# What adjustments are needed to use Newton's Law of Universal Gravity to calculate the force between me and Earth?

Background

Say I weigh $$80 \mathrm{kg}$$, and I happen to fall to the ground with an acceleration of roughly $$9.81 \frac m{s^2}$$, I would, using Newton's second law of motion, $$F = ma$$, come to the conclusion that the gravitational force between me and Earth is roughly $$800 N$$ (rounded for simplicity).

Question

Using Newton's law of universal gravity, $$F = G \frac{m_1, m_2}{r^2}$$, it seems more difficult to conclude the same result, simply because of the way Earth's mass overpowers the calculation. Moreover, it's not obvious to me what the distance $$r$$ is between me and the Earth.

Can this formula be used in a reliable way between two objects with such vastly different masses, and "wonky" distance?

Own thoughts

The reason I call it a "wonky" distance is because

• using a distance of $$r = 0$$ is out of the question. I can't divide by it, and besides, Earth probably doesn't pull with all its force emanating from the point on the surface closest to me

• using a distance of $$r = \text{Earth's radius}$$ also seems like an oversimplification. Presumably it's not just the very center of the planet that's pulling on me, but rather the entire thing, in different directions, with different magnitudes (verification needed)

I assume the formula would make more sense if I were in deep space, and the distance was such that the radius of the planet was almost negligible? Nevertheless, would the formula hold if I used sufficiently precise values for $$G$$ and the mass of Earth? Or is it near impossible because of some other factor?

Similar questions

I submit that it does not answer my question because the second object is the moon, which already has a distance that trivializes the radius of Earth, and my measly mass.

• If you consider the Earth to be point-like and yourself too that the force between you and the earth is given by the universal law. The actual force is not much different from this. Apr 3, 2021 at 0:57
• R is the distance between the centers of mass of the two objects, so in your case it would be rhe radius of the Earth plus a few inches. True, the mass if the Earth is huge, but G is extremely small. It all works out. Note that the acceleration of gravity at the surface of the earth, 9.81 meters per second squared, is the result you get if you divide both sides of the gravitational force equation ny your body's mass. Apr 3, 2021 at 0:58
• “It seems more difficult to conclude the same result.” $g$ is simply $GM/R^2$ where $R$ is Earth’s radius. Try calculating $g$ this way! Apr 3, 2021 at 1:57
• Note that it's really hard to measure G precisely, so we only know it to 5 digits or so. We normally use the standard gravitational parameter, $\mu=GM$, which is much easier to measure. Apr 3, 2021 at 2:06
• As the top comment on that linked question suggests, you should take a look at the shell theorem. Apr 3, 2021 at 2:12

## 1 Answer

Short answer: you use the radius of the earth.

Long answer:

You are correct that it is every particle of the earth pulling on you at a different direction. But, if you use calculus to break up this net force into the combined resultant force of every particle on the Earth pulling on you (and you approximate the shape of the earth as an exact sphere), the net result is that "things far away pull weaker" and "things close pull stronger" exactly cancel out in such a way that, so long as you're not in a tunnel inside of the Earth, the net force on you is exactly the same as if it were the case that the earth was a point mass with all of the planet's mass concentrated in the center. In fact, this result ONLY depends on the spherical symmetry of the matter distribution, so this already takes into account things like "the core of the earth is denser than the crust"

• When you say "use calculus to break up this net force into the combined resultant force of every particle on the Earth pulling on you", are we talking about calculating Earth's center of mass, which will end up more or less at the center anyway? Because that would make a lot of sense.
– Alec
Apr 3, 2021 at 1:10
• @Alec: it would be a different "center", because we're averaging the gravitational force, not the mass. For the case of a perfect sphere, these two things coincide, but they won't for more complex shapes. Apr 3, 2021 at 1:24
• I would have thought they'd be the same, since the gravitational force seems to be directly proportional to the mass of any given particle?
– Alec
Apr 3, 2021 at 1:31
• The two calculations are quite different. Apr 3, 2021 at 1:54
• @Alec Sure, but the centre of mass of a collection of particles is $$\frac1M\sum_i m_i\mathbb{r_i}$$ where $$M=\sum_i m_i$$ which doesn't incorporate gravity's inverse square law. Apr 3, 2021 at 2:02