Gravity above surface level

[Dooey Jo]

Is there a formula for describing the gravitational force and acceleration at heights that are not much greater than the Earth's radius, and are not below surface level?

Newton's law of gravitation may work very well for distant objects, such as the moon, and the sun, as well as objects on the surface of planets and other bodies. But it is inadequate to describe the gravitational accelleration on, say an aircraft 10 000 metres up. According to F = GMm/r^2, the accelleration there would only be some 1.5 m/s^2, or roughly that of the Moon. Now, I've never actually been at a 10 000 metre altitude, but that's hardly a correct figure...

I assume that the error has to be because of the Earth being (roughly) spherical, and approximating it into a point-mass is incorrect at distances comparable to the radius of the sphere. So would the gravity then be some kind of area effect, so that an object which is above ground will be pulled in all directions pointing towards the surface, and not just down (although the resultant acceleration would be downwards)? Could the resulting gravitational acceleration then be some sort of "line of sight" effect, and calculated thusly? (Extension of reasoning would seem to suggest that there is no net-gravity at all inside the sphere, because all directions of acceleration would cancel out; there wouldn't be a line of sight effect)

So does this formula exist, or do I have to derive it myself (do not want!!)?

[WyrdNyrd]

The point-mass approximation isn't an approximation! A perfect sphere does act as a point mass if you're outside of it (and as no mass if you're inside the sphere.)

Check your calcs again, and remember to use the distance between the centers of the objects as r, not the distance to the surface. That should reveal the correct answer.

[Il Saggiatore]

Is there a formula for describing the gravitational force and acceleration at heights that are not much greater than the Earth's radius, and are not below surface level?

Newton's law of gravitation may work very well for distant objects, such as the moon, and the sun, as well as objects on the surface of planets and other bodies. But it is inadequate to describe the gravitational accelleration on, say an aircraft 10 000 metres up. According to F = GMm/r^2, the accelleration there would only be some 1.5 m/s^2, or roughly that of the Moon. Now, I've never actually been at a 10 000 metre altitude, but that's hardly a correct figure...

I hope you did not use the altitude in the formula, but the distance between the centers of mass of the Earth and the plane.
If you consider that the Earth's radius is about 6000 km, adding 10 km to the distance should decrease the gravitational acceleration by about 0.17% from sea level.

I assume that the error has to be because of the Earth being (roughly) spherical, and approximating it into a point-mass is incorrect at distances comparable to the radius of the sphere.

Look up Gauss's law (or theorem).
The only significant effects you should see are tidal effects.

So would the gravity then be some kind of area effect, so that an object which is above ground will be pulled in all directions pointing towards the surface, and not just down (although the resultant acceleration would be downwards)?

And what would be the net effect?

Could the resulting gravitational acceleration then be some sort of "line of sight" effect, and calculated thusly? (Extension of reasoning would seem to suggest that there is no net-gravity at all inside the sphere, because all directions of acceleration would cancel out; there wouldn't be a line of sight effect)

Well, if you look at Newton's law for gravitation, the force acts along the geometrical line connecting the two masses, so you might even call it "line-of-sight".

So does this formula exist, or do I have to derive it myself (do not want!!)?

It is Newton's law for gravitation, and you already used it.

[Dooey Jo]

Ah yes, now I see what I did. Those weren't kilometres, those were thousands of kilometres.
But they were from the centre of the Earth...

Please terminate this thread, unless someone has something significant to add.



Bows head in shame

[Braedley]

The difference in gravitational acceleration is negligable for most aplications dealing with objects from earth's surface up to about 160km or 100 miles. In fact there is a greater difference between some points on the surface of earth then there is between the surface and 50km up.

[TheDarkOne]

The point-mass approximation isn't an approximation! A perfect sphere does act as a point mass if you're outside of it (and as no mass if you're inside the sphere.)

Check your calcs again, and remember to use the distance between the centers of the objects as r, not the distance to the surface. That should reveal the correct answer.

I'm pretty sure that it only acts as no mass if it happens to be a hollow sphere with all the mass concentrated at r=Radius of sphere.

[Kuroneko]

The point-mass approximation isn't an approximation! A perfect sphere does act as a point mass if you're outside of it (and as no mass if you're inside the sphere.)

I'm pretty sure that it only acts as no mass if it happens to be a hollow sphere with all the mass concentrated at r=Radius of sphere.

Not quite; gravitational potential is zero in a spherically symmetric hollow object, but it doesn't matter whether or not the mass it concentrated at a specific radius. As for the spherically symmetric model being an approximation, yes, but one that is more than sufficient for most purposes. Approximating the Earth as an oblate spheroid with the appropriate symmetry may give more precise results. If my intuition is correct, radial freefall from rest would be replaced by a hyperbolic trajectory with a focus slightly away from the Earth's center (the w = sin z conformal mapping should be appropriate here).