Basic Question about Centripetal Force

In summary, The moons orbiting a planet have an inertial velocity of 1.8504e10 meters per second. If the planet's gravity is too tight, the moon will collide with the planet. If the planet's gravity is too loose, the moon will fly away. To calculate the gravitational acceleration needed to "catch" the moons in an orbit of 25pm, one must assume that the velocity is maintained. If the mass of the moons is irrelevant, then the gravitational acceleration can be calculated based on the speed and the planet's radius.
  • #1
shintashi
117
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.
 
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  • #2
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.
 
  • #3
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
 

1. What is centripetal force?

Centripetal force is the force that acts on an object moving in a circular path, always directed towards the center of the circle. This force is required to keep the object moving in a curved path instead of moving in a straight line.

2. How is centripetal force different from centrifugal force?

Centripetal force is the inward force that keeps an object moving in a circular path, while centrifugal force is the outward force that appears to pull an object away from the center of the circle. In reality, centrifugal force is just the result of inertia and is not an actual force.

3. What are some examples of centripetal force in everyday life?

Some examples of centripetal force in everyday life include a car turning a corner, a satellite orbiting the Earth, a ball swinging on a string, and clothes spinning in a washing machine.

4. How is centripetal force calculated?

The magnitude of centripetal force can be calculated using the formula F = mv²/r, where m is the mass of the object, v is the velocity, and r is the radius of the circular path.

5. What happens if there is not enough centripetal force acting on an object?

If there is not enough centripetal force acting on an object, the object will not be able to maintain its circular motion and will either slow down or move in a straight line tangent to the circle. This is known as a tangential acceleration and can cause the object to veer off its intended path.

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