Q:

what is the gravational pull of the sun?

- forman (age 15)

north shore, california

- forman (age 15)

north shore, california

A:

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

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)*

Q:

true or false:

the gravitational pull of the sun has a greater effect in tides than the pull of the moon because the sun has greater mass?

- Wing (age 45)

US

the gravitational pull of the sun has a greater effect in tides than the pull of the moon because the sun has greater mass?

- Wing (age 45)

US

A:

False.

Although the sun is much more massive than the moon it's a lot farther away.

Tidal strain is proportional to the mass of an external object divided by the cube of its distance. Put in the numbers and the moon wins by a factor of about two since it's much closer.

LeeH and Mike W

p.s. You might want to look at this:

Although the sun is much more massive than the moon it's a lot farther away.

Tidal strain is proportional to the mass of an external object divided by the cube of its distance. Put in the numbers and the moon wins by a factor of about two since it's much closer.

LeeH and Mike W

p.s. You might want to look at this:

*(published on 10/22/2007)*

Q:

If I was to stand on a stationary (non-orbiting) platform which was the same distance away from the Sun as the Earth, how much gravity would I feel from the Sun? And what would be the acceleration due to gravity if I dropped a ball from my platform?
Also how fast would the surface of the Earth, (assuming it wasn't rotating), be moving under my feet if it passed me?

- Andrew (age 17)

Greater Manchester, UK

- Andrew (age 17)

Greater Manchester, UK

A:

The answer for the strength of the gravitational field, in your coordinates, would be 5.9*10^{-3} meters/sec^{2}, as in an earlier answer in this thread. That would be the rate at which a dropped ball would fall toward your platform. However, you would not "feel" this gravitational field. In fact, you don't feel any nearly uniform gravitational field, since it would accelerate each part of your body in exactly the same way. That leaves no stresses or strains to trigger any nerve signals. Einstein generalized this fact successfully to form the Equivalence Principle, which states that a uniform gravitational field is undetectable by any physical measurements internal to the measuring system.

In order to keep your platform at a fixed distance from the Sun you need something pushing it away from the sun, say a rocket. What you feel is the push of the rocket-driven platform on the parts of your body in contact with it. If you took the same rocket-driven platform far from any stars, it would feel just the same and the ball would "fall" just the same on it.

Mike W.

*(published on 06/05/2013)*