Economically:
Economic arguments and business plans can range in both honesty and subjectivity, but I'll outline a hypothetical estimate.
This quora post written by Peter Hand, VP of Western Pyrotechnic Association puts the cost of a 20 minute fireworks display anywhere from $20,000 to a million dollars, depending on the level of firework used:
For a professional display, you’re probably looking at $10,000 to 20,000. There’s a lot of work for a lot of people in the setup, and a lot of capital tied up in the mortars and firing systems. On the other hand, professional shells are relatively inexpensive compared to the retail consumer kind, with a case of 64 three inch shells often no more expensive than a box of two dozen consumer shells from a retailer. Most professional displays use larger shells, like six inches and higher, and these can cost upward of 300 [dollars] each. If you spend the money, you’ll get these in your display, otherwise it will just be 3 and 4 inch shells. It takes a lot of these small bursts fired at once to fill the sky, which is rather dismissively called “Chinese carpet bombing”. I wouldn’t call that a “decent quality” firework display.
Some really outstanding 20 minute displays have cost upward of a million dollars, but these often cover a huge area and use many custom shells, some 24 inches in diameter and needing 60 pounds of gunpowder to lift. (emphasis added)
The website ReactionFireworks.co.uk says in its faq (in part):
As a guide an organisation such as a Round Table or Parish Council typically spends between £3,000 to £6,000 for a 15 to 20 minute pyromusical.
Major Guy Fawkes displays as seen in many cities accross the UK over November would typically cost between £10,000 and £16,000 and often have audiences in excess of 20,000.
Larger displays seen in such places as Battersea Park, Edinburgh Castle and the London Eye can range from between £30,000 to £200,000.
The record for the most expensive firework display is held by Kuwait for the 50th anniversary of the ratification of Kuwait’s Constitution in November 2012. The display cost a reported £10,000,000 !
So let's say the spacecraft carried enough "stars" to do 100 shows a year for ten years with one hundred 2 gram shooting stars per show. There are certainly that many fireworks displays happening on Earth, but the question of how many would shift to this new technology is... subjective.
Then suppose that they grossed on average $100,000 per show (ranging from personal single shooting stars to million dollar events) that's a gross income of one hundred million dollars, and that could at least conceivably pay for the design, construction, launch, control and orbital maintenance of a ~70 cm, maybe 500 kg nano satellite for ten years, considering the way costs and technology is evolving and beginning to standardize. It's in the right ballpark at least.
Technologically:
Let's try this using the vis-viva equation:
$$v^2 = GM_E\left(\frac{2}{r}-\frac{1}{a}\right),$$
the equatorial radius of the Earth 6378 km or $6.378\times10^6 \ meters$, and the Standard gravitational parameter of the Earth $GM_E = 3.986\times 10^{14} \ m^2s^{-3}$.
The semi-major axis of the satellite in a circular orbit with an altitude of 500 km is $a_0 = 6.878 \times 10^6 \ meters$. Using the vis-viva equation, that gives an initial velocity $v_0 = 7613 \ m/s$. As a check, the period would be $2\pi a_0 / v_0 = 5677 \ sec$ or about 94.6 minutes.
An educated guess would be that a targeted perigee somewhere between about 80 and 100 km would be guaranteed prompt, localized re-entry. This bars oddly or aerodynamically shaped objects or plane-shaped objects, but small spheres at this perigee and this kind of velocity will re-enter and burn up.
What delta-v is needed to do this? If we assume the orbit of the recently ejected ball has its apogee at 500 km altitude and its perigee at 80 km, then the new semi-major axis is now $a_1 = 0.5 \times (6.878 \times 10^6 + 6.458 \times 10^6) = 6.668 \times 10^6 \ meters$. Plugging the new semi-major axis $a_1$ and maintaining the apogee of 500 km into the vis-viva equation, the new velocity required for a re-entry orbit would be $v_1 = 7492 \ m/s$, a delta-v of 121 $m/s$.
This is a minimum. You may want substantially greater delta-v for improved re-entry targeting accuracy.
Using algebra, one can use $x=\frac{1}{2}at^2$ and $v=at$ to obtain $a=v^2/2x$. To get a velocity of 121 meters per second inside a 50 centimeter nano satellite, you need an acceleration of almost 1500 gees.
The satellite is a gun. It fires deer slugs with a muzzle velocity of at least 400 feet per second, ideally much more. It carries of order ten thousand rounds of ammunition or more, in order to be economically feasable. It can get at you, but it's darn hard for you to get back at it, unless you build a bigger gun, and we know how that story goes...
The projectile velocity relative to you and yours could easily be a few thousand meters per second or ten thousand feet per second depending on the inclinations of the two orbits, and so the impact energy would be enormous if one of these hit anything.
But don't worry, it's "not supposed to" do that.
Still, the fact that the spacecraft is effectively a gun with ten thousand rounds could conceivably make the whole project moot. If it were struck by space debris, it could release ten thousand radar-invisible projectiles into short-term stable LEO orbits above the orbit of the ISS and Taingong-1 and Tiangong-2, and this possibility could make this project highly objectionable to any country associated with any current or future manned space mission.
