# Does gravity pull straight down or down and around?

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Does gravity pull mass straight down, or does it pull it down in a spiral configuration to a central point?

Also what is the smallest unit of mass that can have a gravitational force?

And does gravity affect black holes? Meaning do blackholes also revolve around larger objects due to gravity?

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i can't think of any way gravity could have any lateral effects, it's pretty much a direct attracitve force between particles.

all mass has a gravitational feild, otherwise it wouldn't be a mass.

iirc, the only properties of mass are inertia and gravity. some have even speculated that they are the same thing.

if you look at the universe as though mass were gobbling up space-time, it would be inertia that applies the force. so if an object has inertia, it has gravity.

a black hole is a large amount of matter in a very small area. it can be moved by a speck of dust in much the same manner as you can move a large star with a speck of dust.

the only thing a black hole can be said to orbit would be something heftier, like a bigger star or a bigger black hole.

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yeah, gravity is a direct force, until you get into the very wierd world of relativity with effects like frame dragging and the like. it all gets very complicated and it can look as if its slightly off centre but its not. as i said, its complicated and it warps my mind.

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gravity pulls in a straight line between the 2 objects.

-whether or not it orbits another object before hitting it has to do with the fact that they were going in slightly different paths then.

a really good example of this is to take 2 magnetic ball berings and roll them to each other in a flat surface. roll it so that there paths are parallel to each other. they will roll twards each other then hit each other than rotate.

anything that has mass has gravity by definition, the fact that an object is makes it have gravity.

black holes are an object so they have gravity so they are affected by gravity. again the fact that they are means they are affected by gravity.

google video search and see if you cant find a computer simulated video of 2 black holes or galaxies colliding.

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gravity pulls in a straight line between the 2 objects.

-whether or not it orbits another object before hitting it has to do with the fact that they were going in slightly different paths then.

a really good example of this is to take 2 magnetic ball berings and roll them to each other in a flat surface. roll it so that there paths are parallel to each other. they will roll twards each other then hit each other than rotate.

anything that has mass has gravity by definition, the fact that an object is makes it have gravity.

black holes are an object so they have gravity so they are affected by gravity. again the fact that they are means they are affected by gravity.

google video search and see if you cant find a computer simulated video of 2 black holes or galaxies colliding.

Do black holes have mass? (excuse me if that is a stupid question.)

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Yes, tremendous amounts of it.

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Small caveat:

It's assumed all matter has gravitational pull, otherwise all the theories are wrong. However, it's such a weak force, there's no way you could ever actually measure any effects of the gravitational effect of, say, a single atom. However, you could easily measure how it was effected BY gravity.

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However, it's such a weak force, there's no way you could ever actually measure any effects of the gravitational effect of, say, a single atom.

dont you remember newtons universal law of gravitation? cant you just take gravity out of the equation? so whats left is Fg= m1*m2/r squared

someone tell me how i can write this in (math)(/math) form?

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someone tell me how i can write this in (math)(/math) form?

In math mode it would probably read $F_g = \frac{m_1 m_2}{r^2}$

Note:

- You can see the code by hitting "quote" on my post or by moving the mouse over the equation.

- The equation is wrong (unless G=1).

why is it wrong?

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break down the equation piece by piece and you will see. Force is measured in Newtons, which are kg-m/s^2 now you have to masses over the the square of the distance which would give you kg^2/m^2 which is not a force, a constant is needed to ballance the equation, something which has an acceleration over a mass which would be m/(kg-s^2) which would then give you a proper measure of force. and as Atheist said, unless the gravitational constant G = 1 then the equation would not work.

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In other words, it's just in the wrong units, which is not really important. Just replace the "equals" sign with a "varies as" sign and it's correct.

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ooh ok i think i was thinking faster than i was typing

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In other words, it's just in the wrong units, which is not really important. Just replace the "equals" sign with a "varies as" sign and it's correct.

$F_g = \frac{m_1 m_2}{r^2}$ Hey, Earlier you was saying that you had to change the ='s Sign to a Varies to sign and the G ='s to G='s 1, What do you mean exactly?

Thanks..

Ben.

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Ok.

First, in order for it to be "equal," it would have to be quantities of the same thing. Force is mass*distance/time^2, which can't be directly compared to mass^2/distance^2. This can be resolved in two ways.

First, you can multiply one by some constant such that they end up in the same units. You can multiply the second by some quantity of "distance/mass*time^2," and then they would both be quantities of mass*distance/time^2," i.e. force. Of course, what that quantity actually IS would depend on what units of mass, distance, and time you are using. The constant would be numerically different if you were using, say, meters or feet as your unit of distance.

The other way avoids all that, and just says that the two quantities "vary as " one another, meaning that they are directly proportional. So, they're not equal, but increasing one will proportionately increase the other. So, for example, doubling r will decrease the quantity on the right by 75%, which also decreases the force by 75%.

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