Aeromexico AM4875 Flight Status Today

Domestic flight Aeromexico AM4875 takes off from Gainesville (GNV) United States to Atlanta (ATL) United States. It's operated by Aeromexico. The plane leaves Gainesville Regional Airport at 14:05 America/New_York. The flight is expected to land at Hartsfield-Jackson Atlanta International Airport at 15:30 America/New_York. The flight will last about 1 hours 9 minutes.

AM4875 - AMX4875 Aeromexico (AM)
GNV Gainesville United States
Flight GNV to ATL
ATL Atlanta United States
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on 20-December-2019
Departure Gainesville Regional Airport
IATA: GNVICAO: KGNV
Scheduled 14:05
None None
Arrival Hartsfield-Jackson Atlanta International Airport
IATA: ATLICAO: KATL
Scheduled 15:30
S D31
AM4875 Detail
  • Type: Domestic Flight
  • Flight Duration: 1 hours 9 minutes
  • Flight Distance: 484 kms / 301 miles
Airline
  • Aeromexico
  • IATA: AM
  • ICAO: AMX
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 76
  • Freight capacity: 1.9 tons
  • Passenger classes: Economy, Shuttle, Premium Economy
More Detail
  • Aircraft: CRJ9
  • Callsign: AMX4875
  • Departure Timezone: America/New_York
  • Arrival Timezone: America/New_York
  • Current Time in Gainesville: Sunday 2025-12-28 20:46 PM
  • Current Time in Atlanta: Sunday 2025-12-28 20:46 PM

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Aeromexico Flight AM4875 FAQs

What is the scheduled flight duration for Aeromexico AM4875 flight?

On average, nonstop flight takes 1 hour(s) 9 minutes, with the flight distance of 484 km (301 miles).

What type of aircraft is used for the Aeromexico AM4875 flight?

All Aeromexico AM4875 flights are operated using CRJ9 aircraft.

Which terminal the flight Aeromexico AM4875 is arriving at?

Flight Aeromexico AM4875 arrives in Hartsfield-Jackson Atlanta International Airport at Terminal S.