Gazpromavia 4G4953 Flight Status Today

Domestic flight Gazpromavia 4G4953 takes off from Sovetsky (OVS) Russia to Anapa (AAQ) Russia. It's operated by Gazpromavia. The plane leaves Sovetsky Airport at 10:45 Asia/Yekaterinburg. The flight is expected to land at Anapa Airport at 12:20 Europe/Moscow. The flight will last about 3 hours 35 minutes.

4G4953 - GZP4953 Gazpromavia (4G)
OVS Sovetsky Russia
Flight OVS to AAQ
AAQ Anapa Russia
NO FLIGHTS IN THE LAST 14 DAYS Lasted record on
Departure Sovetsky Airport
IATA: OVSICAO: USHS
Scheduled 10:45
None None
Arrival Anapa Airport
IATA: AAQICAO: URKA
Scheduled 12:20
None None
4G4953 Detail
  • Type: Domestic Flight
  • Flight Duration: 3 hours 35 minutes
  • Flight Distance: 2493 kms / 1549 miles
Airline
  • Gazpromavia
  • IATA: 4G
  • ICAO: GZP
  • Operating Days: NO FLIGHTS IN THE LAST 14 DAYS
Passenger Services
  • Service type: Normal passenger
  • Seats: 98
  • Freight capacity: 4.4 tons
  • Passenger classes: Economy
More Detail
  • Aircraft: SU95
  • Callsign: GZP953
  • Departure Timezone: Asia/Yekaterinburg
  • Arrival Timezone: Europe/Moscow
  • Current Time in Sovetsky: Tuesday 2026-01-20 20:46 PM
  • Current Time in Anapa: Tuesday 2026-01-20 18:46 PM

Flight routes similar to Gazpromavia 4G4953

Airline Flight no Departure Arrival
Gazpromavia 4G731
Gazpromavia 		20/01/2026
16:25 		3 hours 20 minutes
None

Gazpromavia Flight 4G4953 FAQs

What is the scheduled flight duration for Gazpromavia 4G4953 flight?

On average, nonstop flight takes 3 hour(s) 35 minutes, with the flight distance of 2493 km (1549 miles).

What type of aircraft is used for the Gazpromavia 4G4953 flight?

All Gazpromavia 4G4953 flights are operated using SU95 aircraft.