# Centripetal forces effect on weight

1. Feb 18, 2010

### jbrightmatter

So I am building this program for fun. I thought I would combine some different elements of basic physics into it. I am letting people place in numbers to create a custom planet (radius, gravitational acceleration, rotation speed (tangential velocity?).

Anyway, I ran in to an issue. I want them to enter a weight in kg and get back their actual mass. As I started doing the numbers, I noticed there was a force trying to remove masses from the surface of the planet due to revolution of the sphere. This means that if the stand on a scale it is giving their weight - a little bit because of this force (normal maybe?).

CLEARLY I barely passed my Kinematic physics in college. Can someone give me a formula to use when i am given a perceived weight, a radius of a sphere, the gravitational acceleration of that sphere, and the speed of rotation? Is it even possible to get an estimated true weight from those numbers?

Let me break it down. You stand on a scale at home and it says 100 kg. The earth stops spinning and you reweigh yourself. What is the new weight and how do you come by that number? ...and yes, I am aware that is the earth instantly came to a halt there would be some catastrophic repercussions but that is completely irrelevant to my question

2. Feb 18, 2010

### dacruick

Well, I'm not exactly sure about the answer to your question, but there is definitely a change in net force. I guess I'm going to try and work through it here and hopefully we can figure it out.

When you spin something on a string, the radius, and how fast it is going contribute to the acceleration. The acceleration is outwards, which creates the force of tension in this string. I also think that you could include velocity and radius into one angular velocity. You can leave them seperate as well I think. Using your speed of rotation and radius you have acceleration = v²/r. I believe that acceleration would be in the opposite direction of gravity. It makes sense to me that the faster you spin, the more you are prone to fly off. But another thing to consider is, is that acceleration perpendicular to gravity, therefore not affecting your acceleration in that direction. I think the first one is right.

Okay so summary, I think that a = v²/r is your equation, and I think that you should subtract that acceleration from the force of gravity. I also think that the radius of the earth might be so large that the change in gravity may be negligible.

Finally, if you are going for complete accuracy, I believe that how quickly the planet rotates may change its orbit. You should check out some planetary motion articles, but someone else on this site should be able to help you further.

3. Feb 18, 2010

### Bob S

If you are building this program for fun, then set the speed of light at something reasonable, like 10 meters per second. A good tennis serve or pitched baseball would be very relativistic. Get a copy of George Gamow's Mr. Tompkins in Paperback. It is a good read.

Bob S

4. Feb 18, 2010

### jbrightmatter

Well, the acceleration trying to remove you from this planet is indeed "negligible" but remember, I am setting it so you can change the rotational speed so say, take this planet radius and mass. Now spin it ten times as fast. Guess what, that negligible force... not so negligible anymore. I am also allowing the change in mass thus reducing the grav. accel. thus making that force something to take note of... Your equation for the force acting in the opposite direction cannot be correct because it does not take into account the rotation speed, the gravitational acceleration, the radius of the planet and the mass of the object this negative acceleration is acting on.

I realize I am asking a lot.

P.S. What does light have to do with it?

Last edited: Feb 18, 2010
5. Feb 18, 2010

### Bob S

Someone should build a video game/ program that includes elements of basic physics, and includes such effects like relativity, such as what happens when the speed of light affects our daily activities. Many high school kids are fascinated by special relativity and space travel, but have little direct knowledge of it. Mr. Tompkins' escapades in George Gamow's books should be re-introduced in a video game.

Bob S

6. Feb 18, 2010

### dacruick

My point was, if you want to make a program that is physically accurate, if you speed the rotation by ten times, that might have a crazy affect on the orbit. So the point I was making is that it is negligible unless you have relativistic spins, which would make youre entire program even more ridiculous. Furthermore, velocity and radius account for rotation speed. When I say velocity, I mean angular velocity really. I mean that if you were to pick a point on the planets surface, how fast would that point be going with respect to the rotation of the planet. So my equation is correct. Also, something else you will need to take into account is the changing mass of your planet as you increase the speed. Also, gravitational acceleration is a totally different equation, my equation strictly deals with the acceleration caused by rotation. I assume you are aware of how to calculate the force of gravity on the planet's surface.

I suggest you ignore relativistic changes in the system, but without it, your program will be 'incomplete'. And when I say negligible, I mean that it might be hundredths of a m/s of acceleration. if you multiply that by 10, thats a tenth, and if you multiply that by 100 you might spontaneously explode.

7. Feb 18, 2010

### EEJaime

Actually, I don't think that any amount of increase in rotational speed will provide sufficient centripetal acceleration to overcome Newton's Gravitational force F=G(m1xm2)/r^2. Unless you do something like make the planet's mass that of hydrogen or something else with a lot less mass.
Just a thought.

8. Feb 18, 2010

### dacruick

agreed

9. Feb 18, 2010

### Bob S

For any satellite in earth orbit, the centrifugal force equals the gravitational force. In 1957, Russia launched a small satellite named Sputnik that circled the earth in a low-Earth orbit ~100 miles up, and with a ~90 minute orbital period. If the Earth's rotation period were ~84 minutes, people on the equator will be in orbit (but the oceans will be also).

Bob S

10. Feb 19, 2010

### dacruick

sure, but would the orbit of our planet around the sun change at that rotation speed?

11. Feb 19, 2010

### Bob S

The effect is negligible. The sum of our rotational angular momentum (2/5 MRe2ωe) and our orbital momentum around the sun (MRs2ωs) is a constant, but the coupling is extremely weak. Eventually, the Earth's day will lengthen to one year (due to Sun tides), but the length of the year will change by a few seconds.

Bob S

12. Feb 19, 2010

### Lsos

"The acceleration trying to remove you from this planet" is not so negligible...satellites are routinely launched close to the equator to take advantage of it.