# 19 Gigajoules; What’s a Gigajoule?

July 4, 2005

Technology

19 gigajoules was the amount of energy which resulted from the terminal impact of the Deep Impact probe. But how many of us remember what a gigajoule is? Hearing this number reminded me of Doc Brown saying that the DeLorean needed 1.21 gigawatts of power in order to initiate time travel. As we all laughed at Doc Brown and his horrid pronunciation of gigawatt (he said “jig-o-watt”), most of us wanted to know just how much power that really was.

Well, I want to know just how much force was imparted upon Tempel 1 this morning. From computer science (or from studying ancient Greek), we know that giga is the term for a billion. So a gigajoule is a billion joules. And a joule is the work done to exert a force of one newton for a distance of one meter. OK. Now that we have that in our minds, let’s look for some common point of reference. According to the Online Unit Converter,

1 gigajoule = 737562149.3 foot pounds, or
1 gigajoule = 372.506136111 horsepower-hour

So the NASA impactor exerted the same amount of force as over 6700 horses could exert over the course of an hour. The pre-impact estimate was that this impact would result in a crater as big as a professional football stadium.

So why do we fling objects into space and aim them at unsuspecting comets? Well, we want to know whether these things are rocks or just dirty snowballs. We will learn that as the data is analyzed. But until thne, we have learned something very important. Tempel 1 is not shaped like a peanut. It’s more like a banana. 😉

-CyclingRoo- ### 5 Comments on “19 Gigajoules; What’s a Gigajoule?”

1. Dave Says:

You could also try and quantify it in terms of moving objects out of our atmosphere.

http://en.wikipedia.org/wiki/Space_shuttle#Technical_data

Says that the mass of the shuttle at liftoff is 2,040,000 kg. A newton of force is the force that is required to accelerate one kilogram by 1 meter per second per second, so kgm/s^2. A joule is applying that same force over a distance of 1 meter. kgm^2/s^2.

So to move the shuttle up out of our atmosphere, we need to counteract the force of gravity + accelerate it to whatever. The final mass is 1.04 * 10^5, so let’s assume that that’s the mass while in orbit as well (I don’t know how much thrust it requires to get back into Earth’s gravitational field enough to come back to Earth).

If we define the shuttle’s mass as a linear function of distance from Earth, and say that orbit is approximately 200 kilometers, then m(y) = (1.04*10^5kg – 2.04 * 10^6kg)/(200000m) * y + 2.04 * 10^6 kg. So, logically, we can setup a line integral from y = 0m to y = 200000m.

Of course, the force of gravity changes as well, so we need to find g(y) to describe the acceleration of gravity as a function of y.

g(y) = (5.9742 × 10^24 kg*6.67 × 10^−11)/y^2

So if the radius of the earth is 6 378100m, then our function will look like:

g(y) = (5.9742 × 10^24 kg*6.67 × 10^−11)/(6378100m+y)^2

where y is the distance from the Earth’s surface.

Int[((1.04*10^5kg – 2.04 * 10^6kg)/(200000m) * y + 2.04 * 10^6)*(5.9742 × 10^24 kg*6.67 × 10^−11)/(6378100m+y)^2*y, {y=0…200000}]

Should give us our answer ;P 1.41*10^17 Joules of potential energy.

Apparently the shuttle travels at 7.7km/s, so (7700m/s)*1.04*10^5kg*1/2 = 3.08*10^12 J.

So in comparison, 19*10^9J doesn’t put a dent in launching the space shuttle, however assuming the space shuttle were already ‘outside’ of the grip from the gravitational field (at least to the point where its effects are negligible), we could accelerate it to about 604m/s.

Phew. I had hoped that it would be comparable to the space shuttle when I started writing this, but I guess the space shuttle is just too badass.

2. David Says:

Oof! Actually, there’s a factor of x that should not have been in there – after differentiation, it turns to 1.

The integral should be:

Int[((1.04*10^5kg – 2.04 * 10^6kg)/(200000m) * y + 2.04 * 10^6)*(5.9742 × 10^24 kg*6.67 × 10^−11)/(6378100m+y)^2, {y=0…200000}]

which yields a MUCH smaller number 🙂 of 2.05*10^12 Joules.

3. David Says:

Nasa’s JPL says the mass of Deep Impact is about 370kg.

http://deepimpact.jpl.nasa.gov/science/cratering.html

It’s also travelling at about 23,000mph.

23,000*1609/3600 = 10279.7m/s

(1/2)*370*(10279.7)^2 = 1.95*10^9 Joules, which is where the 19 gigajoules comes from 🙂

4. CyclingRoo Says:

Dave,

Thanks for the in-depth analysis. You win today’s “uber-geek” prize. And what do we have for the winner, Carol?

BTW, congrats on you AP scores. What are your plans for collegiate study? Psycology or Chemistry?

5. Dave Says:

I’m planning on studying Physics and Mathematics and possibly Biochemistry. 🙂 Ideally, I’d like to do research (government or private industry).

NASA’s best estimate for the mass of a comet is anywhere from 0.1 – 2.5 x 10^14 kg

If we use their most conservative estimate, and our result of about 19.5 * 10^9 J:

1.95*10^10 J = (1/2)(.1*10^14kg)v^2

v = 0.0004m/s

Of course, I’m not taking into account the energy lost to deformation of the comet, the kinetic energy of the dust ejected from the crater, or anything fancy like that (mostly because it’s almost incalculable, especially from 83 million miles away haha)

‘Q: Will the Impactor knock Tempel 1 out of its orbit, change its orbit or cause any debris to hit the earth?

No, the Impactor will not knock Tempel 1 out of it’s orbit. However, the impact will cause a slight change in the comet’s orbit, but that change is so small that it won’t be noticed. A good analogy is to imagine a small pebble hitting an 18-wheeler truck.

Over time, the changes we introduce might become noticeable except for the fact that this comet also has close approaches to the planet Jupiter which exerts gravitational forces that change Tempel 1’s orbit more than the impact. Dr. Don Yeomans has written a detailed history about Tempel 1’s orbit.

Finally, any material knocked off of Tempel 1 will continue along the comet’s orbit. Tempel 1 has an orbit that is larger than Earth’s so the two orbits do not cross. So debris from Tempel 1 will not hit Earth.’ – http://deepimpact.jpl.nasa.gov/faq3.html