3 ms·

> I'd say there's around 10 orders of magnitude between the momentum of a rifle bullet and the momentum of a communications satellite at escape velocity. That'

> I'd say there's around 10 orders of magnitude between the momentum of a rifle bullet and the momentum of a communications satellite at escape velocity. That's quite an overestimation. Let's be generous and say a rifle bullet is 50 gram (0.05 kg) and fires at 1.1 km/s, and a communication satellite weighs 5 metric tonne (5000 kg) and escape velocity is 11 km/s. We have 5 orders of magnitude difference in mass and 1 in velocity. As momentum is m*v, total difference is 6 orders of magnitude. Kinetic energy is 1/2*m*v^2, so that gives 7 orders of magnitude difference.*

db48x 6y ago
I’m not so sure that it is. The mass of a bullet is more like 5 grams, and the mass of a satellite could be around 500,000 kg (the ISS). A quick search shows the the largest commercial communications satellite is Telstar 19V, with a mass of 7,076kg; in my first estimate I guessed 50,000kg. Finally, the closing velocity of two objects in orbit could be up to 22.4 km/s. rifle bullet: 3×10³ J Telstar 19V, geostationary orbit: 3.3×10¹⁰ J Telstar 19V, geostationary transfer orbit: 3.5×10¹¹ J ISS, LEO: 1.2×10¹³ J hypothetical fast ISS: 1.1×10¹⁴ J 14−3 gives 11 orders of magnitude, but good luck making the ISS go that fast. Also, don’t forget to add in the kinetic energy of the net; it’s not going to be zero. Still, even 7 orders of magnitude qualifies as ”quite a few”. You don’t need very many orders of magnitude before you have too many.
- Denvercoder9 6y ago
  The ISS is not a communications satellite, escape velocity is lower than maximum closing velocity, and kinetic energy isn't momentum. Of course you can make the statement true by changing everything it's about.
  - db48x 6y ago
    Forgive me for contradicting you, but I didn’t change anything; I just typed momentum when I meant energy. I first thought of the formula, not the name, and then typed the wrong name. I'm not sure how I made such a mistake, but mea culpa. Also my old friend bzbarsky mentioned closing velocity, and I stuck with it when I made my estimate. I do agree that the ISS is not usually considered to be a communications satellite (though it does have plenty of communications gear on board, including a ham radio repeater if I recall correctly). But it is a thing that we might want to deorbit one day. Since we were considering the ridiculous closing velocity of two objects that are both on escape trajectories, I figure using the ridiculous mass of the ISS was fair game too. But like I said, I originally guessed 50t, not 400t. I was thinking more about the payload capacities of launch vehicles than the satellites themselves. Whatever mass and closing velocity you consider, 7 or 10 or 11 orders of magnitude all qualify as “quite a few” even if the velocity itself only has 1.3 orders of magnitude of difference.
- stopChanging 6y ago
  I mean you stretched the numbers as much as possible (garbage collecting the ISS? LOL!) and you still didn't get there. And anyway, none of this matters, because the relevant OP specifically said "speed". So the plain and obvious conclusion is: False.

db48x 6y ago

I’m not so sure that it is. The mass of a bullet is more like 5 grams, and the mass of a satellite could be around 500,000 kg (the ISS). A quick search shows the the largest commercial communications satellite is Telstar 19V, with a mass of 7,076kg; in my first estimate I guessed 50,000kg. Finally, the closing velocity of two objects in orbit could be up to 22.4 km/s. rifle bullet: 3×10³ J Telstar 19V, geostationary orbit: 3.3×10¹⁰ J Telstar 19V, geostationary transfer orbit: 3.5×10¹¹ J ISS, LEO: 1.2×10¹³ J hypothetical fast ISS: 1.1×10¹⁴ J 14−3 gives 11 orders of magnitude, but good luck making the ISS go that fast. Also, don’t forget to add in the kinetic energy of the net; it’s not going to be zero. Still, even 7 orders of magnitude qualifies as ”quite a few”. You don’t need very many orders of magnitude before you have too many.

Denvercoder9 6y ago
The ISS is not a communications satellite, escape velocity is lower than maximum closing velocity, and kinetic energy isn't momentum. Of course you can make the statement true by changing everything it's about.
- db48x 6y ago
  Forgive me for contradicting you, but I didn’t change anything; I just typed momentum when I meant energy. I first thought of the formula, not the name, and then typed the wrong name. I'm not sure how I made such a mistake, but mea culpa. Also my old friend bzbarsky mentioned closing velocity, and I stuck with it when I made my estimate. I do agree that the ISS is not usually considered to be a communications satellite (though it does have plenty of communications gear on board, including a ham radio repeater if I recall correctly). But it is a thing that we might want to deorbit one day. Since we were considering the ridiculous closing velocity of two objects that are both on escape trajectories, I figure using the ridiculous mass of the ISS was fair game too. But like I said, I originally guessed 50t, not 400t. I was thinking more about the payload capacities of launch vehicles than the satellites themselves. Whatever mass and closing velocity you consider, 7 or 10 or 11 orders of magnitude all qualify as “quite a few” even if the velocity itself only has 1.3 orders of magnitude of difference.
stopChanging 6y ago
I mean you stretched the numbers as much as possible (garbage collecting the ISS? LOL!) and you still didn't get there. And anyway, none of this matters, because the relevant OP specifically said "speed". So the plain and obvious conclusion is: False.