Pluto's gravitational pull on a person on the Earth's surface? My physics teacher stated that Pluto has a gravitational pull on objects on Earth, namely humans. Is this true? What is the free-fall acceleration of Pluto with respect to being on the Earth's surface (i.e. the Earth's free-fall acceleration is $9.8$ m/s$^{2}$)?
 A: As Emilio points out the gravitational attraction by Pluto is on the order of $10^{-14}\textrm{ m s}^{-2}$.
However this is not an acceleration that can be felt in any meaningful way - in a flat gravitational field movement due to acceleration by gravity is identical to inertial movement.  However as it happens the gravitational field from Pluto is not flat, it's inversely proportional to the square of the distance.  This means that with a suitable reference point it's in theory possible to detect the tidal force that Pluto creates.
For us a suitable reference point would be earth.
We know, from the other answers, that the acceleration from Pluto's pull is 
$$a_\mathrm{me}=\frac{GM}{d^2}.$$
Where $M$ is the mass of Pluto and $d$ is our distance to Pluto.
If we assume that we are on the side of the planet facing Pluto, and on a direct line between the center of the planet and the plutoid. We get that the acceleration of Earth due to Pluto is
$$a_\mathrm{Earth}=\frac{GM}{(d+r)^2},$$
where $r$ is the radius of the Earth.
The difference is thus
$$
a_\mathrm{me}-a_\mathrm{Earth}
=\frac{GM}{d^2}-\frac{GM}{(d+r)^2}
=GMr\left(\frac{2}{d^3+d^2r}+\frac{r}{d^4+2d^3r+d^2r^2}\right)
$$
However the radius of Earth is tiny compared to the distance to Pluto so we can approximate the above to
$$a_\mathrm{me}-a_\mathrm{Earth}\approx\frac{2GMr}{d^3}.$$
Plugging this into WolframAlpha tells us that we can perceive a difference in acceleration of roughly 
$$a_\mathrm{me}-a_\mathrm{Earth}\approx10^{-19}\mathrm{\:m\:s^{-2}}$$
between us and the Earth. For reference, this is about 7 angstrom per day per day, or 2 inches per second in the current age of the universe. How to actually measure accelerations of those magnitudes is left as an exercise for the reader.
A: This is true. Newton's law of universal gravitation says everything attracts everything. To get the free-fall acceleration of some object on Earth towards Pluto, take Newton's law and divide by the object's mass to get $$a=\frac{F}{m}=\frac{GM}{r^2}.$$ Subbing in reasonable values - $M=0.002$ Earth masses, and $r$ between 29 and 49 AU you get something like $10^{-14}\textrm{ m s}^{-2}$.
A: I am not a physicist but this will answer your query; According to Newton's law of universal gravitation every point mass in the universe attracts every other point mass with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.
$F=G\frac{m_{1}m_{2}}{r^{2}}$
for example:
lets say r is the distance between earth and pluto, G the universal gravitation constant, m1 is the mass of pluto and m2 be the mass of a human on earth, we can then calculate the gravitational force exerted by m1 to m2.
you can also refer this site http://www.physicsclassroom.com/class/circles/u6l3c.cfm
