# How do objects reach terminal velocity?

I understand the basics of this principle: the force of gravity acting downward on an object becomes equal to the air resistance acting upward on the object because as the object speeds up, air resistance increases. My question was much more specific: I thought that the force on an object increases as the object falls because gravity between objects increases with proximity. In this case, if both force and drag are increasing, how will objects reach terminal velocity? Is it because the change in force isn't enough to counterbalance the change in air resistance?

• Gravity does not vary appreciably over any reasonable heights above the earth's surface- the gravitational force decreases as the square of the distance from the centre of the earth
– Sid
Commented Apr 22 at 5:40
• So it essentially decreases the farther the object is from the center of the earth? Sorry for the basic question but we have to learn this in class. Commented Apr 22 at 5:53
• Earths gravity is remarkably constant from the surface to the edge of the core. This is because the density increases just fast enough. Above the surface it follows inverse square law. Needless to say, the atmosphere varies much more with height than the gravity. Objects start slowing down after a point in their free-fall because they are encountering denser and denser air. Commented Apr 23 at 2:05
• How do objects reach terminal velocity? They don’t. They approach it. Commented Apr 23 at 5:54
• If you think that the trajectory of a ballistic rocket is a parabola and that water boils at 100°C on Mt. Everest, then falling objects approach a constant terminal velocity. Commented Apr 23 at 11:20

The increase in gravitational attraction as a dropped object gets closer to the earth is extremely tiny, because the total distance fallen is very small compared to the size of the earth. That effect can be safely neglected when computing, for example, the terminal velocity of the dropped object.

Usually in mathematical physics when two functions compete, we need to analyze which one converges faster.

Very roughly gravity with distance increases as $$F_G \propto 1/r^2$$ while drag force, assuming main ingredient of down acceleration is due to gravity,- increases as $$F_D \propto 1/r^4$$. So say object has traveled it's $$0.5~r_0$$ distance, in this case gravity will increase by $$\times4$$, while drag force by $$\times16$$. So drag force has still increased by $$\times 4$$ faster compared to gravity and hence terminal velocity will be reached.

• Thanks! I'm just wondering: how would an object remain at terminal velocity if these factors converge differently? Commented Apr 22 at 14:45
• @user386598 It gets more mind bending. If you're sitting or standing while reading this, you're at terminal velocity with the Earth. Commented Apr 22 at 15:49
• "Since the gravitational force is still increasing as the object nears the ground however...": no. I suggest that you consider dropping an object from the top of Mount Everest, assuming that you have well that does from the summit to mean sea level. Mean sea level is roughly 4000 miles from the centre of the Earth, the summit is 4007 miles from the centre. Now work out the ratio of the gravitational forces, i.e. (4007/4000)^2 at top and bottom... Commented Apr 22 at 23:12
• I'm sorry to say but this answer's explanation is entirely incorrect. The statement that $F_D \propto 1/r^4$ is not true. One way to see it can't be right is to imagine a hypothetical drag force that's proportional to the square root of the velocity instead of the square, $F_D\propto \sqrt{v}$. An object subject to such a force would still reach terminal velocity, but the reasoning used here would suggest that it doesn't. Commented Apr 24 at 17:15
• @AgniusVasiliauskas Of course you can say things about hypothetical forces; most of the mechanics questions on this site are about hypothetical situations. But if you prefer I can point out a specific mistake in the reasoning: the step $v^2\propto (gt)^2$ would be true if the total acceleration were $g$, but it's not: it's $g-F_D/m$. Or in the conclusion: if the object starts near terminal velocity, so $F_G\approx F_D$, then after traveling a distance 0.5r how can the drag force be 4 times as large as the gravitational force? Commented Apr 24 at 20:57

How do objects reach terminal velocity?

They don't. If the object is falling after being released, the velocity approaches, but never quite reaches the terminal velocity: the terminal velocity is a limit. On the contrary, if the object is fired downwards at a speed that exceeds the terminal velocity, the upward resistance exceeds the attraction of gravity, so the object will slow down, but will never quite drop below terminal velocity. (If you fired the object downwards at exactly the terminal velocity, you could expect that speed to be maintained).

• Do you think that's actually true in practice? Or merely in spherical vacuum cow theory? Commented Apr 23 at 14:04
• I suspect in reality that objects and air are sufficiently variable that in reality the object will arrive at and then oscillate erratically around any theoretically measured terminal velocity. Sometimes above it before slowing down; sometimes slowing down again, before speeding back up. Commented Apr 23 at 14:06
• "... theoretically measured ..." Do you do your measurements theoretically? How convenient for you. Commented Apr 23 at 19:05
• 😂 thank you. I did of course mean "theoretically calculated" :D Commented Apr 23 at 19:06
• ... yes? I'm not sure what point you're trying to make? Commented Apr 23 at 20:13

The drag force is proportional to both the velocity squared and the density of the air. The terminal velocity is:

$$v = \sqrt{\frac{mg}{\rho b}}$$

Where b is a constant, $$\rho$$ is the air density at a given height, and g is the gravity at a given height.

At sea level, $$g = 9.8 m/s^2$$ and $$\rho = 1.2 kg/m^3$$

At 8000 meters, $$g = 9.78 m/s^2$$ and $$\rho = 0.4202 kg/m^3$$

The change in gravity is only 0.2% while the change in density is 285.86%. So any increase in terminal velocity due to increasing gravity is dwarfed by the decrease in velocity due to increasing air density.

Terminal velocity is a stable equilibrium. If an object is going slower than terminal velocity, the drag force is lower than gravity and the object accelerates. If an object is going faster than terminal velocity, the drag force is greater than gravity and the object decelerates. Terminal velocity is the stable point in the middle.

In theory as one goes down, gravity is increasing and thus terminal velocity increases as well. It does take time to accelerate to this new terminal velocity, so there's some second order effects going on. On Earth this is negligible because the distance between the object and the center of the Earth changes by an astonishingly small percentage. In theory one could come up with exotic systems where the drag never quite catches up and the idea of terminal velocity falls flat. Off hand I think a tungsten cube falling into a black hole might meet those criteria, but things falling into a black hole are much more complicated (we'd likely need to invoke relativity to do the calculations).

Note that terminal velocity does not mean that an object magically reaches that velocity and stops. It's what we call a "first order effect" which ignores a lot of details, like chaotic airflow. As you progress in your physics career, you will learn when it is reasonable to use these simplifications and when it is not. If you're trying to figure out whether a penny thrown from the Empire State Building will kill someone, terminal velocity is a reasonable tool. If you're exploring the effects of a an airplane trying to get back to an airport without power at nearly stall speeds, you'll want to invoke more precise physics.

### Because gravitational force doesn't change that quickly

Suppose we have a 1kg mass at the surface of the Earth, and then take the same 1kg mass up to the edge of the Karman Line, 100km up. That's a generally accepted position for the top of the atmosphere, being high enough that no aircraft can reach it due to the basic lack of air.

The radius of the Earth is 6357 km at the poles and its mass is 5.972 x 1024 kg. This gives us a force on the mass of 9.86 N at the Earth's surface, 6357 km from the centre. (Note that I'm using the lower value for the Earth's radius because it gives the greatest possible difference on gravity in the calculations later - in other words, to give your concept its best possible shot.)

Now let's go up to the Karman Line (6457 km from the centre of the Earth). The force on the mass due to gravity up here is now 9.56 N. In other words, whilst air resistance changes from nearly zero to its full ground level value over that 100km distance, the force due to gravity changes by only 3%.

Suppose you don't go that high though. Let's say you're skydiving from 15,000 ft which equates to about 5 km up (or 6362 km from the centre of the Earth). The force due to gravity at this altitude is 9.85 N - only 0.1% difference from ground level.

There are certainly situations where this change in gravitational force would be relevant - if you're sending a spacecraft to the Moon, for example. But as you can see, freefalling in the Earth's atmosphere really isn't one of those situations.

• Hypothetically, if an object could fall long enough, would an object begin to slow down? If the air resistance changes so drastically, could it eventually create an unbalanced force pushing upwards on the object? Commented Apr 23 at 0:19
• @user386598 Look up reentry for space craft. They start going very fast , and use air resistance to slow down. Hell, freefall parachutists do when they pop their chute too, because suddenly there's a lot more air resistance. I'm not sure what you mean by "unbalanced" - the force is certainly pushing upwards to decelerate the falling object, but it's proportional to the speed so it can't cause the object to start going back up if that's what you mean. :) Commented Apr 23 at 6:47
• @user386598 There is always some matter and hence some friction, even in interstellar space and certainly in low Earth orbits. Low-flying satellites slow down due to air resistance and crash within years. Commented Apr 23 at 11:28
• Essentially, if an object exceeds terminal velocity, it slows down to that speed? I looked up reentry for spacecrafts, and it makes sense. Yet if an object fell at exactly terminal velocity, and the air resistance increased much more drastically than gravitational force, as demonstrated in multiple answers, would the object still remain at terminal velocity or being to accelerate at a slower rate? Commented Apr 23 at 14:37
• @user386598 There isn't a single terminal velocity though. What you have is terminal velocity at some air pressure, which decreases as your altitude goes down and air pressure goes up. If you come in faster than this, you'll start to get a net force decelerating you. If you're not going too fast and there's enough altitude, maybe you'll slow down to (approximately) some terminal velocity, and track that as you fall the rest of the way. If you're going really fast, of course, you'll scrub off some speed but impact the ground still going fast. Commented Apr 23 at 19:00

The usual mathematical analysis assumes gravity is constant, the air is still and has constant density and viscosity, and air resistance does not change with the object's orientation. In this case the object's speed approaches, but never quite reaches, the terminal velocity.

The answer from Stevan V. Saban has shown that during a fall the increase in atmospheric drag is greater than the increase in the gravitational force. This implies that the terminal velocity is greater earlier in the object's fall, and so the object may reach a speed that is greater than its terminal velocity will be later in the fall. In this case the object will be slowing down later in its fall.

If the speed of the object approaches the terminal velocity there will be a point when the difference is no longer significant. It needs to be remembered also that air is a real gas made of rapidly moving molecules, and eventually the difference will be even smaller that the effects of air currents, and eventually smaller even than changes caused by the impact of individual molecules.