Calculating where dropping a lead ball from orbit would land, if dropped perpendicular to the Earth, under very simplified conditions

Question

This question has a lot of "what if"s, so let's simplify this:

• Assume Earth is perfectly spherical

• Assume no wind or anything, just plain atmosphere that is generally uniform (but will change density as the ball falls through the atmosphere, but lets pretend there's no clouds or air currents)

• The lead ball is pretty heavy and does not degrade in the atmosphere at all because its "Magic lead"

Suppose we're in a spaceship that is maintaining orbit around a particular patch of ground on the planet (say some 1x1 meter chunk on the surface of the planet). In other words, if we shoot out a surface normal from this 1x1 patch, it will point directly at the spaceship if we extend the vector along its normal. Further, assume the spaceship is maintaining the perfect velocity at some height $h$ to keep it constantly in line with this surface normal (which may require it to be constantly providing thrust to maintain this, but I am unsure).

We decide to drop a lead ball out of the bottom of our ship.

Is there a way to calculate where it would land, given some height $h$ which is the distance from the ground to the ship? Would the planet pull it down towards it, or would it stay in orbit? We can assume we didn't push it out of the hatch or anything towards the planet, but rather 'let it go'.

Things I don't know are what the range is such that an object will always be pulled in (instead of orbit), and whether or not the speed of the orbiting ship would leave it in orbit.

It would be cool to know if this could be abstracted to a formula like height $h$ off of the surface, and radius $r$ of the planet, and some constantly decreasing atmospheric density $\delta$ (relative to $h$) that may or may not need to be integrated over to affect the calculations.

Answer 1

We decide to drop a lead ball out of the bottom of our ship.

It is one of these persistent urban myths that you can 'drop' an object from a satellite and that the object will then neatly fall to Earth (presumably at great speed!)

But you can't.

Assume you achieve separation between the satellite and the object, then, due to conservation of momentum, the object would become a new satellite which would orbit alongside the first satellite. At the same height and with the same orbital velocity, in fact.

The only way to make the object approach Earth is to apply relevant thrusters on it. It will not free fall to Earth.

Consider also this: if it was otherwise, then every space walk from a Shuttle or the ISS would end in bloody disaster!

Answer 2

We decide to drop a lead ball out of the bottom of our ship.

Would the planet pull it down towards it, or would it stay in orbit? We can assume we didn't push it out of the hatch or anything towards the planet, but rather 'let it go'.

If you just dropped a lead ball out of the bottom of ship without pushing it or anything, then the lead ball , would not fall towards earth. It will stay where it is.

Remember the videos from ISS, where the astronauts let some object float near their face, and it just stays there ? It does not fall towards the floor of the ISS, right ?

Similarly, if you just let a lead ball go , it will just float there i.e. it would stay in orbit

Answer 3

About orbiting at low altitude:
The International Space Station orbits at an altitude of about 400 kilometers. At that altitude there is almost no atmosphere, but it's not quite interplanetary vacuum. There is a little bit of drag on the space station, and every couple of months small thrusters are fired for a couple of hours, to regain altitude that was lost due to orbital decay.

So if an object would be released from the ISS - low Earth orbit - then over time the orbit of that object will decay.

The re-entry process is a self-reinforcing process. Initially the decay is very slow; in the first months there will hardly be any change. As the object descends: the density of the atmposhere increases with lower altitude, so there is more friction, so the object descends faster, which makes it encounter denser atmosphere and suddenly things are changing very rapidly.

Even if you simplify everything else by making everything else uniform, there is no way around that self-reinforcing process.

Two simulation runs, started with initial conditions very close to each other, are likely to end up at quite different parts of Earth.

For a simulation run you would need a precise expression for the amount of air friction as a function of atmospheric density, atmospheric temperature, and velocity of the object. It may well be that such data are not available.

Rocket builders have (presumably) fairly good drag characteristic data for a specific shape: the shape of the rocket that they are in the business of firing into space, gathering telemetry every flight.

Presumably you will not have good data for the particular object whose trajectory you want to develop a simulation for. So you have to do multiple runs of the simulation, allowing for both lower bounds and upper bounds for the amount of drag that you expect. Those unknowns will throw off any aim.