Find at what rate the orbit radius will grow

[jaksa]

Homework Statement

In one proposed means of space propulsion, a thin sheet of highly reflecting plastic film would be used as a radiation pressure "sail". A plane square sheet $100m^2$ is available, and the mass of the spaceship is $1000kg$. If the spaceship initially travels in a circular orbit of 1AU radius about the Sun. Find at what rate $\frac {dR} {dT}$ the orbit radius will grow.

Relevant Equations

N/A

In order to change the radius, additional energy is required, the total energy of mass m on a circular orbit is given by:
$$E_{total} = - \frac {GMm} {2R},$$
The change in energy between orbits $R$ and $R_{0}$ is:
$$\Delta E_{total} = \frac {GMm} {2} \cdot \left( \frac 1 R_{0} - \frac 1 R \right).$$
For $S(R_{0})=1370\frac W {m^2}$, this will give me the equation as:
$$\frac {GMm} {2} \cdot \left( \frac 1 R_{0} - \frac 1 R \right)=\frac {S(R_{0}) \cdot R^{2}_{0} \cdot A \cdot t} {R^2}$$
Solving this equation for $R$ gives me this:
$$R = \frac 1 2 \left( R_{0} + \sqrt{R^2_{0} + 4 \cdot C(t)} \right)$$
where $C(t) = \frac {2S(R_{0}) \cdot R^3_{0} \cdot A \cdot t} {GMm},$.
the rate:
$$\frac {dR} {dt} = \frac {8S(R_{0}) \cdot R^2_{0} \cdot A} {GMm \cdot \sqrt{1 + \frac {4C(t)} {R^2_{0}}}}$$
The answer to this problem is $\frac {dR} {dt} = 360 \frac m s$ but I don't get such answer, even when I try for $t = 0$.
Is my reasoning for this problem correct or should I aim for different approach?
Thanks in advance for any advice!

 

[haruspex]

What makes you think all of the energy in the radiation will be utilised?
How do solar sails work?

 

[jaksa]

I wrongly assumed that total energy at $R$ will be of form $E_{total} = - \frac {GMm} {2R}$. With solar sail the orbit will no longer be a circle. I will try to solve EOM:
$$m \frac {d^2R} {dt^2} = -F_{G} + F_{S},$$
then get the radial component of velocity, that should be it.

 

[haruspex]

With solar sail the orbit will no longer be a circle.

That depends how it is deployed. I think for this question you can assume it stays circular.

I will try to solve EOM:
$$m \frac {d^2R} {dt^2} = -F_{G} + F_{S},$$

I think you can still use energy. You just have to figure out what fraction of the energy can be turned into orbital gain.
You did not answer my question about how solar sails work
https://en.wikipedia.org/wiki/Solar_sail

 

[jaksa]

What makes you think all of the energy in the radiation will be utilised?

There was a hint in the problem:
"Maximize the power transferred to the spacecraft by radiation pressure."
I also assumed that the sail always stays perpendicular to the radius.

You did not answer my question about how solar sails work
https://en.wikipedia.org/wiki/Solar_sail

Due to radiation pressure, force is exerted on the sail.

 

[jbriggs444]

There was a hint in the problem:
"Maximize the power transferred to the spacecraft by radiation pressure."
I also assumed that the sail always stays perpendicular to the radius.

[emphasis mine]
You know the formula for work, right? Force times perpendicular displacement, right?
And for power? Force times perpendicular velocity, right?

So a force that is perpendicular to the direction of motion transfers energy at what rate?

If you angle a solar sail (think a reflective mirror, as the problem statement suggests), in what direction does the force then act? Does this improve the rate of energy transfer?

 

[haruspex]

There was a hint in the problem:
"Maximize the power transferred to the spacecraft by radiation pressure."

Sure, but that would be maximising within the physical constraints. Maybe it is in the nature of the way solar sails work that not all the energy of the light can be used for gaining KE.
My preference is always to look to conservation of momentum first. If it is clear that that is conserved (as here) it is safer than assuming work is conserved.
As you can see at the link I posted, the radiation pressure comes from momentum conservation.
You will need to assume how much of the light is reflected, but at least that puts you in a 2:1 range.

 

[haruspex]

How about trying to solve it this way…

Calculate the radiation force F.
Let the gain in radius be $\Delta r$.
Ideally, work done is $F\Delta r$.
Set that equal to the increase in PE+KE.

 

[TSny]

I think @jbriggs444 has the right approach. Find the orientation of the sail that yields the maximum rate of energy transfer from the radiation to the ship.

If you angle a solar sail (think a reflective mirror, as the problem statement suggests), in what direction does the force then act? Does this improve the rate of energy transfer?

With this approach, I end up with $\dot r = 3.6$ m/s. I get $\dot r = 360$ m/s if the area of the sail is (100 m)² rather than 100 m².

 

[jbriggs444]

I think @jbriggs444 has the right approach. Find the orientation of the sail that yields the maximum rate of energy transfer from the radiation to the ship.


With this approach, I end up with $\dot r = 3.6$ m/s. I get $\dot r = 360$ m/s if the area of the sail is (100 m)² rather than 100 m².

I got 3.4 m/s and have a huge detailed explanation which is, sadly, out of place here until OP shows their solution.

The optimal sail angle turns out to be close to, but not exactly 45 degrees.

 

[TSny]

I got 3.4 m/s and have a huge detailed explanation which is, sadly, out of place here until OP shows their solution.

The optimal sail angle turns out to be close to, but not exactly 45 degrees.

I get an angle that differs from 45 degrees by roughly 20%. We'll wait for the OP. Unfortunately, I will be traveling over the next 6 days and have little time to check in here.

 

[jbriggs444]

I get an angle that differs from 45 degrees by roughly 20%. We'll wait for the OP. Unfortunately, I will be traveling over the next 6 days and have little time to check in here.

My algebra was faulty and indeed, I now get an angle like the one you report.

 

[haruspex]

Find the orientation of the sail that yields the maximum rate of energy transfer from the radiation to the ship.

I do not understand how this works. Light exerts a pressure on the sail leading to an acceleration. (In the present case, the force is at most $10^{-3}N$.)
That does not instantly translate into a velocity. So is the idea to find a terminal velocity?

If the sail is moving away from the light source at speed v then the maximum of the incident energy E it can utilise is $\frac vcE$. And that is with the sail square on to the source. At https://en.wikipedia.org/wiki/Solar_sail I read "Note that the force and acceleration approach zero generally around θ = 60° rather than 90° as one might expect with an ideal sail."

Also, transference between circular orbits is a tricky matter. A Hohmann transfer takes half an orbit and two tangential thrusts, so not suitable for a solar sail.
https://en.wikipedia.org/wiki/Orbital_mechanics#Orbital_maneuver

https://www.mdpi.com/2076-3417/13/17/9590 looks relevant, but somewhat beyond me.

 

[jbriggs444]

I do not understand how this works. Light exerts a pressure on the sail leading to an acceleration. (In the present case, the force is at most $10^{-3}N$.)
That does not instantly translate into a velocity. So is the idea to find a terminal velocity?

Right. The force does not translate directly into a velocity. But if the force has a component in the satellite's existing direction of motion then it does translate into a rate of gain of kinetic energy. Power.

A reflective sail canted at an angle can produce a net force with a tangential component.

Given a rate at which power is pumped into the satellite, we can calculate how rapidly the orbit is lifted -- how rapidly the craft spirals outward. Orbital mechanics makes that a bit more complicated than just totting up a rate of change of potential energy, but the very first equation in the original post had that part figured out properly.

Yes, I get something less than $10^{-3} \text{ N}$ for a $100\ \text{m}^2$ sail. Multiply by orbital velocity and you get something in the neighborhood of 10 watts. [Photonic drives are quite energy-inefficient for low velocity craft]

The efficiency formula given in the Wiki article is incorrect for the purposes of the problem at hand. We do not care about the magnitude of total thrust. We care about the tangential component.

 

[haruspex]

A reflective sail canted at an angle can produce a net force with a tangential component.

But what it cannot do is cancel the eccentricity at will.
As soon as we apply a radially outward force, the satellite will shift into an elliptical orbit, initially on the outward leg. As the Hohmann transfer illustrates, even a purely tangential force (unachievable with a solar sail) leads to that. Without applying a radially inward force, we cannot reduce the eccentricity until it is at or past apogee.

 

[jbriggs444]

But what it cannot do is cancel the eccentricity at will.

The tangential force can be modulated at will.
The radial force amounts to about 0.01 percent of the sun's gravitational force (for the stated configuration). Or 1 percent (for the intended configuration). It does not cause significant eccentricity, especially if continued over the course of a complete orbit.

At 3.6 meters per second over the course of a year the craft will have managed to enlarge its orbit by 100,000 km or so. Meanwhile, the one time increase in orbital radius due to a continuing 0.01 percent decrease in effective gravity is about 15,000 km or so.

We are not doing a Hohmann transfer (two impulsive burns). We are doing a continuously accelerated outward spiral trajectory. Taking 660 years (stated) or 6.6 years (intended) to reach Mars.

Of course, continuous forward acceleration means continuously decreasing speed due to the way orbital mechanics works out.

Stated = 100 m² sail. Intended = (100 m)² sail.

 

[haruspex]

The tangential force can be modulated at will.

But not, at first, in a way that reduces eccentricity. That only becomes possible near apogee. Seems like you have to decide at the start what new orbit you are aiming for.

The radial force amounts to about 0.01 percent of the sun's gravitational force. It does not cause significant eccentricity, especially if continued over the course of a complete orbit.

Is it not the case that a steady radially outward force would result in an orbit with reduced perihelion, no matter for how long it is applied? How does that allow determination of dR/dt? Should the question have asked for the rate of increase of the semi-major axis?

We are not doing a Hohmann transfer

I know, but such a transfer illustrates the difficulty.

 

[TSny]

I think @haruspex is right. We can calculate the rate at which the radiation pressure increases the total mechanical energy of the ship for a given orientation of the sail. But, in my solution, I assumed the orbit would remain approximately circular. I imagined an orbit that slightly spirals outward when the sail is active. So I took $E = -\large \frac {GMm}{2r(t)}$ as being a good approximation for the energy $E$ of the ship at time $t$, at least for small $t$. But, that could be a bad assumption. I won't have much time to think about this over the next few days.

 

[haruspex]

More thoughts…
Consider the case where the sail is always pointed at the Sun. The satellite starts in a circular orbit at radius R then unfurls the sail.
Since the force on the sail is radial and inversely proportional to the square of the orbital radius, it is like a reduction in G, to G' say. The new trajectory is therefore an ellipse.
Since there is no instantaneous change in position or velocity, the trajectory will return to that initial state, and since in that state the velocity is normal to the radius that point will be the apogee.
The angular momentum remains at $mvR=m\sqrt{GMR}$. That makes the perigee $R\frac{G}{2G'-G}$.
Note that at $2G'=G$ the satellite escapes.

 

[jbriggs444]

Is it not the case that a steady radially outward force would result in an orbit with reduced perihelion, no matter for how long it is applied?

You'd subsequently posted a nice and pithy analysis for a purely radially outward force. When a radially outward inverse $r^2$ force is suddenly turned on, the effect on the orbital path is that the current velocity is suddenly too large for the pre-existing circular orbit. The new central net force is still inverse $r^2$, so the new orbit will be an ellipse and will have its perihelion at the original orbital radius.

With a canted sail, the same logic applies. The new orbital path will be approximately elliptical but will spiral outward from that baseline at a rate that depends on the tangential component of the net force. One might try to circularize the new orbit by modulating the tangential thrust, though there seems little point in doing so.

 

[jaksa]

Right. The force does not translate directly into a velocity. But if the force has a component in the satellite's existing direction of motion then it does translate into a rate of gain of kinetic energy. Power.

A reflective sail canted at an angle can produce a net force with a tangential component.

Given a rate at which power is pumped into the satellite, we can calculate how rapidly the orbit is lifted -- how rapidly the craft spirals outward. Orbital mechanics makes that a bit more complicated than just totting up a rate of change of potential energy, but the very first equation in the original post had that part figured out properly.

Yes, I get something less than $10^{-3} \text{ N}$ for a $100\ \text{m}^2$ sail. Multiply by orbital velocity and you get something in the neighborhood of 10 watts. [Photonic drives are quite energy-inefficient for low velocity craft]

The efficiency formula given in the Wiki article is incorrect for the purposes of the problem at hand. We do not care about the magnitude of total thrust. We care about the tangential component.

With canted at an angle sail I have:

The formula for F:
$$F = F_{0} \cdot cos^2(\alpha),$$
tangential component:
$$F_{t} = F_{0} \cdot cos^2(\alpha) \cdot sin(\alpha)$$
Now I find for what $\alpha$ the $cos^2(\alpha) \cdot sin(\alpha)$ is maximum then calculate the rate of potential energy change?

 

[jbriggs444]

With canted at an angle sail I have:
View attachment 344038
The formula for F:
$$F = F_{0} \cdot cos^2(\alpha),$$
tangential component:
$$F_{t} = F_{0} \cdot cos^2(\alpha) \cdot sin(\alpha)$$

I came up with a different formula following a different line of reasoning and used the angle of the sail rather than the angle of its normal. But I suspect that both yield the same result and that both optimize at the same angle.

Yes, after a little behind the scenes work, both come out the same.

[I used the sine of the angle of the reflected ray (twice my alpha) to get its tangential momentum component and the sine of the angle of the mirror to get the cross section. After some trig and double angle formula evaluation, that comes to $2 \sin^2 \alpha \cos \alpha$. The factor of two comes out of the computation of $F_{0}$. Mine is not doubled. The $\sin$ versus $\cos$ swap comes from the choice of using sail angle rather than sail normal angle].

Now I find for what $\alpha$ the $cos^2(\alpha) \cdot sin(\alpha)$ is maximum then calculate the rate of potential energy change?

You can make the substitution of $1-\sin^2 \alpha$ for $\cos^2 \alpha$. That will give you a simple polynomial in $\sin \alpha$. You can optimize that (solve for a zero first derivative) and then take the arc sine of the result. Then plug that angle into your formula for tangential force.

You will want to solve for $F_0$ either using the figures from the solar sail article or from first principles using $E=pc$.

 

[haruspex]

With canted at an angle sail I have:
View attachment 344038
The formula for F:
$$F = F_{0} \cdot cos^2(\alpha),$$
tangential component:
$$F_{t} = F_{0} \cdot cos^2(\alpha) \cdot sin(\alpha)$$
Now I find for what $\alpha$ the $cos^2(\alpha) \cdot sin(\alpha)$ is maximum then calculate the rate of potential energy change?

Why do you ignore the radial component, $\cos^3(\alpha)$?

 

[TSny]

Why do you ignore the radial component, $\cos^3(\alpha)$?

The instantaneous power delivered to the ship depends only on the component of the force parallel to the velocity. Initially, the velocity has no radial component.

 

[haruspex]

The instantaneous power delivered to the ship depends only on the component of the force parallel to the velocity. Initially, the velocity has no radial component.

But the question asks for the rate of gain of orbital radius. Increased tangential speed only leads to that indirectly. Radial acceleration leads to it slightly less indirectly.
Specifically, if in a circular orbit at radius R speed v and subjected to acceleration $a_r\hat r+a_{\theta}\hat\theta$ then after time dt the radius has increased by $\frac{v}{2R}a_{\theta}dt^3+a_{r}(\frac 12dt^2+O(dt^3))$.

The question as posed cannot be answered. No matter how you angle the sail there is no instantaneous creation of nonzero radial velocity. It only makes sense as an average rate of gain over some preset increase in radius, or perhaps as an asymptotic value.

 

[jbriggs444]

Why do you ignore the radial component, $\cos^3(\alpha)$?

As you have analyzed yourself, the radial component leads to an immediate radial acceleration and a periodic radial velocity. Which is not what is being asked for.

By contrast, the tangential component leads to an identifiable long term radial velocity. The problem asks for a radial velocity. So this is plausibly what it is being asked for. The icing on the cake is the fact that the "correct" answer matches the result of taking this approach.

 

[haruspex]

the tangential component leads to an identifiable long term radial velocity.

That is not entirely clear to me. Do you have an analytic solution? Whatever angle you set the sail at, there will be a radial acceleration, whether you want it or not. That could well lead to periodicity again, putting the long term radial velocity in doubt.
Besides, it also leads to an increasing tangential velocity, so how much of the energy remains lost to that?

 

[jbriggs444]

That is not entirely clear to me. Do you have an analytic solution? Whatever angle you set the sail at, there will be a radial acceleration, whether you want it or not. That could well lead to periodicity again, putting the long term radial velocity in doubt.
Besides, it also leads to an increasing tangential velocity, so how much of the energy remains lost to that?

No, I do not have an analytical solution. I get the same 3.6 meters per second that @TSny got. Which sort of matches the erroneous 360 meters per second that the problem setters got.

You seem concerned over possible variations in the rate of orbital energy gain due to the eccentricity of the roughly elliptical orbit. Just how eccentric do you expect the orbit to be? Since effective $g$ is reduced by less than one percent, I would not expect significant eccentricity (though that assessment depends on exactly what significant effect one is looking for and what precision one requires).

You mention increasing tangential velocity. Over the long term (beyond half an orbit or so), there is no increase to tangential velocity. Over the short term, we have roughly about .0004 N tangential thrust on a 1000 kg craft with an initial tangential velocity of some 30,000 m/sec. It is going to take more than half an orbit to change tangential velocity significantly. By which time, the change will turn out to be a decrease.

Eventually, the tangential velocity decrease together with inverse square attenuation will decrease the available rate of energy gain. But I believe that we are after an average rate of orbital radius increase for only the first year or so. For that sort of time frame, we can accept the rough model of a near circular spiral without losing much fidelity.

You gain some advantage due to the decrease in the steepness of the gravitational potential. Tangential thrust is reduced by an inverse square but so is the slope of the gravitational potential. Meanwhile, orbital velocity is also reduced, so it is not a wash. The rate of increase in orbital radius does get slower the further out you go. Back of the envelope, I think a 10% increase in orbital radius means a 5% decrease in the rate of orbital radius gain.

 

[haruspex]

I produced this plot in a spreadsheet.

GM=1
R=10
F/m (force of radiation per unit mass at distance 1 from Sun when sail square-on)=0.01
dt=0.7
Rate of increase of radius maximises with sail tilted at around 0.617 radians (zero tilt meaning square-on), at which angle it averages 0.0023 in whatever units.
However, the rate of increase of r is sinusoidal, going negative occasionally. The time period I ran it for (about 600 time steps) showed two complete cycles.

In order to map these results to the actual question, I tried to deduce how the average rate of radius increase depends on my three parameters by twiddling the values, but I must be doing something wrong [edit: maybe it's correct, see post #30]. I get $R(GM)^{-1}(F/m)$, which makes no sense dimensionally. Remember that my F/m actually has dimension $L^3T^{-2}$, the same as GM, so the deduced expression is missing a $T^{-1}$.

Glossing over that, it predicts the general result to be $0.0235 R(GM)^{-1}(F/m)$

 

[haruspex]

Belatedly, I realised the logic behind going for as much tangential force as possible is that it maximises the immediate rate of gain of total energy. Whether that also maximises the long term rate of gain is unclear, but it is a reasonable starting point.
So I changed my model to adjust the angle of the sail dynamically to maximise the component of the force parallel to the velocity.

The resulting pattern of energy gain still had a flutter (as did the optimal sail angle), but the rate of energy gain showed a clear decline. I got a very good fit to energy = $A-Be^{-\lambda t}$, suggesting an absolute limit to the achievable radius of orbit.
That fits with the dimensional analysis in post #29: $r_{limit}=r_{initial}f(\frac{GMm}F)$, where the maximum available force on the sail at distance r is $\frac F{r^2}$.

 

[jaksa]

I came up with a different formula following a different line of reasoning and used the angle of the sail rather than the angle of its normal. But I suspect that both yield the same result and that both optimize at the same angle.

Yes, after a little behind the scenes work, both come out the same.

[I used the sine of the angle of the reflected ray (twice my alpha) to get its tangential momentum component and the sine of the angle of the mirror to get the cross section. After some trig and double angle formula evaluation, that comes to $2 \sin^2 \alpha \cos \alpha$. The factor of two comes out of the computation of $F_{0}$. Mine is not doubled. The $\sin$ versus $\cos$ swap comes from the choice of using sail angle rather than sail normal angle].


You can make the substitution of $1-\sin^2 \alpha$ for $\cos^2 \alpha$. That will give you a simple polynomial in $\sin \alpha$. You can optimize that (solve for a zero first derivative) and then take the arc sine of the result. Then plug that angle into your formula for tangential force.

You will want to solve for $F_0$ either using the figures from the solar sail article or from first principles using $E=pc$.

I got $F=0.3849 \cdot F_{0}$ and the approach from my first post gave me formula of:
$$\dot{R} = \frac {2 \cdot 0.3849 \cdot F_{0} AR^2_{0} v} {GMm} \approx 3.5 \frac m s$$