Extreme Geekery: Jupiter Jump

JupiterDid you hear the amazing news?  If you jump on the right day, a planetary alignment of Jupiter and Pluto will cause a decrease in gravity and you will be able to float in the air for five minutes!  It’s definitely true because I read it on the Internet.

Ok, you’ve got me.  This isn’t true.  Not in the least.  It’s 100% grade-A hokum.  Phil Plait, aka The Bad Astronomer, has debunked this more thoroughly than I could.  It did make me wonder though:  What would it take to float in the air for five minutes?  How much would we have to alter the Earth’s gravity in order to accomplish this feat?  Or, likely more accurate to the “Zero G Day” myth, how close or massive would Jupiter need to be to accomplish this?  Consider this a cosmic equivalent to the Mythbusters’ replicating the results.

We can think of a jump using this diagram.


First, our legs push off of the Earth with an initial velocity.  Next, the Earth’s gravity decelerates us bit by bit until we reach the top of our jump where our speed becomes zero.  Finally, gravity’s acceleration pulls us back down to Earth.  (In reality, you might need to calculate other items like air resistance, but we’ll simplify matters and ignore the air for now.)

Since the top of our jump is halfway through it, let’s further simplify matters by not calling this a five minute jump that we are going to attempt.  Instead, we’ll say that it takes gravity two and a half minutes to slow our initial push-off speed to zero.

Let’s start with a normal jump, though. Go ahead and jump. If you are like me, you went up and came back down in about a third of a second.  There’s an easy equation we can use to calculate acceleration (or deceleration) over time:

vf = vi + a*t

In other words, the final speed is equal to the initial speed plus the product of acceleration (in this case, gravity) and time.

We don’t know the initial speed, but we do know the time (half of a third of a second, or about 0.167 seconds), the deceleration (-9.8 meters per second2), and the final speed (zero).  This gives us:

0 = vi – 9.8*(0.167)


vi = 1.617

That’s the initial velocity that our legs have provided us.  But what about our “Zero G Day” jump?  How small would gravity need to be to allow us to push off with the same force and decelerate to zero in 2.5 minutes?


vf = vi + a*t

This time, we know vf (zero again), vi (1.617 from the previous example), and t (2.5 minutes, which is 150 seconds).  This gives us:

0 = 1.617 + 150a

or an acceleration of -0.01078 meters per second2.  Given that normal Earth gravity is 9.8 meters per second2, this would be a mere 0.0011g.

How small is 0.0011g?  Well, according to the Planetary Fact Sheet from NASA, if you were walking on the surface of the Mercury, you’d experience 0.378g.  Maybe we need something smaller like our Moon?  0.166g.  Let’s go to dwarf planets and look at Pluto.  0.059g.  We still haven’t gotten to our “Zero G Day” gravity figure.  Even Pluto has over 53 times the gravity we’re looking for.  We might have luck with one of the smaller asteroids, but it would be a difficult search.

Clearly, were Zero G Day to actually occur, this would be a major disruption.  The effects wouldn’t simply be limited to a cool jump.  But what could cause such a thing?  As Phil Plait pointed out, Jupiter – as massive as it is – doesn’t influence us enough gravitationally.  This is because gravity decreases as objects get more distant.  Specifically:


G is the universal gravitational constant or 6.673 x 10-11 N m2/kg2.  This isn’t to be confused with “little g” which stands for the gravity on Earth.  Don’t worry if you are a bit confused, though, as you’ll see, G isn’t going to matter much.

When you jump in our hypothetical “Jupiter Has Much More Gravity” situation, Jupiter pulling up almost as much as the Earth is pulling down.


In fact, we could say that the force of Jupiter pulling you up is equal to 0.9989 times the force of Earth pulling you down.  (Thus giving us the 0.0011 Earth gravity left over.)

Since we have the force equation from before, we can express this as:


Since G and m1 are both on each side, we can eliminate those:


Plugging in the values for the Earth’s mass (5.98 x1024kg) and distance (6.38 x 106 km) gives us:




Bringing Jupiter’s distance to the other side gives us:


Now we can either solve for distance or mass for Hypothetical Jupiter.  Let’s do one and then the other.  First, we’ll assume that Hypothetical Jupiter is in its current orbit and will calculate how massive it would need to be.  Next, we’ll keep Jupiter as massive as it is right now and just move it as close to the Earth as it would need to be.

Plugging in Jupiter’s minimum distance from the Earth – 588.5 * 106 (source: NASA), we get a mass of 509,108.4 * 1029 kg.  That same website gives the real Jupiter’s mass as being 1,898.3 * 1024 kg.  So Hypothetical Jupiter would need to be 26.8 million times as massive as Real Jupiter.  This is much more massive than the Sun is (1,989,100 * 1024 kg).  Almost 25,600 times as massive, in fact.  This would make our Hypothetical Jupiter, more massive than any star we know of.  Perhaps this would even be massive enough for Hypothetical Jupiter to collapse into a black hole.

On the other hand, we could keep Jupiter the same size and just move it closer to the Earth.  Solving for distance, we get this equation:


With Jupiter’s mass (1,898.3 * 1024), this gives us a distance of a distance of about 113,600 km.  At first glance, this would appear to put Jupiter less than a third of the distance between the Earth and the Moon (384,400 km).  Actually, though, it’s closer than that.  You see, the distance is measured as the distance to the center of Jupiter.  Jupiter has a radius of about 70,000 km so this would actually make Jupiter only 43,600 km from the Earth.  This would be only 11% of the Earth-Moon distance and would be a mere 8,000 km from satellites orbiting in geostationary orbit.

At that distance, our Hypothetical Jupiter would not only let us stay aloft for 5 minutes, but would also pour tons of lethal radiation over the Earth.   Not only that, but the Earth would probably be torn apart and fall into Jupiter.

In short, be glad that you can’t jump and stay aloft for five minutes.  If you ever find yourself able to do this, you might not be around for long to enjoy your air time.

NOTE: The Jupiter image above is from NASA.

Extreme Geekery: Printing A Hard Drive

Hard_Disk_SmallIt’s no secret that I’m a geek.  It’s not just limited to watching shows like Doctor Who and movies like Star Wars.  It’s not just limited to memorizing random facts and having an interest in science.  I also love figuring things out sometimes.  Things that are too hypothetical to be practical.

For awhile, I’ve been  in geeky-love with Randall Munroe’s What If series.  In it, tackles some weird questions like What would happen if you put a drain at the bottom of the ocean? and How many people would a T-Rex let loose in New York City need to eat every day?  Randall delves deep into scientific facts and theories to back up his answers.  Sometimes he’ll use complicated equations used to calculate rocket trajectories and sometimes he’ll chart human growth rates.  Randall’s series has been so popular that he’s putting out a book.  (For disclosure purposes, I wasn’t asked by anyone to plug Randall’s book.  However, if I *was* asked to review it, I’d jump at the opportunity!)

I’ve been inspired by Randall.  For awhile, I’ve wanted to delve into some more geeky topics, but I didn’t want to scare everyone away by going full on mega-geek.  So consider this an experimental, possibly semi-regular series of blog posts.  Translation: As I think of them I’ll blog them but they won’t by any means overshadow normal TechyDad.com postings.

WARNING: Extreme Geekery Ahead!

The first question I’ll tackle is:  If you printed the contents of a full 1TB hard drive, how big would the stack of papers be?

To figure this out, we first need to set a few ground rules.  We could print the actual 1’s and 0’s on the hard drive or hexadecimal (base 16) representation of that data.  Hexadecimal would be shorter.  For example, 11111001 in binary is F9 in hexadecimal.  Also, we’d need to decide on a font size for printing.  Obviously, choosing a huge font size would mean less data per page (and, thus, more pages) than a tiny font size.

We’ll print in binary (1’s and 0’s) since that’s more literally what’s on a hard drive.  Also, normal font size is around 10 point so we’ll go with that.

Next, we need to figure out how much data would fit on a sheet of paper.  By opening up OpenOffice.org, I was able to type in 83 characters per line and 49 lines per page using 10 point Times New Roman font.  This means 4,067 1’s and 0’s would fit on a page.  This leads to the next question:  How many 1’s and 0’s are in 1TB?

1TB, or one terabyte, is 1,000,000,000,000 bytes.  Each byte, in turn, is 8 bits.  Bits are those 1’s and 0’s.  This means that a one terabyte hard drive can store 8,000,000,000,000 1’s and 0’s.  (Theoretically speaking, that is.  Practically speaking, there are issues that keep you from getting all of those 1’s and 0’s on your 1TB hard drive.  We’ll ignore those issues for the purposes of this discussion, however.)

With 8,000,000,000,000 bits and 4,067 bits per page, we wind up with 1,967,051,881 pages.  That’s quite a lot of pages, but how tall would the stack be?

Some Google searching turned up that normal printer paper is 0.1 millimeters thick.  This means that the stack would be 196,705,188.1 millimeters tall.  Of course, 1,000 millimeters is one meter and one thousand meters equals one kilometer so this means the stack is actually, 196.7051881 km tall.  For those who don’t use the metric system often, this translates to 122.227 miles.  (Thank you, Google!)

Now we know how tall our stack of paper is, but what about a frame of reference?  We can imagine driving 122 miles, but that’s horizontally.  I doubt anyone drives vertically upward.  According to the earlier-mentioned Randall Munroe, 100km is the official edge of space.  So wherever this stack takes us will be in space.  Low Earth Orbit begins at 160km so some objects orbiting the Earth might hit into our stack.  Thankfully, Wikipedia says that the International Space Station orbits “between 330 km (205 mi) and 435 km (270 mi)”.  That’s much higher than our stack, so at least we don’t need to worry about the ISS crashing into our stack of papers.  (At least, not until we print out the 2TB hard drive.)

So now we know how big of a stack of papers would result from printing out the contents of a 1TB hard drive.  Aren’t you glad we can fit all of those bits into a relatively tiny casing instead of having to lug around a stack of papers reaching to space?

NOTE: The hard drive image above is by ricardomaia and is available from OpenClipArt.org.

Happy Pi Day

piIt’s no secret that I’m a math geek.  Back in school, I used to solve extremely complex math problems for fun.  I still do this from time to time.  For example, I recently felt compelled to figure out just how tall 20GB of data would be if all the 1’s and 0’s were printed out.  (It would stretch from the Earth to Mars at their closest.)

So it should be no surprise that I love celebrating PI day.

For those of you asking what Pi day is, one of the more interesting numbers in math is called Pi.  It’s an irrational number that is the ratio of a circle to its diameter.  Pi turns out to be very useful for many, many calculations.

Since it is irrational, Pi is infinitely long.  Most people remember Pi as being 3.14, but it actually is 3.1415926535….  People have calculated Pi to billions of digits.  (You can find this online, but it takes a LONG time to download!)  Of course, since March 14th is 3/14 (in the US, at least), we call today Pi day.  (Given that next year will be 2015, next year’s Pi day will be on 3/14/15.  At 9:26 and 53 seconds, the date and time will match Pi to 9 digits.)

To celebrate Pi day, you could do some math, look up some digits of Pi, see if certain word strings are in Pi, or just eat some pie.

How would you celebrate pi day?  With math or pie?  Or both?

Note: The pi image above is "Pi" by bnielsen.  It is available on OpenClipArt.org.

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