Distance between Serov, Sverdlovsk Oblast, Russia and the Tropic of Cancer
4026 km = 2502 miles
During our calculation of the distance to the Tropic of Cancer we make two assumptions:
- We assume a spherical Earth as a close approximation of the true shape of the Earth (an oblate spheroid). The distance is calculated as great-circle or orthodromic distance on the surface of a sphere.
- We calculate the distance between a point on the Earth’s surface and the Tropic of Cancer as the length of the arc of the meridian passing through this point and crossing the Tropic of Cancer.
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Serov, RussiaCountry: Russia
Serov’s coordinates: 59°36′12″ N, 60°34′43″ E
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Wikipedia article: Serov
The Tropic of CancerThe Tropic of Cancer or Northern Tropic is a circle of latitude or parallel on the Earth's surface where the Sun appears in zenith once a year, at the time of the summer solstice.
For the points on the Earth's surface located between the Tropic of Cancer and the Tropic of Capricorn, the Sun appears in zenith twice a year.
For the points on the Earth's surface located to the north of the Tropic of Cancer and to the south of the Tropic of Capricorn, the Sun never appears in zenith.
The latitude of the Tropic of Cancer depends on the tilt of the Earth's axis which changes with time.
The current latitude of the Tropic of Cancer equals 23°26′16″ N.
Wikipedia article: the Tropic of Cancer