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      Distance from Jessup to Villanueva de la Canada

      The shortest distance (air line) between Jessup and Villanueva de la Canada is 3,748.70 mi (6,032.95 km).

      How far is Jessup from Villanueva de la Canada

      Jessup is located in Maryland, United States within 39° 8' 55.68" N -77° 13' 22.08" W (39.1488, -76.7772) coordinates. The local time in Jessup is 15:59 (12.03.2025)

      Villanueva de la Canada is located in Madrid, Spain within 40° 27' 0" N -4° 1' 0.12" W (40.4500, -3.9833) coordinates. The local time in Villanueva de la Canada is 21:59 (12.03.2025)

      The calculated flying distance from Villanueva de la Canada to Villanueva de la Canada is 3,748.70 miles which is equal to 6,032.95 km.

      Jessup, Maryland, United States

      Related Distances from Jessup

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      Jessup to Washington (District of Columbia)21.19 mi (34.11 km)
      Jessup to Queens (New York)192.71 mi (310.14 km)
      Jessup to Boston (Massachusetts)369.70 mi (594.97 km)
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      Villanueva de la Canada, Madrid, Spain

      Related Distances to Villanueva de la Canada