While we usually think of the gravity due to planets when we reside on their surface or orbit around them, it is intriguing to consider the role that gravitation plays inside a large body.
Newton's law of universal gravitation states that all bodies having mass emit a gravitational field and thereby attract all other masses to them. The magnitude of the attractive force (which, by Newton's third law, acts on both bodies) is governed by the mass of each body, as well as their proximity to one another. The force is stronger for more massive bodies, and increases as the gap between them reduces. In equation form, the magnitude of the gravitational force, Fg (in Newtons), that acts on both bodies is given by
Fg = GM1M2/r2 (refer to figure below)
Here, G is known as the universal gravitational constant (6.673 × 10-11 m3/kg s2), M1 and M2 are the masses of each of the bodies (kg), and r is the distance that separates them (m).
You might be wondering about that distance between bodies, r. In the case of the Earth, which you are currently stationed on, are you not simply zero distance away? Well, not really.
As indicated in the figure above, we measure the distance between bodies as the distance between the center of mass of each body. This is because each body is composed of billions of tiny particles, and while the law of gravitation applies to each and every one of these particles on an individual basis, we can assess the net force collectively using the center of mass approach. Taking the Earth as a spherical body with a radially symmetric density variation (good assumption), the center of mass of the Earth coincides with its geometric center. So, for any body on or near the surface of the Earth, we say its radial distance away is one Earth radius (R = 6,378,000 m).
We see then that the gravitational force that the Earth exerts on any body that sits atop its surface is given by
Fg = mg where g = GM/R2
Here, we have replaced three constants associated with this situation by the surface gravity, g. Substituting values for the universal gravitational constant, the radius of the Earth, as well as its mass (which tips the scale at a hefty 5.97 × 1024 kg), we get the familiar surface gravity of 9.8 m/s2.
As we see, determining the gravitational force acting on a body near the surface of the Earth is trivial. It is also easy to assess for a distant orbiting body: in this case, we just need to adjust the parameter r to be the radial distance from the center of the Earth to the orbiting body.
Things get a bit trickier if we examine the gravitational force acting on a body that resides at some location inside the Earth. Admittedly, such a calculation is seldom necessary, but is a neat mental exercise nonetheless.
The first time I ever considered this problem, I immediately reasoned that Fg would simply increase the deeper I dug into the Earth. The logic I employed was that as you get closer to the center of the Earth, r decreases, and Fg increases. This notion is however false; it breaks down the moment you enter a body. From within a body, we need to recall that each particle of the Earth acts on us individually. As we descend, more and more of Earth's mass is above us, and is conspiring to pull us back up. Basically, Earth is fighting over you, and the net gravitational force that acts on you is lower as a result.
If we make the assumption that the entire Earth has uniform density (not a great assumption, but it leads to a neat result), then it can be shown that the gravitational force acting on you when stationed inside the Earth is actually proportional to r, your distance from its center. This is astonishing. If you are a falling body, the gravitational force is proportional to 1/r2 until the moment you enter it at which point it is simply proportional to r.
Let us imagine that there were a direct tunnel through the Earth from the North to the South geometric poles. As you step into this tunnel (at either end), the force that acts on you at the surface is mg, while the force that acts on you at the center of Earth is zero (the gravitational effect of all of Earth's particles cancel out there). Throughout your journey to the center of the Earth, Fg decreases linearly, so it is mg/2 when you are halfway there.
This tunnel is thus one where simple harmonic motion can ensue (your motion would follow a perfect sine function, again assuming uniform planet density and neglecting air resistance and such). If you are Bugs Bunny living in America, and you fall into a tunnel through the Earth, your head does eventually pop up in China, but you then begin your journey through the tunnel in reverse. Just like a pendulum or spring-mass system, you hit your maximum speed in the equilibrium position, where the conservative force responsible for the motion disappears.
(If you have completed a full year of post-secondary physics, now might be a fun time to take a break and check out the following: if the Earth did have uniform density, and motion through it were indeed uniform, (1) determine how long a round yet extremely direct trip from America to China and back would take, and (2) calculate the speed Bugs would have as he passes through the center of Earth. Let me know the answers you found if we ever cross paths)
In reality, the density of the Earth does vary radially, and so Fg is not exactly proportional to r when inside it (Earth is denser and denser the closer you get to its center). Still, the underlying principle that the net force acting on us decreases the more we descend remains true.
If you were to teleport to the center of the Earth, you would feel weightless, and then promptly vaporize as the temperature there exceeds 5000 deg C.
So, there you have it: some fun insight into stuff you will probably never need to know. But, whether we physically embark on such an adventure through our planet or not, tunneling into the depths of our minds is most often a worthwhile adventure.
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