Star trek movie physics (spoilers within...)

[RampantAndroid]

So, to anyone who knows WAY more than me...how "correct" is the scene in the recent star trek movie (into darkness) when

Spoiler

the enterprise falls from ~270,000km (they looked like they were near the moon, though, so maybe 380,000km?) to the earth in a matter of seconds because they lost power?

Now, I understand of course that you do feel the earth's gravity when you're that far out, but you're only feeling 1/3600th of earth's gravity or something, right? (are you more likely to enter orbit around the moon, as it is .16G?)

It's been too long since I took physics (and calc) but how long would an object moving at an initial velocity of 0m/s take to fall to the earth? Realizing that as it grows closer its acceleration will increase.

 

[Anteaus]

You're also talking about ships that bank in the vacumn of space and are able to go faster than the speed of light. These films violate relativity theory constantly so I would imagine Newton and Einstein are going to take a back seat. I think suspension of disbelief is appropriate here.

In the first reboot film a ship in order deployed a ridiculously long chain tethered while in order for the purposes of shooting a laser into the crust. Wouldn't it have been far simpler just to mount it on the side of the ship?

The long and short of it is that JJ Abrams had to play with things a bit to keep the plot moving! 😀 I can't wait to see the new Star Trek.

 

[RampantAndroid]

You're also talking about ships that bank in the vacumn of space and are able to go faster than the speed of light. These films violate relativity theory constantly so I would imagine Newton and Einstein are going to take a back seat. I think suspension of disbelief is appropriate here.

In the first reboot film a ship in order deployed a ridiculously long chain tethered while in order for the purposes of shooting a laser into the crust. Wouldn't it have been far simpler just to mount it on the side of the ship?

The long and short of it is that JJ Abrams had to play with things a bit to keep the plot moving! 😀 I can't wait to see the new Star Trek.

But that's part of my problem with the new star trek movies. Saying we might figure out warp drive someday in the future...well, fine. I'll go with that. Same with somehow moving matter from point A to B. But...magically falling to earth when you lose power is beyond what I'm willing to forgive.

However, the initial question of time it takes to fall to earth from 270000km due to earths gravity stands - with an explanation preferably so I can try to follow along 😀

 

[edcarman]

However, the initial question of time it takes to fall to earth from 270000km due to earths gravity stands - with an explanation preferably so I can try to follow along 😀

If you assume the ship starts with zero velocity and falls directly towards the earth, it forms a radial elliptic trajectory:
http://en.wikipedia.org/wiki/Radial_trajectory#Elliptic_trajectory

You can then use the radial Kepler equation to work out the time for it to fall to the centre of the earth (the distances are big enough relative to ship size and earth size that the ship and earth can be treated as points):

You know the initial position (x0=270,000 km), initial velocity (v0=0) and standard gravitational parameter for earth (μ ~= 400,000 km^3 s^−2), so it's a simple matter of using them in the equation to see it will take about 68 hrs to fall to earth.

 

[rsutoratosu]

Well obviously we can't sit through 68 hours of movie so it got shorten to seconds

 

[Revolution 11]

Don't forget to consider Kirk's historically big head which will change the gravitational attraction between the Earth and the Enterprise. It will be shorter than 68 hours.

Ok, in all seriousness, the shorter falling time is for brevity's sake. I have a far bigger problem with Spock being able to see Vulcan's destruction from Delta Vega. Those are some damn good eyes Mr. Nimoy has.

 

[Matt1970]

Look at how many times they have time traveled into the past. Even on Voyager when Captain Janeway went back and interacted with her previous self.

 

[Anteaus]

There are tons of physical violations that occur throughout all through these movies. I think the biggest offender for me was the time travel thing in 4 where they use gravity from the sun to "slingshot" back in time. I'm not opposed to this as a plot device, but considering gravity is assumed to move at the speed of light and the ship was already at warp speed, any gravitation influence that might have acted on the ship would have been outran and rendered useless.

An analogy would be a starter motor that is used after the engine was running at full throttle. The engine would be turning faster than the starter motor, so it wouln't have any effect.

Oh, and then they did it again....in reverse. Spoiler alert...the whales survived.
Still it's one of my favorites.

 

[RampantAndroid]

If you assume the ship starts with zero velocity and falls directly towards the earth, it forms a radial elliptic trajectory:
http://en.wikipedia.org/wiki/Radial_trajectory#Elliptic_trajectory

You can then use the radial Kepler equation to work out the time for it to fall to the centre of the earth (the distances are big enough relative to ship size and earth size that the ship and earth can be treated as points):

You know the initial position (x0=270,000 km), initial velocity (v0=0) and standard gravitational parameter for earth (μ ~= 400,000 km^3 s^−2), so it's a simple matter of using them in the equation to see it will take about 68 hrs to fall to earth.

Thanks for the explanation! My final question though is, does this account for the fact that as you grow closer to earth, the gravitational pull will gradually increase?

I'm not sure I understand standard gravitational parameter - it doesn't account for distance in this usage, does it?

 

[edcarman]

Thanks for the explanation! My final question though is, does this account for the fact that as you grow closer to earth, the gravitational pull will gradually increase?

I'm not sure I understand standard gravitational parameter - it doesn't account for distance in this usage, does it?

The equation does account for the increase in acceleration as the ship gets closer to earth. The equation with constant acceleration would be much simpler.

The gravitational parameter is a combination of the universal gravitational constant, G, and the mass of the earth. It is a constant and is independent of distance.

 

[Anteaus]

I'm guessing that is in a vacumn? How much time is added due to atmospheric drag on such a large craft. I can imagine terminal velocity being all over the the place as the ship was twisting about it got within 30-40 thousand feet above sea level.

 

[edcarman]

I'm guessing that is in a vacumn? How much time is added due to atmospheric drag on such a large craft. I can imagine terminal velocity being all over the the place as the ship was twisting about it got within 30-40 thousand feet above sea level.

According to NASA, re-entry into the atmosphere occurs at 400,000 feet (122 km). The ship would travel through vacuum for about 68h20min. At this point, it would be travelling at just under 11 km/s. I don't think it will reach anything near terminal velocity before crashing or burning up.

I did a rough spreadsheet calculation, based on dimensions from here:
https://en.wikipedia.org/wiki/USS_Enterprise_(NCC-1701-E)

Assuming the ship behaves like a flat plate when falling, including drag based on altitude and assuming it doesn't burn up, it would take around 11.4s for the ship to fall through the atmosphere. At the point of impact, it would still be travelling at 7 km/s and would be decelerating at about 316g. For comparison, if there was no atmosphere, it would take 11.1s to impact and would be moving at just over 11 km/s.