- 05 Dec '09 07:04So I'm looking thru my old college chem book tonight and I run across one of the first formulas we had to learn: f = m1 * m2/d squared (I can't make it as pretty as the book due to being being typographically challenged regarding exponents). But Suppose an 8 pound bowling ball is 8 feet away from another 8 pound bowling ball. Wouldn't that make {f}, the gravitational force between these two balls equal 1 lb. per square foot? My question is, is that a lot? How big does {f} have to be to get an object with mass 1 to attract an object with mass 2 towards it?
- 05 Dec '09 18:52 / 4 edits

Converting to Kg, 8 pounds is 3.6 kg, and 8 feet is about 2.4 meters. Grav constant is 6.67E-11 and plugging that into your numbers, gives in grams, 1.47 E-12 grams.*Originally posted by PinkFloyd***So I'm looking thru my old college chem book tonight and I run across one of the first formulas we had to learn: f = m1 * m2/d squared (I can't make it as pretty as the book due to being being typographically challenged regarding exponents). But Suppose an 8 pound bowling ball is 8 feet away from another 8 pound bowling ball. Wouldn't that make {f}, th ...[text shortened]... does {f} have to be to get an object with mass 1 to attract an object with mass 2 towards it?**

about 1.5 Picograms. Not exactly an overwhelming force

When you convert the units to Kg and meters, the answer comes out in newtons, which, in this case 1.5E-10 newtons. One newton in terms of acceleration force is the force needed to accelerate one Kilogram one meter per second squared. So using that as a force number, tell me how long it would take for the two balls to touch if carefully placed in space your 8 feet apart, way away from the sun and the only force acting on the balls is the mutual gravitation. So after the balls are placed 8 feet apart, how long does it take to go that 8 feet?

How far would they travel if that acceleration continued, assuming the balls had that accel and didn't crash into one another, how far would they be apart after say, one year? Ten years? How fast would they be going ten years later assuming they accelerated at that same rate for ten years?

How far apart would they be ten years later? - 05 Dec '09 20:25

Well, the force increases as the balls are placed more closely. But in practise friction would stop any appreciable motion and the balls would just stay put quasi-forever.*Originally posted by sonhouse***Converting to Kg, 8 pounds is 3.6 kg, and 8 feet is about 2.4 meters. Grav constant is 6.67E-11 and plugging that into your numbers, gives in grams, 1.47 E-12 grams.**

about 1.5 Picograms. Not exactly an overwhelming force

When you convert the units to Kg and meters, the answer comes out in newtons, which, in this case 1.5E-10 newtons. One newton in te ...[text shortened]... y accelerated at that same rate for ten years?

How far apart would they be ten years later? - 06 Dec '09 00:34

I was saying, the force stays as it starts and the balls pretend go through each other and keep going and a magical force keeps it at the same acceleration. That's all I was talking about. How far would it go in ten years with 1.5 piconewtons of force.*Originally posted by KazetNagorra***Well, the force increases as the balls are placed more closely. But in practise friction would stop any appreciable motion and the balls would just stay put quasi-forever.** - 06 Dec '09 03:15

1.5 piconewtons = 1.5 x 10^-12 Newtons*Originally posted by sonhouse***I was saying, the force stays as it starts and the balls pretend go through each other and keep going and a magical force keeps it at the same acceleration. That's all I was talking about. How far would it go in ten years with 1.5 piconewtons of force.**

F=ma

m = 3.6 kg

a = 1.5 x 10^-12 / 3.6 = 4.166667 x 10^-13 ms^-2

t = 10 years = 3600 x 24 x 365.25 x 10 = 315,576,000 seconds

x = 1/2 at^2

x = 1/2 x (4.167 x 10^-13) x 315576000^2

x = 20,747.54 m = 20.75 km

Although it will be quite interesting to work out a general equation taking into account the force increasing as objects gets closer. I figure it will be a differential equation. Will have a go at it tomorrow if someone else hasn't already, it is getting late here. - 08 Dec '09 00:22

OKay, so this equation is saying that my 2 bowling balls have a gravitational attraction to one another (however small), right? So, left to their own devices, just sittin' in my attic, eventually, they will move TOWARD one another? See, this is why I regret never having taken physics. I took a lot of Biology and Chemistry courses, (and even one Psych, to meet the "sociology,religion,and/or humanities" requirement). But physics and astronomy are the ones that fascinate me.*Originally posted by sonhouse***I was saying, the force stays as it starts and the balls pretend go through each other and keep going and a magical force keeps it at the same acceleration. That's all I was talking about. How far would it go in ten years with 1.5 piconewtons of force.**

---thanks for the explanation, by the way

And I DID consider what a previous poster mentioned: friction. But I decided that would make it too complicated, so I'll re-phrase my question to---If my attic were in a vacuum, and my 2 balls are in there 8' apart, would they eventually move toward one another? - 08 Dec '09 01:10 / 1 edit

My thought is no, they would never touch because they are on Earth not in space and as such will be resting on some object and thus subject to friction and other electrical attractive forces plus the gravity of the Earth itself deforming the surface of the object they would be resting on therefore making a small, even if only a few atoms deep, dent, creating a valley out of which the application of a few piconewtons would be insufficient to break free. Only in space away from other large masses would the tiny force of mutual gravity be the majority force acting on those bodies.*Originally posted by PinkFloyd***OKay, so this equation is saying that my 2 bowling balls have a gravitational attraction to one another (however small), right? So, left to their own devices, just sittin' in my attic, eventually, they will move TOWARD one another? See, this is why I regret never having taken physics. I took a lot of Biology and Chemistry courses, (and even one Psych, t acuum, and my 2 balls are in there 8' apart, would they eventually move toward one another?** - 08 Dec '09 02:07

A force is a force, the force acting on the balls would be orthogonal to the gravitational force(unless they are initially very far apart) and a net force causes an acceleration......so if your attic was in a vaccum and your balls were resting on a frictionless floor in a totally theoretically ideal environment then I say yes, given enough time the balls would collide.*Originally posted by sonhouse***My thought is no, they would never touch because they are on Earth not in space and as such will be resting on some object and thus subject to friction and other electrical attractive forces plus the gravity of the Earth itself deforming the surface of the object they would be resting on therefore making a small, even if only a few atoms deep, dent, creatin ...[text shortened]... arge masses would the tiny force of mutual gravity be the majority force acting on those bodies.**

Btw. I would like to see how the differential equaton is derrived for the continously changing force between the balls as they approach each other...I tried to figure some of it out, but I know very little about differential equations so i just became frustrated and lost very quickly. - 08 Dec '09 11:37

F = ma is already a differential equation;*Originally posted by lausey***1.5 piconewtons = 1.5 x 10^-12 Newtons**

F=ma

m = 3.6 kg

a = 1.5 x 10^-12 / 3.6 = 4.166667 x 10^-13 ms^-2

t = 10 years = 3600 x 24 x 365.25 x 10 = 315,576,000 seconds

x = 1/2 at^2

x = 1/2 x (4.167 x 10^-13) x 315576000^2

x = 20,747.54 m = 20.75 km

Although it will be quite interesting to work out a general equation taking into account th ...[text shortened]... ion. Will have a go at it tomorrow if someone else hasn't already, it is getting late here.

in 1D:

F = m dx²/dt²

-Gm²/x² = m dx²/dt²

-Gm/x² = dx²/dt²

I think this would be in the frame of reference of the left ball, but not quite sure if I've been thorough enough. - 08 Dec '09 12:56

Both balls accelerate, so just double the acceleration.*Originally posted by KazetNagorra***F = ma is already a differential equation;**

in 1D:

F = m dx²/dt²

-Gm²/x² = m dx²/dt²

-Gm/x² = dx²/dt²

I think this would be in the frame of reference of the left ball, but not quite sure if I've been thorough enough.

A hint to solving the equation - multiply both sides by dx/dt, and both sides become an exact derivative with respect to t.