Let's say you were drifting along in a hot air balloon at a ground speed of 50 knots at an altitude of roughly 10,000' MSL/AGL. The airmass is uniform, moving at the same speed at all altitudes. In your hand is a streamlined, very dense object. Picture a 5 pound chunk of lead shaped like a torpedo.

If you were to drop this object (over an unpopulated area of course!) would it continue to track along the ground at the exact same speed as the airmass/balloon, or would the vertical acceleration due to gravity overcome the horizontal inertia, slowing the groundspeed and causing it to arc towards a trajectory more perpendicular to the ground?

In other words, would the lead impact the ground directly beneath the hot air balloon, or at some point "upwind"?

My gut tells me that it would fall almost straight down immediately, with only some slight sideways drift, but I don't know how to quantify that hunch.

FYI, this question has its origins in the discussion here: Force Balance From Wind

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    $\begingroup$ demonstrations.wolfram.com/TrajectoryOfABomb $\endgroup$ – Ron Beyer Oct 25 '19 at 19:59
  • $\begingroup$ Neat little graphic, but irrelevant and false. I will explain when I have more time. $\endgroup$ – Michael Hall Oct 25 '19 at 20:15
  • $\begingroup$ Draw the force vectors first (and trust them). A vertical force vector will not act on a horizontal one. Relative to the air mass, there is no horizontal motion anyways. The groundbased trajectory is based only the horizontal velocity of the balloon (steady state, not accelerating), and the vertically accelerating falling object, producing an increasingly downward (ballistic) path. If the "balloon" is out of the atmosphere and travelling fast enough, the dropped object will orbit. $\endgroup$ – Robert DiGiovanni Oct 25 '19 at 21:23
  • $\begingroup$ @Michael Hall Yep, for a powered plane (luckily for it) the "bomb" will fall behind it. $\endgroup$ – Robert DiGiovanni Oct 25 '19 at 21:27
  • $\begingroup$ I felt it was safe to presume a uniform airmass because of the obvious effect, but for posterity I will edit. Thanks for the answers, turns out my gut was wrong! And thanks to Ron for the video; by recognizing immediately that it was wrong it helped me to look at this scenario from another angle, and to understand where my misconception originated. Paradigm adjusted. $\endgroup$ – Michael Hall Oct 26 '19 at 15:23

This answer assumes that the wind is uniform in direction and speed, from the balloon all the way down to the ground.

The object will impact the ground directly below the balloon, not "upwind" of the balloon.

As the falling object accelerates vertically, the trajectory of the falling object as viewed from the ground reference frame continually becomes more and more perpendicular to the ground. Yet the horizontal components of airspeed, force, and acceleration are all zero throughout the entire fall, so the falling object remains directly under the balloon.

As viewed from the airmass reference frame, the trajectory of the body is purely vertical, i.e. completely perpendicular to the ground, throughout the fall.

The horizontal component of the object's groundspeed is constant throughout the fall, and equal to the wind speed.

Now imagine simultaneously dropping a low-density object and a high-density object from the balloon. Both would be directly below the balloon at the moment of impact, but they would impact at different times.

For more, see Force Balance from Wind

Also, note that saying the object will tend to fall straight down relative to the earth, rather than relative to the balloon and the airmass, is essentially saying that the earth is a privileged reference frame, which goes against the laws of physics.

  • $\begingroup$ I thought you might chime in! ;) Looking forward to hearing from some dissenters... $\endgroup$ – Michael Hall Oct 25 '19 at 19:47
  • $\begingroup$ @MichaelHall, the only thing to dispute here is the mention of density as you can have two objects of the same density, but still with very different drag (e.g. a solid rod and a wide open cone). And of course it only applies as long as Earth curvature is negligible over the range, but at scales where it isn't, we would be talking about space reentry and balloons wouldn't be involved. $\endgroup$ – Jan Hudec Oct 27 '19 at 21:42

Newton's first law of motion states:

An object in motion stays in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

The instant you let go of the object, it has 50 kt sideways motion, and 0 kt vertical motion.

First, let's assume this happens in a vacuum, so only gravity applies. This is essentially a classic projectile motion problem, where you are firing your object at 50kt from a height of 10,000 feet. The gravity applies perpendicular to the sideways motion, so it has no effect on that initial velocity. The two velocities can be considered separately. Using the equations of motion for the vertical component:

$10000 ft = 0.5 * 32.17405 \frac{ft}{s^2}*t^2$

Solving this for $t$, it takes about 25 seconds to fall from 10,000 feet. To find the final velocity:

$v = 32.17405\frac{ft}{s^2} * 25s$

This gives a speed of 802 ft/s, or 475 kt. In that time, it has continued to move at 50kt along the ground, same as the balloon.

$s = 84\frac{ft}{s}*25s$

This puts the impact point about 2100 feet away from where it was directly above when it was dropped.

But this is assuming there is no drag, and this is Aviation.SE after all. Let's see what the drag might be. Let's assume the drag coefficient for a streamlined object is 0.05 and the reference area is 0.5in2.

$D = 0.05 * .002373 \frac{slug}{ft^3} * 802\frac{ft}{s}^2 * \frac{1ft^2}{2*24^2}$

We end up with 0.066 lbf of drag from the vertical speed at the bottom of the fall. To find how that affects the acceleration of the object:

$a = \frac{0.066 lbf}{0.155 slug}$

This results in 0.427 ft/s2 which is only 1.3% of the acceleration due to gravity. For such a streamlined object, the drag will have only a small effect of slowing the fall. If we assume the wind is blowing at 50kt with the balloon and is constant with altitude, the object will continue to move with the wind, experiencing no drag in this direction, and will remain directly below the balloon until impact.

You'll notice the drag depends on the shape of the object, and the acceleration from that drag force depends on the mass of the object. Changing either one will have an effect on the trajectory of the object. An object with lower mass or higher drag will be affected more by drag, taking longer to reach the ground. If the object is moving with the wind, there is no drag acting against the lateral motion, so regardless of the time taken to fall, it will remain directly below the balloon.


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