# For Newton's gravitation equation, how do you account for planet size?

My daughter was doing a physics problem where she had to calculate the gravitational force of an object that was 250 km above the earth... She used the equation:

$$F=\frac{G\cdot m_1 \cdot m_2}{r^2}$$

For $$r$$, she used the distance between the two centers of gravity: 250 km + the radius of the earth (6371 km)

But I don't think this is correct. While it is true that the center of gravity of the earth is in the middle of earth, I would imagine that the effective center of gravity is actually closer...

Consider the same problem where your weight is a 1 kg weight 250 km above 'earth', but replace earth with two 1 kg weights spaced 6371*2 km apart, which are inline with the object floating above it. Note that the new earth has the exact same center of gravity as the old earth.

So, now using the equation:

$$F=\frac{G\cdot m_1 \cdot m_2}{r^2}$$

we get:

$$F=\frac{G\cdot 1\,\mathrm{kg}\cdot 2\,\mathrm{kg}}{(6371\,\mathrm{km}+250\,\mathrm{km})^2}$$

which resolves to: $$3.04423\times 10^{-24}\,\mathrm{N}$$

But, if instead we sum the forces of the two weights individually, we get:

$$F = \frac{G\cdot 1\,\mathrm{kg}\cdot 1\,\mathrm{kg}}{250\,\mathrm{km}^2} + \frac{G\cdot 1\,\mathrm{kg}\cdot 1\,\mathrm{kg}}{(12742\,\mathrm{km}+250\,\mathrm{km})^2}$$

which gives you: $$2.06801 \times 10^{-21}\,\mathrm{N}$$, which is not the same thing at all. So, I'm wondering how one would actually calculate the force of gravity on a 1 kg (point object) 250 km above the earth...

• The site standard for displaying mathematics is Mathjax. The basics are not too hard and it helps your question get answers if you use it as it makes it much easier to read and look well presented. It's also the standard on the Mathematics SE and some other sites here. May 19, 2021 at 15:40
• This sounds like a question that has been asked many, many times before. Has it? May 20, 2021 at 13:03
• May 20, 2021 at 13:12

In general, you cannot treat arbitrary mass distributions as if they were concentrated at the center of mass for the sake of calculating the gravitational force, as your example shows. You can, however, do this for spherically symmetric mass distributions if you're outside that mass distribution (even if the radial mass distribution is not uniform) by the shell theorem1. So for a point outside the (roughly) spherically symmetric earth your daughter's calculation is correct, but for a more general setup, one would need to do it more explicitly.

1. The same theorem also includes the interesting fact that if we have a hollow shell, the gravity inside that shell caused by the shell's mass is zero, while outside of it, it acts the same as if all its mass was concentrated in the center.

• I feel like we've maybe confusingly shifted from solid-ball sphere (like the Earth) to hollow-shell sphere halfway through this answer now? From the link: "A corollary is that inside a solid sphere of constant density, the gravitational force within the object varies linearly with distance from the center, becoming zero by symmetry at the center of mass." May 20, 2021 at 12:28
• @DanielR.Collins Reading it again, I agree. I kept the point about having to be outside, but moved the empty shell stuff to a footnote.
– noah
May 20, 2021 at 12:42
• I heard for sattelite trajectory simulation a multipole expansion can be used. Not sure about that though. May 20, 2021 at 18:00
• @lalala the real-world problems include gravity varying in magnitude and direction by location for various reasons including the Earth not being a spherical ball (it is oblate), as well as local density and mountains. This is substantial enough to affect the accuracy of GPS satellites and others May 21, 2021 at 15:36
• @lalala A monopole approximation is good to maybe three decimal places of accuracy at the 250 km altitude mentioned in the question. That isn't nearly good enough for even an amateur level satellite trajectory simulation. At a minimum, accounting the Earth's oblateness is a must. There are several satellites in low Earth orbit that take advantage of the Earth's oblateness, the sun synchronous satellites. May 22, 2021 at 16:53

So, you're right that it can matter.

In this case it doesn't, because of a marvelous cancellation which may seem less marvelous after you see the “Gaussian pillbox” approaches of a classical electromagnetism course.

In this cancellation, we discover that if you spread out mass evenly across a spherical shell, first off, the gravitational force inside that shell is zero no matter where you are inside of it. (At least, due to that shell. Other masses could be inside or outside of the shell and could be exerting forces on you of course.)

The “Gaussian pillbox” arguments say essentially that “all of the field lines describing the force have to go out to infinity, because we do not have negative mass, and they can't go into the sphere in any easy spherically-symmetric way without flowing into a negative mass at the center of the sphere. So because we don't have a negative mass they must just all flow outward in a spherically symmetric fashion.”

That is the other part of the argument, it turns out that the gravitational force of the sphere on anything outside of the sphere, is mathematically exactly the same as if you crunched all of the mass into a single point at the center of the sphere. Other mass distributions don't happen to do this, but the spherical shell does.

In turn, any mass distribution that's kind of like an onion—nested concentric spherical layers of homogeneous density, can be built out of these spherical shells and have basically the same property. The field inside the onion is the field from all of the layers underneath you, as if they were compressed to a point mass at the center; the layers above you cancel themselves out. The field outside the onion is the field that you would get if you compressed all of the onion to a point mass at the center.

In particular a homogeneous sphere has a force inside of it that goes like $$r^3/r^2\propto r$$ as the volume of “the layers beneath you” goes like $$r^3$$ and the force laws goes like $$r$$, so if you could tunnel through the Earth, and it was approximately homogeneous and not rotating, you would experience simple harmonic motion between the point you stepped in and its antipode, same as the spring force.

These sorts of field line arguments only work for these $$1/r^2$$ force laws that you happen to see in both gravity and electromagnetism. Change the fouce law and you don't get this nice behavior. But it's really a wonderful cancellation and you would not have expected such a complicated integral to have such a simple result if you hadn't known to look at the force lines.

As others have said, she is correct.

In general, you can’t just substitute a shaped object by two or three objects with the same centre of gravity. That's why your suggestion fails. The gravitational effect of earth on an object is obtained by adding up the vast number of tiny gravitational effects of every last particle that takes up earth, towards every last particle that makes up that object.

Adding up a near-infinite number of near-zero items, is the realm of calculus (integration), and so in a real-world situation, you'd have to integrate the gravitational force, over the volume of the bodies that attract each other.

However in simple homework, and for many many real-world situations, we can simplify this by using two mathematical results.

1. In many cases, we can represent a relatively small or distant object, by a point object - it’s not precise but it’s virtually the same answer. So the gravitational force of earth pulling on a mountain, or sun pulling on earth, we can often replace by a point mass at the centre of gravity for simplicity. That's usual in homework-land, and also in many real situations.
2. The shell theorem, which says that any time you have a spherically symmetrical "shell" of matter (think the "skin of a tennis ball"!) and a force that works according to an inverse square law (like gravity does), then when you integrate the total gravitational force from it, you find remarkably, it has no net effect on any point inside it, and has the same net effect as the same point mass at its centre, on any point outside it. Since many objects in the universe are spherically symmetrical, we can treat objects like the earth, moon, and sun, as well as balls, like point masses at their centres. But that's only because they are spherically symmetrical; it’s not generally true. So it will not work for your two 1 kg objects 6371 x2 km apart. They do not act the same as a 2 kg object at its centre.

These two results simplify many problems, and explain why your daughter is right.

Consider the same problem where your weight is a 1kg weight 250 km above 'earth', but replace earth with two 1kg weights spaced 6371*2 km apart, which are inline with the object floating above it. Note that the new earth has the exact same center of gravity as the old earth.

The Earth is not a pair of 1kg masses, nor does it resemble this model in any way. If your calculations had resulted in a similar effect I would have been shocked. Two masses separated by distance will produce a very different result than a single mass, even if the center of mass of each system is the same.

The reason we can calculate the Earth's gravity as if it were a point source is that the math works the same for point masses, spherical shells, spheres of uniform density and layered spheres. As long as the mass distribution is spherically uniform (no big concentrations of mass in the body) then the math works out the same.

One way to demonstrate this is to simulate a shell with a cloud of smaller point masses distributed uniformly (or as uniformly as possible) on the surface of the shell. The higher the number of point masses the closer the result resembles the result for a single point mass. We can extend this to a layered collection of such shells, and as long as the total mass remains the same the answer is the same.

Of course in the real world the Earth is neither a uniform sphere nor a layered sphere, since the Real World is not very tidy. In reality the gravitational field of the Earth is a complex structure because of non-uniform mass distribution and so on. You just have to look at the Geoid map to see it.

Fortunately for the sanity of physics students the world over, we don't insist that they calculate the actual values for an object at a particular spatial location relative to Earth's Geoid. Not until long after they've already understood the basics anyway. We're not monsters.

• +1 For the mention that although the gravitational field of the Earth is approximately spherically symmetric enough for some purposes, we've actually measured the deviations from spherical symmetry quite well. May 21, 2021 at 18:16

Your daughter is correct. Let's not confuse her.

The mistake you are making is common enough.

While it is true that the center of gravity of the earth is in the middle of earth, I would imagine that the effective center of gravity is actually closer...

Your model of two objects of different mass is not the same as single object of the same total mass in the middle of the two.

$$F_{correct}=GM_0\left( \frac{M_1}{(r+d_1+h)^2} + \frac{M_1}{(r-d_1+h)^2} \right)$$

Your incorrect caluclation is the same as :

$$F_{incorrect} = GM_0\frac{M_1+M_1}{(r+h)^2}$$

You don't have to understand much math to see that these are completely different in form and will not in general be equal.

In effect what you did was prove that the model you made (two parts) is not the same as the single object of the same mass.

• What if you turn the model with two object into a model with one object, by connecting the two objects by an infinitely thin rod? It seems to me it's irrelevant whether it's two objects or one, but rather that mass distribution is of importance.
– towr
May 20, 2021 at 9:17
• The asker already knows this is what they proved and is asking why it doesn't apply to Earth, since it is not a single object, but many objects called atoms, whose total mass is the Earth's mass. This answer doesn't demonstrate why you can treat the Earth as one object at its center, and further, it seems to indicate you cannot. May 20, 2021 at 12:00
• @Vaelus I've simply pointing out to the OP that the model they selected cannot be equivalent to a single massive spherically symmetric object. The point being that they did good physics in comparing the results but reached the wrong conclusion. Note that the approach to this (using calculus for a spherically symmetric sphere) works precisely because it does not impose a new model for a sphere as two seperate spheres but treats a single sphere's parts correctly individually. The point is that changing the underlying model is the fault. May 20, 2021 at 15:41
• @towr This won't change the mass distribution (an infinitely thin rod of finite lenght has no mass, you can try it as an exercise for a non-trivially thin rod of some density) nor give it the spherical symmetry required for general equivalence. It is possible that for certain values of $d_1$ to make the model produce the same value but I think that equates to moving the parts so that the point being affected by them is in the middle of them. May 20, 2021 at 15:51

Another way of arguing that the radius of the Earth doesn't matter is to use Gaussian surfaces. These are generally introduced in the context of electrical charges, but they also apply to gravity. The gravitational force on a sphere around the Earth depends only on the mass of the Earth, not the distribution of mass of the Earth. In the hypothetical you mention with the mass distributed into a barbell shape, the distribution of gravitational force on the Gaussian surface will be different, but the total flux will be the same. If the distribution of the mass is spherically symmetric, then the distribution of flux on the Gaussian surface will also be spherically symmetric.

Actually your daughter is correct. For Newton's gravitational force you can consider the earth as a point mass having the whole mass (although you will encounter problems where e.g. explicitly the inner gravitational potential of the earth is calculated, but for these calculations of an object in the earth's orbit the assumption is fine).

If the two one-kilogram point masses (the ones you use in the new calculation) would each be point masses containing half the mass of the Earth, their center of mass lies exactly on the middle of the line connecting them. Their "effective center of gravity" (the point on which you can concentrate the two Half-Earths) lies somewhere on that line too but it doesn't coincide (except for two situations) with their center of mass. That is because the effective center of gravity is dependent on where you put the 1-kilogram point mass relative to the two Half-Earth masses. If the 1-kilogram mass is aligned with the Half-Earth masses, then the effective center of gravity will lie closer to the Half-Earth mass that is closer to the 1-kilogram mass. When you rotate the 1-kilogram mass around the center of mass of the Half-Earths (keeping the distance to the center of mass to it fixed, while the three masses lie in one plane), the effective center of gravity will move to and fro wrt the center of mass. They only coincide two times in one whole rotation of the 1-kilogram mass (when the 1-kilogram mass has equal distances to both Half-Earth masses).
You can say that the effective center of gravity is a relative point (only to be defined if, in this case, a third mass is present) while the center of mass is an absolute point. If the mass distribution is spherically asymmetric, as for the two Half-Earths point masses, the effective center of gravity will change if you change the position of the 1-kilogram mass. Only for a spherically symmetric mass distribution (like the point mass emerging if you make the two Half-Earths coincide at one point or the mass distribution of the real Earth), the center of mass and the center of effective gravity will always coincide, no matter where you place the 1-kilogram mass (even if you place it inside the mass distribution).

So, even though the half of the Earth closest to the 1-kilogram mass contributes more to the gravity "experienced" by the 1-kilogram mass, the effective center of gravity will reside exactly in the middle of the Earth (in contrast to the effective center of gravity of two Half-Earth point masses if they are aligned with the 1-kilo mass).
An even better way to see this is by considering the contribution of the two halve pieces of the Earth that are at equal distance to the 1-kilogram mass (perpendicular to the halves close and far from the mass). In that case, the effective center of gravity lies clearly in the middle of the Earth (like was the case two times when the 1-kilogram mass rotated around the two Half-Earth point masses, namely when that mass had the same distance to both Half-Earth point masses). Because there can only be one effective center of gravity (for a given position of the 1-kilogram mass), the center of effective gravity for the halves of the Earth that are close and far from the 1-kilo mass must lie in the middle of the Earth too (so not closer to the Earth's surface).

So you are making the mistake to assume that the center of effective gravity doesn't lie at the center of the Earth. That is, your daughter is right.