# Orbiting magnetic balloons

I recently got the following question:

Let’s say I have a 500kg balloon floating in the stratosphere at fixed altitude with solar cells collecting 10kw from the sun, then my computation shows that if this energy can be converted to horizontal magnetic propulsion by repelling against the earth’s magnetic field at 100% efficiency then it could reach escape velocity in about one month. This is possible because at this altitude the air resistance is quite small so it is almost like pushing at an air hockey which does not require much force to get it to speed up horizontally. … My question for you here is that in reality how close to practicality is the design of this ‘spacecraft?’

Launch costs are one of the big drivers in the space industry, and the propellant required to get a spacecraft up to orbital speed is a major part of that cost. If we could use some sort of “propellant-less” means to get a vehicle into orbit, we could revolutionize the whole space industry. In fact, this is an idea that my grad school research group once brainstormed about during a lab meeting: push on the Earth’s magnetic field. If we start pushing from high altitude, where air resistance is small, then we just have to wait long enough to accelerate our spacecraft up to at least low Earth orbit speed (about 7 km/s). Launches might take a long time, but they would be far cheaper and easier.

As long as we can push on the spacecraft with a net force in the direction of its velocity, then it will accelerate. So, the first question we come to is this: how much drag force do we need to overcome? That force will provide us with an estimate of the minimum force our electromagnetic device needs to produce.

Air resistance causes a force in the opposite direction to an object’s velocity. For a sphere moving through the air, this force has a magnitude equal to 1.1 d A v2, where d is the air density, A is the cross-sectional area of the sphere (pi r2), and v is the object’s velocity. Let’s suppose we mount our 500 kg spacecraft on a high-altitude balloon that can get all the way up to 30 km altitude before we engage the magnetic propulsion device. At that altitude, the atmospheric density is in the ballpark of 0.02 kg/m3. (I’m reading off of the 1962 US Standard Atmosphere graph on Wikipedia, since I can’t look at NASA’s web resources. Thanks, Tea Party!) Now we have d.

Next question: how big is the balloon? Way back in Ancient Greece, when Aristotle had the original “eureka!” moment, he realized that objects float in a fluid when they displace a weight of fluid equal to their own weight. (Equivalently, they displace a mass of fluid equal to their own mass.) So, our 500 kg balloon-based vehicle has to displace 500 kg of air – and if it’s floating at a level where the air pressure is 0.02 kg/m3, then that means the balloon takes up a volume of at least 25,000 m3. That’s a sphere 36.3 m in diameter. (Note that here I’m assuming that the mass of the vehicle includes the mass of the balloon and of the gas we pumped in to inflate the balloon. What finally gets to orbit will be less than 500 kg.) So: A is about 1035 m2.

Now we have an estimate for the drag force magnitude on our electromagnetic launch vehicle at 30 km altitude, of about 22.77 v2. If we start our electromagnetic devices pushing, the spacecraft will start to move – but it will eventually settle on a steady-state speed at which the drag force and propulsive force balance each other. Here’s the bad news: even though the atmosphere is not very dense 30 km up, that v2 in the drag equation will really get us as we reach higher and higher speeds. If the balloon gets going at 1 m/s, the drag force will be 22.77 N. If we reach 10 m/s (about normal human sprinting speed), the drag force is 2,277 N. If we tried to accelerate the balloon all the way up to 7 km/s at this altitude, putting the vehicle in orbit, then the drag force will get to over one billion newtons! It’s not feasible to build a compact device that could push on the Earth’s magnetic field and generate this kind of force.

You might get the idea that as we accelerate, we can also gradually increase the balloon’s altitude. After all, if the air gets less dense, that drag force will decrease. With less resistance opposing our spacecraft, we don’t have to work as hard to accelerate it.

There are two problems we’ll run into if we follow this idea. First, while going up in altitude makes our spacecraft encounter less atmospheric density, it also has a weaker magnetic field to work with. At these high altitudes, atmospheric density is very much like an exponential decay. But the magnetic field from a dipole (like the Earth’s) falls off with distance from the dipole as 1/r3. How do the two functions compare? This is good news. While at first, the magnetic field is lower than the density, eventually we come to a point where the magnetic force will be stronger than the drag force for fixed velocity. (This makes sense, because some spacecraft use magnetic forces to orient themselves when they are well above the levels of appreciable atmospheric density.) Suddenly, this idea doesn’t seem so crazy.

The second, problem, though, is tougher. Remember buoyancy? Once we get up to about 34 km altitude, according to that graph, the air will be about half as dense – which means our balloon will need to take up twice as much volume in order to stay afloat. The higher up our spacecraft goes, the bigger than balloon has to be. Eventually our balloon is going to need to be kilometers in diameter, since we won’t yet be up to orbital speed and gravity will just pull the spacecraft down unless we keep our spacecraft buoyant. (This is why high-altitude balloons always eventually pop!)

Because our vehicle has to solve both problems simultaneously – staying afloat and accelerating – I don’t think it’s feasible to get a large satellite into orbit this way.

However, if we move to a size scale where some of the physics behave differently – say, if we make our spacecraft very small – then perhaps we won’t run into this problem with the balloon. A few years ago, one of the researchers in my old lab took a look at some of these very questions of drag and magnetic forces on tiny spacecraft, though not with the goal of launch in mind. But one could, theoretically, make tiny spacecraft capable of accelerating to high speed by interacting with a planetary magnetic field. One could also, theoretically, make spacecraft tiny enough to flutter down through an atmosphere unharmed. Combining and reversing these ideas would be an interesting long-term research challenge!

## 9 thoughts on “Orbiting magnetic balloons”

1. Walabio says:

I found an error:

?r^2

should be

(?/2)r^2

¡? drools! ¡? rules!

http://TauDay.Com

The radius defines the circle. A long time ago, people thoughtlessly measured across and around a circle and called that the circle-constant, but that leads to continuously multiplying by 2, which leads to errors (forgetting to multiply, dividing when one should divide, dividing when one should multiply, forgetting to divide, et cetera).

Many ?ists, rather than admitting their mistake, point to the area of a circle with glee. On closer examination, this is the nail in the coffin:

In calculus, one can derive the area of a circle in 2 ways:

* The sum of the area of an infinite number of triangles cutting the circle like a pizza.
* Turning the circle into an onion with a infinite number of layers, peeling the layers, ponding them flat, and staking the layers into a triangle.

The area of a triangle is

(hb)/2

Height times base over 2.

In terms of ? the area of a circle is truly:

(2?/2)r^2

The 2s cancel, leaving ?:

?r^2

¡2?=?!

¡? was their all along, but one could not see it because the 2s cancelled!

¡Quod Erat Demonstrandum!

Post Scriptum:

A blogPost about Tauism (not to be mistaken for Taoism) would be interesting article.

2. Walabio says:

¡Your blog does not speak Unicode!

The ¿Question-Marks? are either Tau or Pi.

3. Joseph says:

I’m not sure I understand. The area of a circle is pi*radius^2; that is the formula I wrote and used in my calculations.

In any case, a factor of 2 doesn’t really impact the back-of-the-envelope nature of the scratch calculations I did here.

4. Walabio says:

Joseph says:
11 October 2013 at 23:09

> “I’m not sure I understand.”

I shall rewrite in ASCII:

I found an error:

pir^2

should be

(tau/2)r^2

¡Pi drools! ¡Tau rules!

http://TauDay.Com

The radius —— ¡not the diameter! —— defines the circle. A long time ago, people thoughtlessly measured across and around a circle and called that the circle-constant, but that leads to continuously multiplying by 2, which leads to errors (forgetting to multiply, dividing when one should divide, dividing when one should multiply, forgetting to divide, et cetera).

Many piists, rather than admitting their mistake, point to the area of a circle with glee. On closer examination, this is the nail in the coffin:

In calculus, one can derive the area of a circle in 2 ways:

* The sum of the area of an infinite number of triangles cutting the circle like a pizza.
* Turning the circle into an onion with a infinite number of layers, peeling the layers, ponding them flat, and stacking the layers into a triangle.

The area of a triangle is

(hb)/2=area

Height times base over 2.

In terms of pi the area of a circle is truly:

(2pi/2)r^2

The 2s cancel, leaving ?:

pir^2

¡2pi=tau!

¡Tau was their all along, but one could not see it because the 2s cancelled!

¡Quod Erat Demonstrandum!

Post Scriptum:

A blogPost about Tauism (not to be mistaken for Taoism) would be interesting article.

1. Joseph says:

Okay, you can use whatever constant you like instead of pi, just so long as all the geometric expressions come out mathematically equivalent. But there is no physical or mathematical “error,” as you put it, in the statement that the area of a circle with radius r is pi*r^2. If the 2s in your derivation cancel, they cancel, and neither of them is there any more.

5. Walabio says:

Well, it does matter:

One defines a circle as all points equidistant from a point on an Euclidian Plane. The mistake of defining the Circle-Constant as circumference rover diameter (c/d) is an easily understood mistake because one might measure across a circle, but this leads to pædological problems and 2pi showing up everywhere, leading to errors in forgetting to multiply or divide by 2 or going the wrong way. The fact is, we should use tau (circumference over radius (c/r)):

Let us look at trigonometry:

pi/3= 1/6 of a circle. This leads to errors and is difficult to learn. in terms of tau, it is:

tau/6 =1/6 of a circle. This is less error-prone and easier to learn.

I can use this to make my point:

e^ipi=-1

Pi only takes us halfway around the circle.

e^itau=1

Tau takes us all of the way around the circle.

Evidently, pi is only half the circle-constant, but tau is the circle-constant.

¡Quod Erat Demonstrandum!

1. Joseph says:

QED? You haven’t proved anything…you’ve proposed an alternative convention. There is still no error in my statement.

The most important thing is for everyone to agree on the convention. Some conventions are more elegant in some situations than others, certainly. (For instance, in many equations in physics, pi comes with all sorts of numeric factors in front of it. Your choice of making 2pi the fundamental constant makes some of those less elegant!) But there is substantial value in getting everyone to speak the same language. That’s the value of pi: you can do the math without having to launch into this argument every time!

If you’re going to pick on any convention in physics or math, this is a rather pedantic one. Might I suggest the choice to associate negative charge with electrons as a convention with greater impact on understanding?

6. Walabio says:

Well, we could define the circle-constant as 23 and three quarters, but then, we would have irrational, indeed transcendental factors in every equation involving circles. If we use a circle-constant, we might as well chose the best circle-constant. In terms of tau, all of the the math is easier to learn except the circle-area because of a hidden 1/2. When students go on to the third dimension, tau shows its strength:

If area of a circle is pir^2 then volume of a sphere is pir^3. Only it is 4/3pir^3. This throws off many students. In terms of tau, it is 2/3taur^3 One simply must remember to add 1 to both the numerator and denominator and the power:

1/2taur^2 plus 1 to everything in the equation equals 2/3tau^3.

¡Easy Peasy!

7. Ben says:

A launch of thousands of chipSat-balloons would look even sweeter than those floating-lantern festivals.

And Joe, don’t feed the trolls.

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