
An arc of \[1\] degree on earth’s surface is equal to how many kilometers?
Answer
560.7k+ views
Hint: Earth's perimeter is the distance around the Earth. Estimated around the posts, the outline is \[40,007.863km\left( {24,859.734mi} \right)\]. Estimated around the equator, it is \[40,075.017km{\text{ }}(24,901.461mi)\]. Estimation of Earth's perimeter has been essential to the route since old occasions. The main known logical estimation and figuring was finished by Eratosthenes, who accomplished an extraordinary level of exactness in his calculation. Treated as a circle, deciding Earth's periphery would be its single most significant measurement. Earth veers off from round by about $0.3\% $, as described by leveling.
Complete answer:
The value of the equatorial radius of the Earth is \[6378.1km\].
The difference is \[21.3km\]. The polar radius is \[6366.8km\].
At latitude\[{\theta ^o}\dfrac{N}{S}\], the radius is \[6378.1 - \left( {\dfrac{\theta }{{90}}} \right)\left( {21.3} \right)km\],
Nearly,
An arc of 1 radian is equal to the radial distance of the latitude circle, from the Earth’s center
\[{1^\circ} = \dfrac{\pi }{{180}}\]radian.
Thus, the length of the arc =\[(\dfrac{\pi }{{180}})\left( {6378.1 - \left( {\dfrac{\theta }{{90}}} \right)\left( {21.3} \right)} \right){\text{ }}km\],
At the equator, \[\theta = 0\]and this arc length =\[111.32km\], nearly.
At Baltimore, \[\theta = {39.48^\circ}\] and the length is \[111.16km\], nearly.
At the poles, \[\theta = {90^\circ}\] and the length =\[110.95km\], nearly.
Going through a little circle, the length is less. On the off chance that this little circle is the scope circle, apply a factor cos (latitude).
At the equator, ${1^ \circ }$ length of the arc =\[111.32km\], almost. At Baltimore (latitude=\[{39.48^\circ}N\]), this length is \[111.16km\], almost. At the shafts, it is \[110.95km\], almost.
Note: In current occasions, Earth's periphery has been utilized to characterize essential units of estimation of length: the nautical mile in the seventeenth century and the meter in the eighteenth. Earth's polar outline is extremely close to \[21,600\] nautical miles in light of the fact that the nautical mile was expected to communicate one moment of scope, which is \[21,600\] parcels of the polar boundary.
Complete answer:
The value of the equatorial radius of the Earth is \[6378.1km\].
The difference is \[21.3km\]. The polar radius is \[6366.8km\].
At latitude\[{\theta ^o}\dfrac{N}{S}\], the radius is \[6378.1 - \left( {\dfrac{\theta }{{90}}} \right)\left( {21.3} \right)km\],
Nearly,
An arc of 1 radian is equal to the radial distance of the latitude circle, from the Earth’s center
\[{1^\circ} = \dfrac{\pi }{{180}}\]radian.
Thus, the length of the arc =\[(\dfrac{\pi }{{180}})\left( {6378.1 - \left( {\dfrac{\theta }{{90}}} \right)\left( {21.3} \right)} \right){\text{ }}km\],
At the equator, \[\theta = 0\]and this arc length =\[111.32km\], nearly.
At Baltimore, \[\theta = {39.48^\circ}\] and the length is \[111.16km\], nearly.
At the poles, \[\theta = {90^\circ}\] and the length =\[110.95km\], nearly.
Going through a little circle, the length is less. On the off chance that this little circle is the scope circle, apply a factor cos (latitude).
At the equator, ${1^ \circ }$ length of the arc =\[111.32km\], almost. At Baltimore (latitude=\[{39.48^\circ}N\]), this length is \[111.16km\], almost. At the shafts, it is \[110.95km\], almost.
Note: In current occasions, Earth's periphery has been utilized to characterize essential units of estimation of length: the nautical mile in the seventeenth century and the meter in the eighteenth. Earth's polar outline is extremely close to \[21,600\] nautical miles in light of the fact that the nautical mile was expected to communicate one moment of scope, which is \[21,600\] parcels of the polar boundary.
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