There’s a Nissan commercial currently airing in which a guy is sitting in his car watching the sun drop below the horizon. He then floors it and is apparently driving so fast that the sun rises back above the horizon so he stops the car and watches the sun set again. I know this is probably impossible to do in a car but how fast would you need to go in order to accomplish this?
Well, it depends on where you are on the earth. Assuming that you’re at the equator, the circumference of the earth is 24900 miles around. The earth rotates once in 24 hours, therefore a point at the equator moves approximately 1037 miles in one hour, so you’d need to go that fast in order to keep up with the sun.
Jman
It depends on your latitude, for one thing, and the time of year, too. At the south pole, you could go zero miles an hour for about six months out of the year and never see the sun set.
Trying to make all things equal, we’ll assume you’re driving along the equator during the equinox. The earth is about 25,000 miles around, the sun goes around that in 24 hours, so you’d have to go about 1050 miles per hour to keep up with the sun.
…All of which makes Brendan Fraser running ahead of the line of sunlight in the new “Mummy” movie incredibly hilarious.
It’s even funnier that everybody around me glanced over in irritation at my sarcastic laughter, apparently unable to do the simple (25,000 divided by 24) math in their head.
:rolleyes:
Yeah, it’s a stupid summer action movie. But, come on.
Actually, all you have to do is jump straight up and come to a dead stop.
Peace,
mangeorge
High enough, of course, so that a mountain or building or something doesn’t come along and smack you.
Peace.
mangeorge
I take it its one of those “Had to have been there” sort of situations. (I haven’t seent he movie yet.)
I’m looking at the sunlight (unreflected) on my floor, and it ain’t a’movin’ all that fast. Out running it would be pretty easy in this situation.
I guess if the sunlight in the movie was reflected and the source of the reflection was moving, (IE: increasing the speed of which the beam of light travels across the ground.) then I could see having an increased level of difficulty in staying ahead of it…
Cervaise:
I’ve seen the movie too, and at first, I was also rather skeptical about it. However, if the sunlight is occluded (such as in DoubleClick’s example) then the light wouldn’t move all that fast. In the movie, I figured (warning, slight spoiler) “Hey, mebbe that large mountain in the way would work like that, since they’re in a valley.” and my mind was put at ease.
>> Trying to make all things equal, we’ll assume you’re driving along the equator during the equinox
It makes no difference what time of year it is at the equator.
In fact, it make NO difference what time of year it is anywhere else. Assuming you are traveling along a constant latitude, all you have to do is complete the small circle in 24 hrs and the sun will always maintain a constant hour angle.
so, we start out at noon, with Local Hour Angle (LHA) = 0 and we travel west at a speed of 15 degrees of longitude per hour. The LHA remains zero.
What ground speed is 15 degrees of longitude per hour? it depends on your latitude as it is 60*15 NM * cos(lat).
At the equator cos(lat)=1, at the pole cos(lat)=0 so, the linear speed would be somewhere between 900 NM/hr (1035MPH) and zero.
Time of year has nothing to do with it in this case.
We can propose other scenarios where you always travel a great circle or where the LHA is different from zero etc. We can make a million different scenarios but I have illustrated the simplest case
sailor
Yeah, that’s basically right, of course. I mentioned the time of the season to throw in the comment about zero mph at the south pole. I kept it in there for the equator part because I didn’t stop to think that, while the length of daylight can change at the equator, the “twilight zone” is always traveling at the same rate, for the most part. It can still vary slightly, however (theoretically, anyway. I haven’t bothered to do any actual figuring with it; the difference may be extremely negligible, but it’s still a slight difference). Imagine travelling west along the equator with the sunset. If the days are getting longer, your speed can slack off a little; but if the days are getting shorter, you’ll have to pick up the pace a bit. Either way, of course, the answer is still a little over 1000 mph.
This is possible in an ordinary car, but only if you’re watching the sun set over a hill or something; you could drive to the other side of the hill and see the sun set again on the far horizon.
Cabbage, nope. Neglecting the equation of time issue, all days are the same length at the equator. only as you go north or south do days get longer and shorter with the seasons.
If I start out in my vehicle at the eqautor when the sun is on my meridian, and travel at 900 NM/hr for a year, the sun will only wander a bit back and forth due to the equation of time but the LHA will never be greater than 17 minutes of time (or whatever it is). OTOH, the sun would go from north to south of me oscillating between the two tropics.
30 minutes, 49 seconds (total variation).
sailor, yeah, you’re right, I ain’t thinking straight today; sorry about that. :o
MrDeath, that is the total difference but the Sun is fast by a max of 14 mins in mid Feb and late by a max of 16 mins in late Oct. Never more than 17 mins.
Actually, as the Sun rose, the top of the pyramid should have been lit first.
Also, the place where he was was surrounded by mountains, as you may recall. I therefore postulate he was outrunning the decreasing length of the mountain shadows as the SUn rose. The speed was actually about right, except for one thing: the shadow line was moving the wrong way.
I have a complete description of this (including my math!) at my website where I reveiw the (small amount of) astronomy in the movie: http://www.badastronomy.com/bad/movies/mummyreturns.html.
Occam’s Razor: The landscape in the commercial isn’t flat as it appears, but a long, gently sloping hill (or series of same). Wouldn’t the concept work that way?