Well, the strength of a gravitational field of a spherically
symmetrical object is inversely proportional to the square of the
distance you are from the center (that is, if you are outside -- I hope
you're outside of the sun!). So your answer depends on where you are.
If you're standing on the photosphere of the sun -- the "surface",
the gravitational strength of the sun will be about 27.9 times that of
the Earth, if you were standing on the surface of the Earth. In metric
units, on Earth, the acceleration due to gravity is 9.81 meters/sec^2,
so on the Sun, that would be 273.7 meters/sec^2.
If you're out here at our Earth's orbit, that gravitational strength gets multiplied by the factor
(r_sun/r_orbit)^2, where r_sun is the radius of the sun (6.96E5
km), and r_orbit is the radius of the Earth's orbit (1.5E8 km), for a
total gravitational acceleration of 5.9E-3 meters/sec^2, or 0.0006
times the Earth's gravitational force on the surface.
Nonetheless, this keeps us in orbit (it takes a whole year to get
us to go around in a circle, however). The variation in the sun's
gravitaitonal field strength from one side of the earth to the other is
partly responsible for the ocean tides (the moon's responsible for
(almost all of) the rest).
Tom
(published on 10/22/2007)
