My Blog for Physics in Film

Where to begin…

1. First of all, the professor wrote WAY too neatly on the chalk bored, and also everybody knows that scientists are NOT artists.

2. Besides that, would a bird really crack the windshield of car? Birds are relatively light, would they really have the momentum to break a windshield that is designed to withstand much heavier objects.

3. The magnetic field does NOT protect us against microwaves.

4. When the laser drills through the dirt, where does all the dirt go?

5. One word…Unobtainium

6. IF the rays can penetrate lead, shouldn’t they cause immediate cancer in the dude walking behind the lead?

7. I think it is stereotypical of the movie to show Europeans watching soccer.

8. Stone does not conduct electricity that well. And why would the lightning carve the street; wouldn’t the lightning at least strike the tall buildings. I am sure some have metal antennas.

9. There was the masterpiece scene in which they struggled against the unknown force during freefall when they should havebeem weightless.

10. It seems very unlikely that microwaves could destroy the Golden Gate Bridge and most of San Francisco.

11. Would those heat suits even work?

12. Would those giant crystals in the geode shatter that easily, they aren’t rock candy.

13. Unless my knowledge of air conditioning is highly mistaken, wouldn’t it be impossible to thoroughly cool the air in that ship?

14. Why would you trust autopilot on the first journey to the center of the Earth?

15. IF DESTINY stopped it the first time without all the volcanoes erupting, shouldn’t they try it again first before sending all those people down to the depths.

16. Don’t the fuel rods have to be Part of the mass of the bomb in order to contribute to the bomb?

17. And how are they communicating through miles of rock. Secondly, wouldn’t there be just a little bit of delay in the transmission.

I had a couple more than this but I couldn’t read my handwriting…no big loss.

My scenerio is the classic asteroid t-bones the earth. So, we have this asteroid coming towards us, it is traveling at 11,000 m/s and it is the size of a Midwestern state. I will use the asteroid Ceres 1 as a reference (wiki). It has a mass of around 9.43e20 kg. So the asteroid is coming perpendicular to Earth. Therefore I feel that it would only be necessary to slow the asteroid a little bit so that Earth will “be late” for its disaster appointment. So, my goal is to see how much time we would need to slow down the asteroid using a 150 megaton warhead. So, the earth orbits the sun at a velocity of about 29 km/s (source: http://hypertextbook.com/facts/2000/ IlanaEpstein.shtml). The earth is roughly 12,700 km across (wiki). Therefore it would take the Earth 437 seconds to move out of the way of the path. Is a 150 megaton warhead enough to give us that much time?

So let us see how much the warhead would slow the asteroid. 150 megatons is about 6.276e17 J. The asteroid current has (1/2)( 9.43e20 kg)(11,000m/s)^2 or 5.7051500e28 J of kinetic energy. If all the energy of the bomb goes towards decreasing the amount of kinetic energy of the asteroid, then the asteroid would then only be traveling with 5.7051499e28 of kinetic energy. Solving for the new velocity we get 10999.99986 m/s which equates to roughly 1 m/s less speed (and that Is pushing it but there has to be some change) at 10,999 m/s. So let us say that the asteroid is 1 year (31,556,926 s) from impact. This equals a distance of 3.4712618e11 m using the initial velocity of 11,000 m/s. At 10,999 mi/h, it would travel that distance at In 31,559,795 s, this is 2,869 later than it would be without the blast. This is good news for Earth. Even if the change in velocity were less, we could save ourselves from certain doom barring human and/or calculation error. We just have the get a nuke out there as soon as possible and just give it a nudge, the farther out, the better. Problem solved

I chose the scene in Eraser when the bumbling goon picks up John’s shotgun, hence causing the wielder of the rail gun to tag him as John. The bullet sends him flying backwards into the necessary glass window. I used the following site for conversions: http://www.translatorscafe.com/cafe/units-converter/velocity/calculator.

I will start this discussion with the momentum of the bullet, the star of the show. It was mentioned that the bullet was an aluminum slug traveling near light speed. A bullet usually weighs around 8 g or so. However, those bullets are usually of a heavier metal, so I will say the bullet masses around 5 g, or 0.005 kg. Lightspeed is approximately 3.8E8 m/s. Ignoring relativity, and denoting the path of the bullet to be the positive direction, the momentum of the bullet, or Pb= (0.005 kg)(3.8E8 m.s) = 1.9E6 kg*m/s.

Next, the victim’s momentum. Being the average goon, he likely has a mass around 90 kg. So what should his velocity be when he is hit by the bullet with a momentum of 1.9E6 kg*m/s. Since momentum is conserved his momentum, Pg, g for goon must be equal to Pb because he will be going in the same direction, assuming all the momentum of the bullet is transferred. So given that, the Mg*Vg=1.9E6 kg*m/s, so Vg= (-1.9E6 kg*m/s)/90 kg, kilograms cancel and his new velocity is 21,111 m/s or 47,223 mi/h. Actually, according to Dr. Fragile’s site, escape velocity for the earth is only 11,000 m/s. So the bullet should actually put him into orbit (not necessarily literally). However, the bullet did not transfer all its momentum to him as the bullet passed through him with relatively little visual evidence slowing down (but it’s traveling at the speed of light, how do you see it?). So let’s say the bullet only transferred 0.01 % of its momentum. He would still be going about 2.1 m/s. This would be more reasonable…if the bullet could even be fired at lightspeed. Hence the next subject is below.

The gun and the gunner MUST have the same momentum (albeit negative) of the bullet. Let us say the total mass of the gun combo is 100 kg (10 kg for the gun). Assuming the gunner is attached to the gun, with something like a heavy duty harness or by the sheer grip of his hands, the gun and him would be traveling at the same velocity, and in the opposite (negative direction) of the bullet. So –Pb = Mc*Vc, c for “gun combo”, so Mc = 100 kg and –Pb = -1.9E6*m/s, therefore Vc = 19,000 m/s. Enough to also send the gunner into orbit. The other scenario would be if the gun flew off on its own, transferring no momentum to the gunner. The 10 kg gun would then probably burn up in atmosphere as it slowly drifts towards the horizon at 190,000 m/s.

The three scenes I chose are when the captain falls off the ship, when the cruise ship hits a small sailboat, and the dock collision scene.

For the first scene, I am trying to see how high the deck is from which the captain fell. The quantities I need are the time it takes for him to fall and his acceleration. He fell for about 5 seconds and his acceleration is 9.81 m/s^2­, which is that due to gravity.

In the second scene, I want to know how much momentum the cruise ship lost when it hit the small sailboat (which subsequently exploded). I will need to know the speed and mass of the cruise ship and the mass of the small sailboat. The cruise ship was going around 20 knots at this time, and since one knot is equal to about .514 m/s, the ship was going about 10.28 m/s. An approximate mass of the ship was hard to obtain as most of the data for ships care mostly for the capacity. Anyway, I found that on one site that the weight of the Carnival Destiny is 51000 tons which converts to 4.62E7 kg. For the small sailboat, I looked around the internet for similar boats and estimate the boat is about 10000 pounds, which would be 4,536 kg.

The third scene requires the mass of the boat, the time, and the length of the dock. The mass of the boat is already done. The time was about 3 or 4 minutes but I will stick with 3 minutes, which would be 180 s. Lastly is the length of the dock. After the boat came to a stop, it was completely in the dock, which means the length of the boat is approximately the length of the dock. According the Wikipedia, the ship that was featured in speed was 135 m long, so that must be the length of the dock.

PS: I meant to put that I wanted to see if the dock used would be sufficient to stop the boat. for what it is worth.