could be a rubble pile, barely holding together, surrounded by a layer of powdery regolith about 330 feet (100 meters) thick.

“The funny thing about the result is that it shows Phobos has a kind of mildly cohesive outer fabric,” said Erik Asphaug of the School of Earth and Space Exploration at Arizona State University in Tempe and a co-investigator on the study. “This makes sense when you think about powdery materials in microgravity, but it's quite non-intuitive.”

An interior like this can distort easily because it has very little strength and forces the outer layer to readjust. The researchers think the outer layer of Phobos behaves elastically and builds stress, but it’s weak enough that these stresses can cause it to fail.

Mars’ Moon Phobos is Slowly Falling Apart, NASA

Scott Manley goes into the "pile of rubble" issue, and other issues with turning asteroids into space stations, in his video about Spinning Asteroids To Make Space Stations.

  • $\begingroup$ Thank you for the valuable contribution. I thought the interior of Phobos would be more like a rubble pile held together by frozen water. $\endgroup$
    – Cornelis
    Sep 16, 2018 at 22:14

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