Utter baloney! I wonder if it's not meant as a joke.
Interesting fact: The tidal force of objects of the same density is proportional to their apparent area in the sky, measured by steradians, or square degrees if you like.
Note the sun and moon have the same apparent size, but the density of the sun is less than that of the moon, and it has a proportionately smaller tidal effect on the earth.
Likewise, Pluto's tidal effect may be estimated by its apparent angular area. It has an angular diameter of about 2000km/6 billion KM, so it's angular diameter is about 1/3billion, compared to the moon's 1/100. So the tidal effect of Pluto on the earth is about 1/30 million squared, or about 10-15 times the lunar tidal effect.
Doh!
I mis-recalled my back-of the envelope calculation, which uses MG/R^2 X d/R as the force, where d is the distance on the earth from the earth’s center. Of course, there’s a tensor involved to give you the direction of the force, but we’re just concerned with the magnitude here.
Then we just use M = rho r^3, where rho and r are the density and diameter of the moon or planet exerting the tidal force, and we see we have rho G (r/R)^3 X d. So it’s the CUBE of the angular size, not the square. So make that 10^-23 instead of 10^-15.
... but the point remains the same.