How fast must a plane fly along the earth's equator so that the sun stands still relative to the passengers?

How fast must a plane fly along the earth's equator so that the sun stands still relative to the passengers? Radius of the earth = 6400km.





Please show how you got the answer, thanks!|||divide the circumference of the Earth by 24 hrs





2 pi r / 24 = ___km/h|||The circumference of the earth is 24901.55 miles(2piR=40,192km using 6400 as the radius) so in 24 hours so 1,037.56 miles per hour(1674.66km/hr). Of course this is ground speed so the actual plane would be flying a smaller indicated speed. Also the plane would have to fly faster westbound since prevailing winds are usually out of the west at 40 thousand feet(~12,000m) I would give a rough wag of about 450 MPH(725km/hr) indicated with a tailwind to get a groundspeed high enough. I found a cool true airspeed calculator you can mess with here: http://www.paragonair.com/public/aircraf鈥?/a>