VATICAN CITY, MAY 9, 2003 (VIS) - The itinerary for Pope John Paul's June 5-9 visit to Croatia, his 100th foreign apostolic trip, was published today.
On Thursday, June 5 at 3:30 p.m. the Holy Father will leave Rome for Rijeka, Croatia where he will arrive at 4:45. After the welcome ceremony at the airport on the island of Krk, he is scheduled to meet the president of the Republic in the archdiocesan seminary. The following day he will travel to Dubrovnik where he will celebrate Mass and beatify Sr. Marija Petkovic.
On Saturday, June 7, the Pope is scheduled to travel to Osijek where he will celebrate Mass, followed by a private visit to the cathedral of Djakovo. On Sunday, June 8, after presiding at a Eucharistic celebration and reciting the Angelus in the Rijeka Delta, he will meet with the bishops of Croatia and, later in the afternoon, with the prime minister.
On Monday, June 9, John Paul II will say Mass in private in the archdiocesan seminary of Rijeka after which he will travel by plane to Zadar where he will preside at a Liturgy of the Word. Following this he will go to the airport where there will be a farewell ceremony before he departs for Rome at 1:15 p.m.
JPII-ITINERARY;TRIP CROATIA;...;...;VIS;20030509;Word: 210;
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