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Radar-like Solar System

You know how all the planets around the Sun move at different speed and distances from the Sun? Would it have been possible, by luck or extension of how the solar system was made, that all the planets would always be at the same angular position? Would it be possible that all the planets would always be aligned perfectly?

 

In the previous equation, the velocity of an object is its angular velocity times the radius. If we want all the planets to be aligned, T (how long it takes to do a complete revolution) must then be constant.

r (the radius, or the distance between the Sun and the body) is the variable here, which gives us the body’s speed through space.

Is that all there is to it?

 

Kepler’s Third Law of Planetary Motion indicates that

meaning that the time taken for one orbital period, squared, is proportional to the distance from the sun, cubed.

For the Earth, (149,598,261,000m)^3 / (31,556,736s)^2 = 3.362E9. For Mars, (227,939,100,000m)^3 / (59,354,294.4s)^2 = 3.362E9. Same thing for the other planets.

And that means that, for a given orbital period (say, one julian year), a planet can only be at a specific distance from the Sun.

 

So, no, a planetary system can’t have all its planets constantly aligned. But that would be eerily cool!

Categories: Science

Gravitational Pull of Distant Objects

I had a bit of fun today trying to explain to a friend how a star millions of kilometers away could have a gravitational pull on us.

It’s pretty simple, and still relies on Newton’s Law of Universal Gravitation.

 

 

Gaia, Mother Earth

Our planet has a mass of 5.97E24 kilograms, and a mean radius of 6.37E6 meters. Let’s assume a human mass of 75 kilograms (165 lbs) for the equations on this page.

The Earth pulls a 75kg person with a force of 736.22 N. One Newton is equal to an acceleration of one meter per second per second, for a weight of one kilogram. If we divide the force by our mass of 75 kg, we get Earth’s gravity, 9.81 m/s^2.

 

Sirius, the Brightest Star in the Sky

It’s actually two stars, Sirius A and Sirius B. Since Sirius A is twice the mass of the Sun, and Sirius B is about the same mass as the Sun, they have a combined mass of 5.96E30 kilograms. The binary star system is about 8E16 meters away.

All this means that Sirius pulls you with a force of 4.66E-12 N. If all the universe was to disappear, except for Sirius, we would initially fall towards it at a rate of 6.21E-14 m/s^2.

 

Polaris, the North Star

The north star is six times heavier than our Sun, and is 433 light years away. With values of m = 1.19E31 kg and r = 4.10E18 m, we can calculate that Polaris is pulling you with a 3.54E-15 N force.

 

Proxima Centauri, our Closest Neighbor

With a mass of 2.45E29 kg and a distance of 4.01E16 meters away, Proxima Centauri pulls you with a force of 7.62E-13 N.

 

The Andromeda Galaxy

Andromeda is 2.40E22 meters away (it takes light 2.5 million years to reach it), and has an estimated mass of 2E42 kg (one trillion Suns). It pulls you with a force of, wait for it, 1.74E-11 N. It pulls you stronger than Sirius, which is a million times closer.

 

Last but not Least, the Moon

The moon has a mass of 7.3477E22 kg, and a mean distance of 3.84399E8 meters from the center of the Earth.

If the moon happens to be exactly overhead, it pulls on you from a distance of 3.84399E8 m – 6.371E6 m (the radius of the Earth) = 3.78E8 m, exerting a force of 2.57E-3 N.

If, on the other hand, the moon would be on the opposite side of the Earth, its distance from you would be 3.84399E8 m + 6.371E6 m = 3.91E8 m, exerting a force of 2.40E-3 N. This is actually a big difference of around 7%!

Even at its strongest, we would not consciously feel the moon. After all, Earth’s pull is almost 300,000 times stronger!

And before you ask, even on top of Mount Everest wouldn’t change anything.

Categories: Science Tags: ,

Multiplying Recurring Decimals

Let’s get to the heart of it right away, and pick a totally non-arbitrary number, 7, just because it’s my favorite one-digit number. The reciprocal (or multiplicative inverse) of 7 can be written in decimals as the following shows:

The bar on top of the 142857, also called a vinculum, marks these digits as being recurring (see repeating decimals), meaning that 

an that, to infinity (or for eternity, if you actually try to calculate it).

 

Do you see something strange about 142857? It starts with 14, then adds 28 (14×2), then it almost has 56 (14×4). But try to calculate the next number, and you get 112 (14×8). If you add the first few numbers, you get the following:

which is summed as

 

As you do more and more iterations, more digits are fixed and the result gets closer and closer to the expected value (it converges towards that value). For example, the next iteration (…3584) will fix the 39 at the end of the sum to the correct 42. And so on and so forth.

 

Are there any other numbers that have the same property? After all, this seems like a pretty weird coincidence.

Actually, all numbers’ reciprocals behave as such. Hn hnnn, it’s true! More to follow.

 

Are there any other sum of numbers that will yield the recurring 142857? To be honest, the previous addition seems like luck, or like there’s something alien about it.

Actually, it would seem alien to anyone who has not studied number theory, I guess. There are an infinity of such sums that give you a recurring 142857. I need to repeat that: an infinity of such sums exist.

 

To make the rest of the post easier to read, let’s define these sums. In the previous example, you begin with 14, and then multiply it repeatedly by 2, moving your position (precision) by two each time. We can “encrypt” that precision (position moving on each iteration) in the starting value: 14 means you move by two, and 014 means you move by three. I could then write the above sum as 14××2 (I couldn’t find a suitable, untaken character for this).

So, what other sums can give you 142857 for infinity? How about 142××6? Try it out! Then try 1428××4, then 14285××5, and 142857××1. Also, don’t forget to check 1××3, and 1428571××3. Right there, you can see that you can have an infinity of possible sums.

Although they’re based on the same numbers, it’s still interesting, right? No, it gets better. A lot. Try 13××9. Honestly. Pick a piece of paper and try it. Here, let me help you with it:

 

Granted, it converges slowly, but you can easily see that, given some more iterations, you get the expected value. (In fact, the farther you are from ××1, the slower it converges.)

Now, try 12××16, and 11××23. You can already see a trend, right? Down to 01××93.

Let me also add 2××-4, and 15××-5. Up to 28××-96. Then I could pick random ones, like 087××391, or 1431××-17.

Yes, all of them converges to a recurring 142857.

 

If you list them all, it’s pretty easy to find a general formula:

where k is the precision, v is the arbitrary integer number you want to start with, and 0 < v < 2×10^k/7. It’s that easy! For example, if you want a number with three digits of precision (k), you can take any initial value v that is between 1 and 285 (2×1000/7). v = 0 results in a division by zero, and other values make the series diverge. Actually, you can do the same with negative values of v, except that the result is negative.

Remember when I talked about numbers different than 7 having the same property? You probably noticed (or wondered) if the 7 in the general formula above can be replaced by i, which would be the source of the reciprocal. Let’s define the new general equation more formally:

 

And what’s incredible with that, is that every reciprocal can be calculated that way. You have your i value at the start (the reciprocal you want),  then arbitrarily choose any precision (k) and multiplier (v), and you have an equation that matches a set of recursive numbers to infinity.

Categories: Science, Thinking Tags:

Orbits of Different Mass Bodies

The scenario from last post (Orbits of Same Mass Bodies) was a very simple one. Lets kick it up a notch, and change one of the masses.

Two orbiting bodies of different mass.

 

The first thing to notice is that the center of mass is no longer in the middle of the system. This point is called the barycenter, and can be calculated with the following formula:

M is the total mass of the system, and m and r are the mass and position of the individual bodies. This gives the position in any number of dimension, from any origin.

For example, if you have a 1 kg mass orbiting 1 m away from a 2 kg mass, this formula tells us that the heaviest mass is 0.33 m from the barycenter, while the lightest is 0.67 m.

 

Since the two bodies have different masses, this changes the centripetal force a bit:

 

Unintuitively, the acceleration of a body in orbit depends on the mass of the other body, not its own. The Earth accelerates towards the moon at 3.33E-5 m/s², while the moon accelerates towards the Earth at 2.70E-3 m/s². About a hundred times faster.

 

For the velocity, the formula could be simplified to this:

 

Once again, it is worth noting that the velocity of a body doesn’t care about its own mass. The Earth’s velocity around its orbit is around 113 m/s, while the moon’s is about 1019 m/s.

 

With these two formulas, we can calculate how two bodies orbit each other, if the orbit is stable.

 

There is no solution for a three-body system, though I may present the problem in a future post…

Categories: Science, Thinking

Orbits of Equal Mass Bodies

When you have two objects, they influence each other through gravity. Take, for example, two marbles. If we assume a weight of 5 grams each, and a distance of 10 centimeters between the centers, we can calculate how much force they exert on each other.

where G is the gravitational constant, 6.67E-11 N m² / kg².

So the attraction between the two marbles is 1.7E-17 N (0.000000000000000017 N). That’s tiny compared to the force of friction that keeps the marbles on the floor. If the marbles are on a wood floor, the friction would be around 0.1 N, meaning the marbles would need almost 6 quadrillion times more force than their mutual attraction. To give some point of reference, the attraction between the marble and the entire Earth is 0.5 N. Between you and the entire Earth? 740 N. How about the earth and moon? 2E26 N.

 

To better understand orbits, let’s imagine the simplest system. There are two bodies of equal mass in a stable orbit around each other, in a universe that contains absolutely nothing else.

Bodies of equal mass in orbit.

What forces are in this system? The easiest one to guess would be gravity, towards the other body, equal to the formula at the beginning of the post. Since the two objects are of equal mass, we can simplify the equation to

We can also calculate how fast they move towards each other, using Netwon’s second law of motion,

Solving for the acceleration, we find that

There is not only gravity, because the two objects would just go straight for each other and collide. To be a stable orbit, there should be a speed, perpendicular to the acceleration, and can find it from the centripetal force formula:

In our case, the gravitational attraction is the centripetal force, and, solving for the velocity, we find that

Forces acting on one of the bodies.

 

If there are two marbles in a stable orbit around each other in an otherwise empty universe, their mutual attraction is still 1.7E-17 N, and, assuming they would be one meter away from each other, they would accelerate towards the other marble at 3.3E-16 m/s². Calculating their velocity in the orbit results in a speed of 5.8E-7 m/s. We know that a circle with a diameter of one meter has a circumference of 2Π meters (6.3 m), so our marbles would do one complete revolution in roughly 4 months.

Right now, if we look at the formulas to calculate the acceleration and the velocity, we know that the distance and mass are the only things that can have an effect on the system.

In the next post, we will calculate what happens if the two bodies are not the same mass.

Categories: Science, Thinking