Will ''Austin Powers'' set a box office record? | EW.com


Will ''Austin Powers'' set a box office record?

Will ''Austin Powers'' set a box office record? Mike Myers' ''Goldmember'' should be the biggest debut yet for the shagadelic star

Mike Myers, Michael Caine, ...

(Austin Powers in Goldmember: Melinda Sue Gordon)

Can ”Austin Powers” 3 open bigger than ”Austin Powers” 2? Yeah, baby, yeah!

But by how much? Three years ago, ”Austin Powers: The Spy Who Shagged Me” premiered with an astonishing $57.4 million, more than the first ”Austin Powers” grossed in its entire run. Since then, Mike Myers’ creation has become an even bigger phenomenon, causing celebs like Katie Couric, Britney Spears, Kevin Spacey, and Gwyneth Paltrow to clamor for cameos in the newest installment, ”Austin Powers in Goldmember,” which costars Destiny’s Child singer Beyoncé Knowles and Michael Caine in featured roles.

But in this summer when ”Men in Black II” failed to sell more tickets in its opening weekend than the original ”MIB,” ”Goldmember” isn’t guaranteed to shatter Myers’ personal best, but merely to top it by a few meeeeellion dollars. In all, look for ”Goldmember” to earn a shagadelic $60 million this weekend.

The week’s other new entry is the Disney flick ”The Country Bears,” based on a popular Walt Disney World attraction. Lately, family films haven’t been performing too well – witness ”Stuart Little 2”’s mediocre opening – but the Disney name could power this no-star production to a double-digit debut. Should ”Stuart” fall the requisite 35 percent from its $15.1 million premiere, both it and the ”Bears” could gross $10 million.

That would push last week’s No. 1, ”Road to Perdition,” down to fourth place, which seems likely considering ”Stuart” had been beating ”Road” midweek. Look for ”Road” to drop 40 percent to $9 million.

And rounding out the top five will be ”Men in Black II,” with about $8 million. But with the advent of Mel Gibson’s ”Signs” next weekend, ”MIB” will be vanishing from the top five more quickly than a wayward alien.