Basic Question about Centripetal Force

Click For Summary
SUMMARY

The discussion centers on calculating the gravitational acceleration required to maintain the orbit of satellites around a planet with a radius of approximately 25 picometers. The satellites are moving at an inertial velocity of 1.8504e10 meters per second. The key formula mentioned for this calculation is a = v²/r, where 'a' is the gravitational acceleration, 'v' is the velocity, and 'r' is the total radius of the orbit (50 picometers). The conversation also touches on the relationship between mass and gravitational force, emphasizing that mass does not affect the fall rate in a vacuum.

PREREQUISITES
  • Understanding of centripetal force and gravitational acceleration
  • Familiarity with the formula a = v²/r
  • Basic knowledge of quantum physics concepts
  • Proficiency in scientific notation and unit conversions (e.g., picometers, picoseconds)
NEXT STEPS
  • Research gravitational acceleration calculations in astrophysics
  • Explore the implications of superluminal speeds in quantum physics
  • Study the relationship between frequency and mass in particle physics
  • Learn about the gravitational constant (6.67300 × 10^-11) and its applications
USEFUL FOR

Students and professionals in physics, astrophysics, and quantum mechanics, particularly those interested in gravitational calculations and the behavior of particles at quantum scales.

shintashi
Messages
117
Reaction score
1
Hi. I've been having some difficulty with calculating this equation, mainly because I've forgotten most of the equations over the years, and this is sort of a reverse algebra form of an equation I scarcely remember. (i.e., I'm screwed)

Ok, here goes.

Assume you have objects, as per satelites(sp?) around a central field. (wewill call this central field a "planet"

This planet has a radius of aproximately less than 25pm, or to say, 25 picometers. Some prefer this planet have a radius of 24.2 picometers, but who really cares, right ?

The satelites or "moons" passing by this planet have an inertial velocity of
1.8504e10 meters per second. (for the moment, we are just going to have to suspend that whole "C" thing to work out the math)

So the question is, what would the gravitational acceleration, have to be, in order to "catch" these moons in an orbit of 25pm. Obviously we don't want much more or less than this number.

Buddha says " If planet's gravity is too tight, moon collide with planet; but if planet's gravity is too loose, moon will fly away"


one of the things that bugged me about this was that when gravity accelerates, you normally only count the first 50% of distance travelled, but count 100% of the speed from acceleration.

if its easier on you, you can use picometers and picoseconds, I started to, before it got too confusing.

As far as mass- well, a bowling ball and a feather both fall the same in vacuum, so it doesn't matter what the mass of the "moons" are. All I am ask for the aceleration rate in meters, or picometers, or kilometers, or whatever, at 25 picometers radius, each second.
 
Physics news on Phys.org
well, assuming it maintains its velocity, you can just say that a=v^2 /r where r is the orbit length plus the planets radius: 50pm, and v is 1.804e10 m/s. i think that makes sense.
 
ahh, there I am :smile: I wondered where I got moved to. I think this is an ok move, and I will happily punch in the data of pperkins.

I put it in Quantum physics originally, because its an excerpt from that area, although the question itself is pretty basic *whoo*

I was attempting to calculate how much actual force/kinetic energy/etc. was really in an electron, assuming a cymatic oscillation of @1.2e20 hz and a 2.42e-11 m radius. Something caught my eye about an old theory postulated concerning the actual Spin of an electron, and its superluminal paradox. As a side note, some how, when punched in the numbers I came up with for its mass (based upon sqrt Gravitational constant, etc.) was virtually identical to the frequency, which was really weird.

For one, you wouldn't think that you could get within decimal points of the scientific notion of 100 quintillion hertz, and you would probably be equally suprised if the calculator was telling you, that THAT new number, was also the "mass" in what you can only conclude is kilograms.

I thought this was odd, but there must have been something done right, since I did not use the 6.67300 × 10-11 in my original numbers, yet it still produced a value identical... i.e., the equation was saying

"frequency = mass"

But that's besides the point. Just thought I would explain why my brain told me to put this part of the bigger equation in the quantum physics instead of general area.

Thank you thank you pperkins, and anyone else who has a fun and simple way of answering the question. :D
 

Similar threads

  • · Replies 8 ·
Replies
8
Views
4K
  • · Replies 15 ·
Replies
15
Views
4K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 17 ·
Replies
17
Views
7K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K