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Outracing the sun 
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#1
Jul410, 11:41 AM

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Me and a friend have been wondering if it would be possible for a person to outrace the sun. more accurately, in a plane.
One of us is argueing that we would need to outrace the earths rotational velocity, the other is argueing that we could calculate the velocity of the sun by assuming that the sun travels the diameter of the earth in 24 hours, thereby resulting in a necicary speed of ~330 MPH. i was just wondering if someone could shine some light on this? how fast do we need to go (In MPH) to outrun the sun, and stay in daylight for as long as the fuel would last us 


#2
Jul410, 12:30 PM

P: 27

0 mph.



#3
Jul410, 12:37 PM

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P: 953

In order to keep the sun in the same position in the sky by flying due west you need a speed of 1035 mph times cosine(latitude). That means it takes the full 1035 mph at equator and only half, around 500 mph, at 60 deg north or south (cos(60) = 0.5). Since 500 mph is around the top speed at altitude for most civilian airliners, such jet would be able to keep up if it flew at around latitude 60 deg or more north.



#4
Jul510, 10:50 PM

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Outracing the sun
So, if the circumference is 0, such as at the poles.... 


#5
Jul810, 06:01 PM

Sci Advisor
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I flew from Paris to San Francisco years ago in late May. We took off 10 minutes before sunset, and landed 10 minutes after sunset. We followed the sunset around the world. I had a window seat and watched this 10hour sunset. Had we travelled slightly faster, shaving 20 minutes off the flight time, we would have matched the speed of the terminator.



#6
Jul1110, 09:34 PM

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#7
Jul1110, 09:47 PM

P: 3,014

Actually, you won't outrace the Sun, but outrace the Earth. If you keep travelling due West with a sufficiently high speed, then your angular velocity around the Earth's axis would become equal by magnitude, but opposite in direction from the angular velocity of Earth's rotation around its axis. Then, relative to the Sun, you will not have any angular speed and the Sun will be in the same position on the sky.
But, your trajectory is a circle with a different radius, depending on the latitude where you travel. You may convince yourself, that if the Earth's radius is R, then the radius of a circle perpendicular to the Earth's axis at a latitude [itex]\lambda[/itex] is: [tex] r = R \, \cos \lambda [/tex] The necessary speed is then: [tex] v = \frac{2 \, \pi \, r}{T} = \frac{2 \, \pi \, R}{T} \, \cos \lambda = v_{e} \, \cos \lambda [/tex] where [tex] v_{e} = \frac{2 \, \pi \, R}{T} = \frac{4 \times 10^{7} \, \mathrm{m}}{8.64 \times 10^{4} \, \mathrm{s}} = 463 \, \frac{\mathrm{m}}{\mathrm{s}} [/tex] is the necessary speed on the Equator. Notice that this is higher than the speed of the sound and corresponds to a Mach number of about [itex]M = 1.3[/itex]. However, because the cosine is a monotonically decreasing function with angle, at higher latitudes (both North and South of the Equator), the necessary speed decreases. Question: What is the necessary speed at the Poles and when can you observe this effect? 


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