# If I weigh 150 pounds on earth, does the earth weigh 150 pounds on me?

Weight is mass times gravity. My gravitational force is a very small amount compared to the earth. So if I weigh 150 pounds on earth (my mass * earth gravity), then shouldn’t the earth weigh 150 pounds on me (earth’s mass * my very small gravity)?

Many people said that it is so, but, using math, I've only been able to disprove it.

To work with 2 simpler objects and not a human and earth, let's say we have two spheres, one with a mass of 10^24 kg, the other with a mass of 50 kg, and both with a density of 5 g/cm^3.

$$m_1 = 10^{24} \textrm{kg}\\ m_2 = 50 \textrm{kg}\\ p = 5 \frac{\textrm{g}}{\textrm{cm}^3}$$

Using this information, we can calculate the radius of both objects.

$$p = \frac{m}{v}\\ p = \frac{m_1}{v_1}\\ v_1 = \frac{m_1}{p}\\ v_1 = \frac{10^{24} \textrm{kg}}{5 \frac{\textrm{g}}{\textrm{cm}^3}}\\ v_1 = 2 \times 10^{26} \textrm{cm}^3\\ v = \frac{4}{3} \pi r^3\\ r_1 = \sqrt[3]{\frac{3v_1}{4 \pi}}\\ r_1 = 3.628 \times 10^8 \textrm{cm}\\ v_2 = \frac{50 \textrm{kg}}{5 \frac{\textrm{g}}{\textrm{cm}^3}}\\ v_2 = 10^4 \textrm{cm}^3\\ r_2 = \sqrt[3]{\frac{3v_2}{4 \pi}}\\ r_2 = 13.365 \textrm{cm}$$

Next, we can use the masses and radii of both objects to figure out their gravity.

$$g=\frac{Gm}{r^2}\\ g_1 = \frac{G \times 10^{24} \textrm{kg}}{(3.628 \times 10^6 \textrm{m})^2}\\ g_1 = 5.070 \frac{\textrm{m}}{\textrm{s}^2}\\ g_2 = \frac{G \times 50 \textrm{kg}}{(0.13365 \textrm{m})^2}\\ g_2 = 1.868 \times 10^{-7} \frac{\textrm{m}}{\textrm{s}^2}$$

Finally, we can figure out what each of the objects weighs on each other using the information that we have already.

$$w = m \times g\\ w_{\textrm{2 on 1}} = m_2 \times g_1\\ w_{\textrm{2 on 1}} = 50 \textrm{kg} \times 5.070 \frac{\textrm{m}}{\textrm{s}^2}\\ w_{\textrm{2 on 1}} = 253.5 \textrm{N}\\ w_{\textrm{1 on 2}} = m_1 \times g_2\\ w_{\textrm{1 on 2}} = 10^{24} \textrm{kg} \times 1.868 \times 10^{-7} \frac{\textrm{m}}{\textrm{s}^2}\\ w_{\textrm{1 on 2}} = 1.868 \times 10^{17} \textrm{N}$$

Unfortunately,

$$w_{\textrm{1 on 2}} \neq w_{\textrm{2 on 1}}$$

Was my method wrong? Were my calculations themselves wrong? From everything I've done, I can't get the wights to equal each other.

• You are taking the distance to the interface, not to the center of mass. Gravity is a body force, it applies to each part of the body. Either approximate with the center of mass, or perform the full integral.
– Emil
Commented Oct 1, 2022 at 22:11
• (I am assuming they are rigid objects)
– Emil
Commented Oct 1, 2022 at 22:18
• Commented Oct 2, 2022 at 0:22

The forces are equal and opposite. This can be deduced from Newton's third law, which would preclude e.g. a force law proportional to one mass times the other mass squared. Of course, compared with Earth's gravity all other gravitational forces on you are negligible, but the Earth experiences many gravitational forces from other objects, which add and largely cancel.

In the formula for the gravitational force between two spherical objects $$F = \frac{GMm}{r^2},$$ the distance $$r$$ is not the radius of either object, but the distance between their centers. This will be the same in both force calculations since the distance from the center of $$m_1$$ to $$m_2$$ is the same as from $$m_2$$ to $$m_1.$$ The proper distance to use in the gravitaional force equation is $$r = r_1 + r_2.$$ Since $$r_2$$ is tiny compared to $$r_1,$$ we can use $$r = r_1.$$ This results in $$g_2 = \frac{G\times{}50\,\textrm{kg}}{(3.628\times{}10^8\,m)^2} = 2.5354\times 10^{-26}\,\textrm{m/s}^2$$ $$w_\textrm{1 on 2} = m_1 \times g_2 = (10^{24}\,\textrm{kg})\times(2.5354\times 10^{-26}\,\textrm{m/s}^2) = 253.54\,\textrm{N} = w_\textrm{2 on 1}$$

While it is true that according to Newton's Third Law, you do exert an equal and opposite force to your weight on the Earth, it may be inappropriate to call it a weight. Why? It's merely a matter of semantics, as when one typically raises such a question, they assume themselves to be in the interior of the planet.

In particular, you are on the surface of the Earth as you consider your weight to be $$w = mg$$, where, $$g = GM/R^2$$. Here, $$G$$ is the gravitational constant, $$M$$ is the mass of the earth, $$m$$ is your mass and $$R$$ is the radius of the Earth. Notice now, that the Earth is not on you and moreover, $$g_m = Gm/R^2$$ is very small. Thus, the equal and opposite force due to you acting on the Earth is primarily by virtue of it's own mass since $$F = Mg_m$$.

Thus you weigh on Earth due to its gravity but the Earth doesn't weigh on you as your gravity's effect on it is negligible.

Assume that all the mass of the earth is at the center of the earth and you are at distance $$r$$ from this center.

Assume the equivalence of gravitational mass $$m$$ and inertial mass $$M$$.

Let $$g=\text{acceleration of you}$$ and $$h=\text{acceleration of earth}$$.

$$F=\frac{G m_{earth} m_{you}}{r^2}$$

$$F=M_{you} g = m_{you} \frac{G m_{earth}}{r^2} \rightarrow g = \frac{G m_{earth}}{r^2}$$

$$F=M_{earth} h = m_{earth} \frac{G m_{you}}{r^2} \rightarrow h = \frac{G m_{you}}{r^2}$$

Because we are assuming that $$M_{you} = m_{you}$$ and $$M_{earth}=m_{earth}$$.

$$\frac{h}{g}=\frac{\text{acceleration}_{earth}}{\text{acceleration}_{you}} \approx \frac{1}{100000000000000000000000} = 10^{-23}$$

Notice that the force on both objects is given by $$F=\frac{Gm_{you} m_{earth}}{r^2}$$. This is the weight that you measure on your scale. But how the forces are experienced by each are not the same because their masses are not the same.

If you get hit by a truck, sure the forces are equal and opposite, but you are less massive than the truck so the force has more impact on you.

• I wouldn't write this uncountable (by an average reader) number of zeros on this site. Commented Oct 3, 2022 at 8:51