# Travelling to the moon.

1. Nov 23, 2004

### IcedB

A solar sail is orbitting the earth. It needs to get to the moon by increasing its orbit energy through solar propulsion. How many orbits will it need to make and how long will it this take? Considering it starts in a Geostationary Transfer Orbit and has an acceleration of 0.5 mm/sec2.

2. Nov 23, 2004

### tony873004

That's a tough one. The sail is not always going to be oriented for prograde propulsion. And certain times of the year it will cross the nodes and be subject to a blackout once per orbit as it passes through the Earth's shadow. Is this a homework question where you can ignore stuff like that?

3. Nov 23, 2004

### IcedB

I could certainly just say that I haven't taken black outs into account.

4. Dec 5, 2004

### pervect

Staff Emeritus
Interesting problem, I'm still working on the details, but I've got a rough outline of how to proceed

If you define the problem so that the satellite is always accelerating in the direction of its orbit, and you make the additional assumption that the orbit is always nearly circular, and if you ignore the gravity of the moon, I think you can solve it. The assumption that the orbit is always nearly circular seems reasonable at least for a first pass, as the accleration of the moon towards the eart is about 2.7 mm/s^2, about 5x as great as the radial acceleration, so the departure of the acceleration vector from radial won't be severe even at the moon's orbit. this should probably be checked more closely.

First you write down the expresion for the total energy E = .5*m*v^2 - GmM/r, with the provisio that the orbit is circular so that GmM/r^2 = mv^2/r

This gives you an expression for v(r), the orbital velocity as a function of r, and the expression for E(r), the total energy as a function of r. The latter is -.5*GM*m/r

Now, the work done in one orbit will be force*distance

dWork = acc * 2*pi*r(t)

and the time it will take is

2*pi*sqrt(r^3/GM) (keplers law)

Now we need to get clever about the choice of variables. The simplest thing to do is to write energy as a function of time, E(t), and to compute the value of r from the value of E

Because E = -GM/2r, we can say that r = -GM/2E (E is of course negative)

Then in one orbit

dE = acc*2*pi*r = acc*2*pi*(-GM/(2*E)) = - acc * pi * GM / E

which takes a time

t = pi * GM * sqrt(-1/(2*E^3)

dE/dt = acc*sqrt(-2*E), if I did all the math right (no guarantees!)

I suppose I ought to wrap this up nice and neatly, and work out a number, and go look for some math errors - but I'm getting hungry - maybe later.

5. Dec 5, 2004

### tony873004

I don't know how to do the math, but I can simulate it.

This assumes a constant acceleration in the prograde direction of 0.5 mm/s^2.

It will spiral out, slowly at first. It will complete 17 orbits before a gentle tug from the Moon adds additional velocity. Still accelerating at 0.5 mm/s^2, on the 48th day of its journey from geosynchronous orbit (42,170 km) it will cross the Moon's orbit after 17.5 orbits. It will not complete its 18th orbit before escaping the Earth system and heading into interplanetary space.

http://orbitsimulator.com/gravity/sd/sp1.GIF
http://orbitsimulator.com/gravity/sd/sp2.GIF
http://orbitsimulator.com/gravity/sd/sp3.GIF

*** edit ***
Trying it again, but setting the Moon's mass to 0 gives very similar results.
http://orbitsimulator.com/gravity/sd/sp4.GIF

Last edited: Dec 5, 2004
6. Dec 5, 2004

### pervect

Staff Emeritus
I'm getting 17.6 orbits, and 5.8 million seconds (67 days), ignoring the moon's gravity, for it to reach the moon's orbit - assuming the orbit stays circular, with a .5mm/s^2 constant acceleration in the direction of orbit, and ignoring the moon.

It looks like this is in the same ballpark as the simulated answer, which is reassuring.

If the moon's gravity were actually included, these assumptions would fail sometime before the ship reaches the Earth-moon L1 point.

The final equations I got were

E(n), energy/unit mass as a function of number of orbits

dE(n)/dn = -GM*pi*acc/E

giving E*dE = -GM*pi*acc * dn, which can be integrated to solve for n

E(t), energy/unit mass as a function of time

dE(t)/dt = sqrt(2E)*acc

Giving dE/sqrt(2E) = acc * dt, which can be integrated to solve for t

GM = .3968e6 km^3 / sec^2, from

Eini = -4.727 Efin = -.519 (that's in megajoules/kg, compatible with distance in km, seconds)

corresponding to geosynch orbit at 42164 km, and lunar orbiat at .384e6 km

7. Dec 6, 2004

### tony873004

Maybe it's the assumption that its orbit stays circular that accounts for the difference. It's a spiral, so it never will be perfectly circular, although it is close for the first 14 orbits. But the 15th - 17th orbits are far from circular.

Looking at the diagram (picture 4 with moon mass = 0), it's a little past the 17.5 mark (it started at the 9:00 position when the simulation date was 0/0/0 0:0:0), so I'd say 17.6 is right on the money.

And it crossed the Moon's orbit on day 49, not 48 like I said earlier, so this closes the gap between the two methods a little.

8. Dec 6, 2004

### pervect

Staff Emeritus
If the first answer matches, the second should too. <checking>

Yes, the equation I had was right, but the final calculation was off

dE/sqrt(-2*E) = acc * dt

Integrating this out gives t = (sqrt(-2*Eini) - sqrt(-2*Efin) ) / acc

Thats 4.11 megaseconds, 47.5 days, close enough for me.

9. Dec 6, 2004

### BobG

There's one big problem with solar propulsion. If the orbit is aligned with the ecliptic plane, you won't get constant acceleration. Half of your orbit, you're moving away from the Sun and the acceleration is positive. Half of your orbit, you're head toward the Sun and making your orbit smaller. You can minimize that by changing the orientation of the solar sail so eliminate or reduce the effect of solar pressure while headed towards the Sun, but it won't be constant acceleration.

If you orient the orbit perpendicular to the ecliptic plane so you're never headed towards the Sun, you might do better. At least some of the pressure is going to change the plane of the orbit instead of adding to the size of the orbit, but that's a good thing. If you want to stay perpendicular to the ecliptic plane, you want the orbit's plane to rotate in synch with the Earth's orbit.

Unfortunately, the Moon's orbit is separated from the ecliptic plane by only about 5 degrees and that's the orbit you need to match. Since the energy is external to the spacecraft and is only coming from one direction, your orbit will become more elliptical instead of remaining circular. As long as your increasing eccentricity doesn't decrease perigee faster than you're increasing the semi-major axis, that's a good thing. A small change in the semi-major axis will yield a bigger change in how close to the Moon you get.

Technically, you also need to keep track of how far the satellite is from the Sun. In geo orbits, the difference in solar pressure on the side of the orbit nearest the Sun and the solar pressure on the far side of the orbit is so small you can treat them the same. By time you reach the moon the difference between the near side and far side (relative to the Sun) is getting up to almost 1/2 a percent. Small enough if you only need a rough estimate.

10. Dec 6, 2004

### tony873004

Yours may be exact.
My simulation is the sum of about 1,000,000 approximations :yuck: (time step of 4 seconds for ~48 days),
and timing is usually the first thing that deviates from reality.