Radiation Pressure & Gravity on Solar Sail at 1AU

[will mick]

Homework Statement

I am trying to work out the force on a solar sail at 1AU due to radiation pressure from the Sun, and also the force on it due to the Sun's gravity. I know G=6.67x10^-11; the mass of the sun M_s=2x10^30 kg; the distance from the Earth to the Sun r=1.5x10^11 meters; the speed of light c=3x10^8 and the intensity of sunlight at 1AU is I=1368 w/m^2.
The surface area of the solar sail A is unknown as is its mass m and i need to find a relationship between A and m for the sail to escape the Sun's gravitational pull.

Homework Equations

The Force due to gravity is F_g=mGM_s/r^2 where m is the mass of the solar sail
The Force due to radiation is F_r=2IxA/c because the light is being reflected.

Are these equations correct??

The Attempt at a Solution

When I work through the sums and plug in the numbers above, I am left with m<1.54x10^-3 A. But that does not seem right? The mass has to be tiny and area massive for this to work. When I plugged in the mass and area of IKAROS, the solar sail in space at the moment, A=200m^2 and m=315kg, it seems to fail to escape the Sun's gravitational pull by a factor of 10^3 but IKAROS is working right now.

Any ideas where I may be going wrong?

Thanks

 

[gneill]

Trying to leave the solar system by flying directly outwards is a very energy expensive proposition. Better to begin with an orbit and gradually accelerate, enlarging it over time. If there's no time constraint, you can use as little power as you want (less power = more time = smaller solar sails or ion engines, or what have you).

Remember, it takes *no* energy to simply stay on a given orbit.

 

[will mick]

I hadn't seen it that way..
So say the solar sail starts with an elliptical orbit around the Sun, even the smallest force would allow it to escape the Sun's gravitational pull?
If I wanted to incorporate the radiation pressure and initial orbit into an equation to show how a different surface area and mass of the sail could effect the rate at which the sail escapes how would I do that? I am trying to get my head round it..
I have the standard equation for an elliptical orbit as r=h^2/u(1+e cos(theta)) but can't see how to put it all together..

 

[gneill]

I hadn't seen it that way..
So say the solar sail starts with an elliptical orbit around the Sun, even the smallest force would allow it to escape the Sun's gravitational pull?

Well, assuming that the small force was directed appropriately, yes.

If I wanted to incorporate the radiation pressure and initial orbit into an equation to show how a different surface area and mass of the sail could effect the rate at which the sail escapes how would I do that? I am trying to get my head round it..
I have the standard equation for an elliptical orbit as r=h^2/u(1+e cos(theta)) but can't see how to put it all together..

Integrating the equation of motion for a constantly applied, but varying magnitude acceleration doesn't sound like an easy job. What you might do is look at the total mechanical energy of the orbit. When it's negative, the object is bound and the free-fall orbits are elliptical or circular. If it's zero then the orbit is parabolic (escape!) and if it is positive the orbit is hyperbolic (also escape!).

The specific mechanical energy (energy per unit mass) of a body in orbit is

$$\xi = \frac{v^2}{2} - \frac{\mu}{r}$$

Multiply though by the orbiting body's mass to get the actual values.

The semi-major axis for the bound orbits, is

$$a = \frac{-\mu}{2\xi}$$

So it sounds like you're adding energy at a rate that depends upon the size of the solar sail, its distance from the Sun, and its angle with respect to the Sun which will also determine the direction of the applied force vector -- you might want a gentle spiral outwards, so you'd want to keep the osculating orbit (literally the "kissing orbit", or the orbit that would obtain if the accelerating force was curtailed at a given instant and the body went into un-powered free-fall) nearly circular. Or, you might not care if the trajectory takes you back inwards for a while as long as you come back out with a higher velocity and larger semi-major axis.

 

[will mick]

But as the distance from the Sun changes so too does the Force on the solar sail so how would that affect the equation?

 

[gneill]

But as the distance from the Sun changes so too does the Force on the solar sail so how would that affect the equation?

The equation applies to any point on a free-fall trajectory. The r is the current distance from the Sun. Note that this is the mechanical energy equation for the free-fall (un-powered) trajectory with a given energy.

What you're doing is adding energy at some rate that depends upon r (probably inverse square of r), so at any given time this will provided the energy details for an osculating orbit at that instant. How much energy is gained as a function of time would depend upon the details of the orbit shape (since that determines how quickly r changes with t). The orbit shape will depend upon the history of how the thrust was applied. So it sounds like some sort of differential equation is going to need solving. It may be more amenable to integration by program (simulation) than by manual analysis.

I can't think of any other way forward at the moment.

Is this a homework question? If so, perhaps we're reading too much into it, and all that's wanted is the ratio of sail size to mass that just (over)balances the forces, and to show how this ratio looks at different radii. In other words, assume the worst possible trajectory -- straight outwards.

 

[will mick]

It is part of a project I have to do at Uni but I'm stuck on this part!

I have gone back to the beginning to try to understand it more..I have the force produced by the sun's intensity (for a perfectly reflecting surface area A) is:
$$F_r=\frac{\2<I>A}{c}$$
where <I> is the time-averaged intensity (I don't see how the 'time averaged' part affects anything?)
and that
$$I=\frac{P}{4\pi\times{r^2}}$$
is how the intensity changes as you move closer or further away from the sun..so putting these together I get that
$$F_r=\frac{2PA}{4\pi\times{r^2}\times{c}}$$
But when I equate this to the gravitational force out from the Sun when I am trying to escape from the Sun radially with no starting orbit, the r^2 will cancel, so the force required to balance out the gravitational force will not depend on distance from the sun?!
$$F_r=F_s\\ \frac{2PA}{4\pi\times{c}}=GMm$$
If the sail does have a starting orbit I thought I could find the escape velocity as a function of position which is apparently
$$V_e=\sqrt{\frac{2GM}{r'}}$$
where r' is the distance from the sun corresponding to the escape velocity, then use the equation F=ma to effectively go back in time to when the force was first applied when the sail was in a circular orbit..
but putting it all together is still confusing me!

 

[will mick]

I have worked through some sums, got a bit further and got stuck again:
The escape velocity at a position r' is:

$$v_e=\sqrt{\frac{2GM}{r'}}$$

Say the sail is in circular orbit. Then Newtons 2nd law for centripetal force at a radius r'' is:

$$F_c=\frac{mu^2}{r''}$$

This force is only provided by gravity so we can equate F_c and F_G:

$$F_c=F_G$$

$$\frac{mu^2}{r''}=\frac{GMm}{r''^2}$$

$$u=\sqrt{\frac{GM}{r''}}$$

So it's initial speed will be u. It needs to accelerate from u to v_e to escape and the sail provides the acceleration. But the acceleration is not constant and depends on r.

$$a=\frac{F}{m}$$

But is the force just the force provided by the sail or does it have to include the centripetal force too? If not,

$$a=\frac{2PA}{cmr^2}$$

where P is the power of sunlight, c is the speed of light. But this is where I get stuck..I am trying to show a relationship between the area of the sail, A, its mass m and how quickly it can escape but can't put the equations together just because the acceleration is not constant..if it was constant it would be easy!
I think i want to show how long it takes to accelerate from u to v_e with an acceleration a=F/m that is not constant..but then the acceleration will be radially out, not in the direction of travel...

 

[gneill]

I have worked through some sums, got a bit further and got stuck again:
The escape velocity at a position r' is:

$$v_e=\sqrt{\frac{2GM}{r'}}$$

Say the sail is in circular orbit. Then Newtons 2nd law for centripetal force at a radius r'' is:

$$F_c=\frac{mu^2}{r''}$$

This force is only provided by gravity so we can equate F_c and F_G:

$$F_c=F_G$$

$$\frac{mu^2}{r''}=\frac{GMm}{r''^2}$$

$$u=\sqrt{\frac{GM}{r''}}$$

So it's initial speed will be u. It needs to accelerate from u to v_e to escape and the sail provides the acceleration. But the acceleration is not constant and depends on r.

Not only that, but your sail can't supply thrust directly along the velocity vector for a circular orbit, because the Sun-Sail line is perpendicular to it. Any thrust it provides will tend to de-circularize the orbit (which is to be expected because you want to spiral outwards).

$$a=\frac{F}{m}$$

But is the force just the force provided by the sail or does it have to include the centripetal force too? If not,

$$a=\frac{2PA}{cmr^2}$$

where P is the power of sunlight, c is the speed of light. But this is where I get stuck..I am trying to show a relationship between the area of the sail, A, its mass m and how quickly it can escape but can't put the equations together just because the acceleration is not constant..if it was constant it would be easy!
I think i want to show how long it takes to accelerate from u to v_e with an acceleration a=F/m that is not constant..but then the acceleration will be radially out, not in the direction of travel...

By angling the sail you have some control over the direction of the thrust, but changing the angle also affects the sunward profile of the sail, and thus the energy captured.

It's really not a simple problem, as there are many variables.

For example, suppose that the craft was initially in a circular orbit at some distance R. The sail is deployed so as to catch the maximum sunlight (directly facing the Sun at all times). Then the thrust will always be radially outward. Can you write the differential equation of motion for this case? What would the trajectory look like? Would it always grow further from the Sun, or like an elliptical orbit, spend some time closer and some time further from the Sun as the overall size of the orbit grows?

 

[will mick]

If the only acceleration is radially, can i do this:

$$a=\frac{dv}{dt}$$ and $$v=\frac{dr}{dt}$$

so $$dt=\frac{dv}{a}=\frac{dr}{v}$$
$$vdv=adr$$

$$\int{vdv}=\int{adr}$$ with the limits on dv being u and v_e and the limits on dr being r'' and r' (i'm not sure about the limits)

$$\frac{v_e^2-u^2}{2}=\frac{2PA}{cm(r''-r')}$$

where $$v_e^2=\frac{2GM}{r'}$$ and [tex]u^2=\frac{GM}{r''}

and working through this gives:

$$\frac{2GMr''}{r'}-3GM+\frac{GMr'}{r''}=\frac{4AP}{m}$$

putting in r'=r'' (i.e. it starts at escape velocity) implies A=0 which would be true

But I'm not sure if what I've done can be applied or whether the limits on dv need to be the radial velocities, ie 0 and something else?

 

[will mick]

I can see three different ways of thinking about it with completely different answers and I don't know which one (if any) are correct:

1) the force pointing into the sun is

$$F=\frac{mv^2}{r}$$

The force out from the sun provided by the thrust from the sunlight is

$$F=\frac{2PA}{4\pi r^2 c}$$

So the total force on the system is

$$F_tot=\frac{2PA}{4\pi r^2 c} - \frac{mv^2}{r}$$

So the force needed to reach v=v_e, the escape velocity, is

$$\frac{2PA}{4\pi r^2 c}=\frac{m(v_e)^2}{r}$$

where
$$v_e=\frac{2GM}{R}$$

Substituting this in though cancels out the r's and gives

$$\frac{PA}{cm}=4\pi GM$$

but it doesn't make much sense that the r's have cancelled..

2) Using the same equation for the total force and equating that to ma gives

$$\frac{2PA}{4\pi mcr^2}-\frac{v^2}{r}=a=\frac{dv}{dt}=v\frac{dv}{dr}$$

working this through gives the differential equation

$$rv^2-vr^2\frac{dv}{dr}=\frac{2PA}{4mc\pi}$$

I tried to solve this but the constant on the right threw me..I would know how to solve it if it didn't have the constant on the right as it would be homogeneous.

3) Using the method I detailed in the previous post, but I did it wrong when I substituted the limits on dr in i should have got:

$$\frac{v_e^2-u^2}{2}=-\frac{PA}{2mcr'\pi}+\frac{PA}{2mcr''\pi}$$

working this through, multiplying by r''r'/r''r' etc gave me

$$\frac{PA}{2mc\pi}(r'-r'')=GM(2r''-r')$$

which then gives

$$\frac{PA}{2GMmc\pi}=\frac{2r''-r'}{r'-r''}$$

but if I put in the numbers and plot

$$y=\frac{2b-x}{x-b}$$

where y=A/m, r''=b a constant, and x=r'...it gives a graph which does not look right..its a 1/x relationship with discontinuities at y=b and x=b and y is negative after y=2b which doesn't sound right to me

Are any of these methods right?

 

[gneill]

Here are some thoughts. Take them any way you wish; I haven't "solved" this problem by any stretch of the imagination!

The differential equation of motion for the trajectory of the craft is going to look like:

$$\ddot{\textbf{r}} + \left(\mu - \frac{2 P A}{4 \pi \;c \;m}\right) \frac{\textbf{r}}{r^3} = 0$$

Where $$\mu$$ is the usual gravitational parameter for the Sun, and assuming that the sail is pointed directly sunwards.

Note that this looks just like the usual Newtonian equation of motion with an altered value for $$\mu$$, namely, if

$$\mu' = \mu - \frac{2 P A}{4 \pi\; c\; m}$$

Then the usual equation of motion comes out:

$$\ddot{\textbf{r}} + \mu' \frac{\textbf{r}}{r^3} = 0$$

In other words, the effect of the solar sail is to reduce the effective gravitational parameter in the equation of motion! What's more, this will apply at any distance because both gravitation and power received vary as the inverse square of the distance.

You should now be able to use the usual equations pertaining to orbits (especially the total mechanical energy) to predict the shape of the orbit (as long as the sail orientation remains pointed directly at the Sun), including whether or not it is on an escape trajectory (total mechanical energy greater than or equal to zero). If the craft is currently on orbit with specific mechanical energy $$\xi$$, then you can determine what the new energy of the orbit will be if the sail is raised at that moment.

If the sail orientation is not directly sunwards, then all bets are off. There will be non-radial components to the force, and you'll end up with a set of second order non-linear differential equations to solve. At that point, I'd resort to numerical integration of the equations (simulation).

How's that?

 

[BobG]

A slightly different tack using some assumptions to simplify things.

The force due to solar radiation is inversely proportional to the square of the distance, just as the force of gravity is. Likewise, acceleration due to solar radiation will be inversely proportional to the square of the distance, just as acceleration due to gravity is. In other words, they share a common denominator and decrease at the same rate. The proportion between the two stays constant.

gneill gave you the equation for specific energy of an orbit. When the specific energy is 0, you have an escape orbit.

Calculate the specific kinetic energy of for an object in the same orbit as the Earth (29.7 km/sec and a radius of 1.5 x 10^8 km). If your effective potential energy equals your kinetic energy, then you have an escape orbit.

In order for the potential energy to equal the kinetic energy, the effective heliocentric gravitational constant would have to be lower. In other words, something would have to counter the acceleration due to the Sun's gravity in order for the spacecraft 's kinetic energy to carry the spacecraft into a parabolic trajectory.

Calculate what the effective heliocentric gravitational constant would have to be in order to turn your orbit into an escape orbit. Calculate the acceleration a heliocentric gravitational constant of that magnitude would give you.

Your solar sail has to give enough acceleration to reduce the actual gravitational accleration due to the Sun to a low enough acceleration to match the imaginary effective heliocentric gravitational constant. It's probably a good idea to convert the acceleration to meters/sec^2 instead of leaving it in km/sec^2.

It's better to know the solar pressure in microNewtons per square meter (9.15 uN/m^2 at Earth radius for a perfectly reflective surface or, more realistically, about 4.57 uN/m^2(1+q), where q is the relectivity) than the Watts per square meter. Dividing your necessary acceleration by solar pressure gives you the necessary Area to Mass ratio.

Note: The 4.57 uN/m^2(1+q) is essentially the same equation used in Wertz's Space Mission Analysis and Design, except he uses 4.5 uN. I've noticed that there isn't much consistency in solar 'constants'. For example, he also uses 1353 W/m^2 instead of the number you used.

It's essentially the same solution gneill gave. Reduce the effective acceleration of the Sun's gravity.

And it does seem to require a very large area to mass ratio.

By the way, the IKAROS satellite isn't leaving the solar system (i.e it's not escaping the Sun's gravity). It's demonstrating that a combination solar sail/solar power generator can be used to both accelerate a spacecraft and to generate electricity that will power an ion engine. This is conflicting goals, since reflected light doesn't generate electric power, but it is true that solar panels tend to reflect a lot of light and convert a little light to heat and a little light to electric power. Yes, the solar sail was a success and it did accelerate the spacecraft .

The same approach could be used to design a solar sail for a trip to another planet further away from the Sun within the solar system, except the total specific energy would match whatever was necessary to establish your semi-major axis. This would obviously reduce your area to mass ratio (but probably not to a .667 ratio).

 

[will mick]

Cheers guys, I'm finally getting my head round it, I think I was trying to over complicate it by trying to find trajectories etc. I followed through these steps and got an answer of

$$\frac{A}{m}=\frac{1.355\times10^3}{1+q}$$

is this what you got?

If not, here are the steps I followed:

$$\varepsilon_k=\frac{v^2}{2}$$
and
$$\varepsilon_p=\frac{\mu}{r}$$
so
$$\varepsilon=\frac{v^2}{2}+\frac{\mu}{r}$$

The trajectory is an escape trajectory when $$\varepsilon=0$$
so when

$$\frac{v^2}{2}=-\frac{\mu}{r}$$

$$\frac{1}{2}(2.97\times10^4)^2=-\frac{\mu}{1.5\times10^{11}}$$

$$\mu=-6.62\times10^{19}$$

but

$$F=ma$$

$$\frac{\mu}{r^2}=a$$

$$a=\frac{\mu}{2.25\times10^{22}}$$

so using $$F=pA=ma$$,

$$\frac{A}{m}=\frac{a}{p}$$

$$\frac{A}{m}=\frac{2.03\times10^{-16}}{1+q}\mu$$

so substituting in $$\mu=6.62\times10^{19}$$ gives

$$\frac{A}{m}=\frac{1.355\times10^3}{1+q}$$

does that sound like good maths?

 

[gneill]

Cheers guys, I'm finally getting my head round it, I think I was trying to over complicate it by trying to find trajectories etc. I followed through these steps and got an answer of

$$\frac{A}{m}=\frac{1.355\times10^3}{1+q}$$

is this what you got?

If not, here are the steps I followed:

$$\varepsilon_k=\frac{v^2}{2}$$
and
$$\varepsilon_p=\frac{\mu}{r}$$
so
$$\varepsilon=\frac{v^2}{2}+\frac{\mu}{r}$$

Usually, the total specific energy is defined as:

$$\varepsilon=\frac{v^2}{2}-\frac{\mu}{r}$$

 

[will mick]

Thanks, my mistake I wrote it down wrong, I corrected it a bit further down when I plugged it into the $$\frac{A}{m}=\frac{a}{p}$$ equation

 

[will mick]

I have finally cracked it!
Thanks for all the help gneill and BobG

After some algebra I found that the ratio

$$\frac{A}{m}=\frac{4\pic}{P(1+\kappa)}(\frac{v^2r}{2}-GM)$$

where A=Surface area of sail, m=mass of sail, P=irradiance or power of the Sun (constant)=$$3.846\times10^{26}W$$, $$\kappa$$=reflectivity, v=speed of sail in orbit=$$2.978\times10^4$$, G=gravitational constant=$$6.67\times10^{-11}$$, M=mass of the sun=$$2\times10^{30}$$

which all gives an answer of $$\frac{A}{m}=\frac{657}{1+\kappa}$$

whereas if the sail were not starting from an orbit, v=0 and the ratio would then be $$\frac{A}{m}=\frac{1310}{1+\kappa}$$

I was trying to make it too complicated before by finding equations of trajectories etc..