View Full Version : Gravity
Ukitake Jyuushirou
May29-06, 11:33 AM
i was doing some reading on newton's law of universal gravitation. the equation is given as mass of 2 objects divided by distances square and then multiply by gravitation constant. using 50kg and 100 kg mass and a distance of 2 metres i get a value 8.34e-8 N
this a small force, which sets me thinking. how big a force is needed between 2 objects before we start to feel the effects of gravity?
maverick280857
May29-06, 12:06 PM
i was doing some reading on newton's law of universal gravitation. the equation is given as mass of 2 objects divided by distances square and then multiply by gravitation constant. using 50kg and 100 kg mass and a distance of 2 metres i get a value 8.34e-8 N
this a small force, which sets me thinking. how big a force is needed between 2 objects before we start to feel the effects of gravity?
As you can see from Newton's law, the force between any two masses of normal size (human beings, cars, swimming pools, etc) is quite small despite the dependence on the product of the masses. The gravitational constant goes roughly as 10^{-11} (in SI units) so in order to get what you call an appreciable force, the product of the masses should be at least as large as 10^{11}. But there is another important factor here: the inverse square dependence on distance...so the smaller the distance the better.
You can plug in values of the sun and earth to compute the approximate force of gravitation as predicted by Newton's Law.
Ukitake Jyuushirou
May31-06, 08:42 AM
As you can see from Newton's law, the force between any two masses of normal size (human beings, cars, swimming pools, etc) is quite small despite the dependence on the product of the masses. The gravitational constant goes roughly as 10^{-11} (in SI units) so in order to get what you call an appreciable force, the product of the masses should be at least as large as 10^{11}. But there is another important factor here: the inverse square dependence on distance...so the smaller the distance the better.
You can plug in values of the sun and earth to compute the approximate force of gravitation as predicted by Newton's Law.
in other words, an object needs tremendous mass b4 there is a noticeable gravitational pull?
thanks :)
Of course, which is why the earth attracts us, and in turn is kept in orbit by the sun , which is of course affected by the gravitational pull of other massive objects as blackholes and other stars .
This is also the reason why we don't find ourselves sticking to each other or other objects of comparable mass .
arildno
May31-06, 01:30 PM
Well, you could think of it like this:
Consider 2 balls of constant equal density, radius r, and let F be the magnitude of a "noticable" force.
Furthermore, let the balls touch each other.
Then, the density must satisfy the equation:
\rho=\sqrt{\frac{F}{G}}\frac{3}{2\pi{r}^{2}}
where G is the gravitational constant.
Given some particular radius, you'll get a measure of how huge the density will have to be in order for a noticeable force to be felt.
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