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标题: How to Calculate Earth Rotating Speed? [打印本页]

作者: Dinesh Vishwakarma 时间: 2019-04-02 14:19
标题: How to Calculate Earth Rotating Speed?
Edited by Dinesh Vishwakarma at 2019-04-02 11:53

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[p]How fast is our Earth rotating? More specifically, how fast are you moving due to the Earth's rotation on its axis? The answer always seems surprising, since as you observe the Earth from your residence, objects on the Earth don't seem to be moving that fast. In fact, some of the ancient astronomers (including Aristotle) believed that the Earth was stationary since there seemed to be no direct evidence and this was complicated by the fact that many believed that the Earth was at the center of the known universe and that everything, including the Sun, orbited it.

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[p]Of course, we know now that the Earth is not stationary, is not at the center of anything, including our own Solar System, and does indeed rotate on an axis. We see this rotation every single day as the Sun, the Moon, planets and stars appear to move from east to west in our local skies every single day.

[p]The Earth's gravity keeps us rooted on the ground, thankfully!

[p]We need only two values to calculate the rotational speed of the Earth at its equator (0 degrees latitude). The first value is the Earth's equatorial radius, which can easily be obtained from the astronomy texts or the internet:

[p][strong]R[sub]eq[/sub] = 6,371.14 km[/strong]

[p]The second value is the Earth's sidereal rotation period. We cannot use the widely used 24 hours, since that is not the Earth's true rotation period. The "solar day" (24 hours) is based on the time required for the Sun to appear in the same part of the sky. This is caused not only by the Earth's rotation but by the Earth's orbit, which carries the Earth approximately one degree per day. This small angle will certainly change the Sun's apparent position in the sky as observed from the Earth's surface.

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[p]The sidereal rotation period (or "sidereal day") is based on the time required for the stars (which can be construed as being an infinite distance from the Earth) to appear in the same part of the sky as observed from the Earth's surface. The Earth's sidereal day can easily be looked up:

[p][strong]T[sub]rot[/sub] = 23 hours 56 minutes 4.1 seconds[br]T[sub]rot[/sub] = 23.934472 hours[/strong]

[p]A spot on the Earth's equator traces out a full circle every sidereal day. The circumference of this circle can easily be calculated by using the basic equation learned in high school:

[p][strong]C[sub]eq[/sub] = 2πR[sub]eq[/sub][/strong]

[p][strong]C[sub]eq[/sub] = 40,031.1 km[/strong]

[p]The spot on the Earth's equator travels this full circumference every sidereal day. Therefore the speed of a person standing on the equator can be calculated by using the following equation:

[p][strong]v[sub]eq[/sub] = C[sub]eq[/sub] ÷ T[sub]rot[/sub][/strong]

[p][strong]v[sub]eq[/sub] = 1,673 kilometres per hour (km/h) = 465 metres per second (m/s)[/strong]

[p][strong]v[sub]eq[/sub] = 1.35 the speed of sound (!!!)[/strong]

[p]Wow! That's fast! a person standing on the equator is actually moving that fast. The only reason why we cannot feel it is because the Earth's rotational speed is quite constant (thankfully) and therefore has a negligible acceleration and the Earth's gravity keeps us well grounded.

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[p]What is the rotational speed of a person not on the equator? For this calculation, we will use the perfectly circular Earth as the ideal case. Since the Earth is nearly spherical, the values will be very close to the actual.

[p]The maximum speed will be at the equator, since the circle traced there is the largest. In other words, at the equator, the maximum distance is traced in the sidereal rotation time.

[p]The size of the circle at a latitude (λ) other than 0 degrees (the equator) can be calculated using the following equation:

[p][strong]C[sub]λ[/sub] = 2πR[sub]eq[/sub]cosλ[/strong]

[p]At the equator, cosλ is 1. At the north pole (+90 degrees) or the south pole (-90 degrees) cosλ is 0. At a latitude of 45 degrees (or -45 degrees), cosλ = 0.707. Therefore:

[p][strong]C[sub]45 deg[/sub] = 28,302 km[/strong]

[p]and the rotational speed will definitely be slower at this latitude:

[p][strong]v[sub]45 deg[/sub] = 1,182.5 km/h = 328.5 m/s = 0.957 the speed of sound[/strong]

[p]So, at the equator, you are whipping around at above Mach 1, but at 45 degrees latitude (north or south), you are travelling just under the speed of sound.

[p]Below is a table of the equatorial rotation speeds of the major Solar.

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[h2][span style="font-weight: bold;"]Thank-you![/span][/h2]

作者: jk885884 时间: 2019-04-02 14:50
Nice information

作者: Ateur 时间: 2019-04-02 15:14
good share

作者: Thinesh 时间: 2019-04-02 18:24
Nice, Thanks for sharing

作者: reddykrishnayad 时间: 2019-04-02 23:39
good share

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