# Gravational force

firewall
What is gravational force? Is that every object with certain number of mass will have gravational force?

Related Other Physics Topics News on Phys.org
Any object with mass will experience a gravitational force from any other object with mass. It doesn't matter how much mass either object has.

Gravitation is an inherent attraction between objects with mass, supposedly due to some space-time curving neat stuff that someone more qualified will elaborate for you.

I think gravity is one of many things that we have been able to describe but not fully understand. Why does gravity exist? What really causes it? There is the basic Newton equation to calculate it, but for me there is still so much mystery to it.

Just wondering...our planet Earth has gravity and every objects on Earth fall towards the Earth. So, it that possible we will know more gravity if we can study or explore the center of the earth?

the newtonian equation is $$F=-G\frac{m_1m_2}{r^2}$$ where G is the gravitational constant and r is the distance between objects. if you want to get g, then the equation is $$g=-G\frac{m}{r^2}$$ where m is the mass of the object and r is it's radius. i see no advantage in going to the center of the earth when it comes to studying gravity.

edit: crap, this forum doesn't appear to have the same LaTeX system as http://scienceforums.net

Last edited:
Thinktank--not only does our Earth cause a gravitational field, but every mountain, rock, piece of dirt, molecule, and massive particle making up our Earth does as well. So not only does every object on Earth fall towards Earth, the rest of the Earth is falling (albeit very slightly) towards every object (in the universe, not just Earth).

We know from observation that the gravitational force is attractive and acts radially from a source. That is to say that the gravitational force of the Earth pulls apples down towards the center of the Earth, not side-ways or pushing away.

We also know from observation that the gravitational field from a spherically symmetric object (such as our Earth and things like bowling balls and oranges) is also spherically symmetric. That is to say that it should look the same if we rotate the object. Why is this? Well, imagine you had a perfectly spherically symmetric object. If you rotated the object in any way, it would look exactly the same. Ineed, we would expect the physics of the object to be exactly the same. Thus, the gravitational pull of the object should itself be spherically symmetric.

What happens at the center of the ball bearing? If you "look" in any direction from the center of the ball bearing, the mass distribution is exactly the same. (By the spherical symmetry of the object.) That is, if I look in any one direction, I see so-and-so length of material of so-and-so density, and if I look in any other direction, I see the same thing. Thus, we expect the physics to be the same in every direction (if it weren't, what could have caused one direction to behave differently from another?). Thus, at the center of the Earth, what do you think the gravitaitonal field will look like?

-Flip

rbj
i see no advantage in going to the center of the earth when it comes to studying gravity.
you might be able to easily light your cigi's there. no need to bring your butane lighter.

maybe we wouldn't need rides in the "vomit comet" to test out weightlessness if we could go to the center of the earth.

edit: crap, this forum doesn't appear to have the same LaTeX system as http://scienceforums.net
it's different than wikipedia, too.

r b-j

rbj said:
maybe we wouldn't need rides in the "vomit comet" to test out weightlessness if we could go to the center of the earth.
wieghtlessness at the center of the earth? i don't think so. it obeys the INVERSE square law. that means, the shorter the distance, the less force. as the distance approaches zero, the force approaches infinity. $$\lim_{r\to0}\mathbb{F}_{grav}=\infty$$

krab
Approaches infinity only for a point mass. IOW, you need infinite density to get an infinite force. As you approach the centre of the earth, your weight decreases to zero, in a sense because all the other mass above you is pulling the other way, cancelling the force that's pulling you down.

yea, it would be zero. if you place yourself in the center, you would need to find the net force. you split the earth into an infinite number of spheres that are tangental to the center and have a radius of .25 the radius of the earth. net force would be zero. i wasn't thinking about it that way. way different answers, wow.

edit:i should have known that, too. i once calculated the force of a ring of black holes and found that there is a place in the exact center of the ring where there is no force.

Last edited:
Gravitational Force is the result of the Gravitational Field and is not the inverse. Any object with mass will hinder spacetime around it with its Gravitational Field and when another object enters or is kept in this field , it experiences a push toward sthe first object an dthis push is called the Gravitational Force and is a result of the field.

BJ

rbj
wieghtlessness at the center of the earth? i don't think so. it obeys the INVERSE square law.
not inside the planet. you need to apply Gauss's Law and, assuming a constant density of mass, the strength of the gravitational field is directly proportional to r, the distance to the center.

anyway, it seems like someone else set you straight.

r b-j

Thanks for sharing. I also wondering is it possible to create an object with gravity? If we can create an object with huge gravity, does it mean we can create a black hole?

theoretically, yes. For example if you could make the radius of the Earth close to 3 inches. You would have a black hole. Keep in mind that you should not change the mass. Practically we can't make a black hole.(as far as I know)

By the way, everything that is done by human has a mass, and therefore has a gravitation field... So your first question is kinda weird...

i wasn't aware that Guass's Law applied to gravity as well. what is it for the gravitational force?

Mmm..let me rewrite my first question...is this possible to create an object with a huge gravitation field

Danger
Gold Member
Thinktank said:
Mmm..let me rewrite my first question...is this possible to create an object with a huge gravitation field
If you mean some kind of artificially high field without an increase in mass, no. Even a black hole still has only the total gravitational field of the original collapsed object, but it's extremely concentrated. If you could compress the Earth to black hole status, it would still keep the moon in orbit, but wouldn't start sucking Jupiter toward it.

Gauss' Law for gravity is almost identical to the law for electrics. Its just a flux integral. You can find more info in it http://physics.smsu.edu/faculty/broerman/fall03/phy203/gauss1.htm [Broken].

Last edited by a moderator:
Sorry about i have so many questions....i want to know more about physics..however, i don't have a chance for the education...hope you guys can answer my silly questions...

Question: The gravational force base on mass of the obect. If both Earth and Jupiter become black holes, will the (jupiter) black hole suck the earth (black hole)? Or both forces will pull each other and expand the space between them?

they will affect each other the same way they do now

rbj
i wasn't aware that Guass's Law applied to gravity as well. what is it for the gravitational force?
it's just like Gauss's Law for electrostatics when you make the corresponding substitutions. mathematically, Gauss's Law applies for any inverse-square field.

$$\oint_A \mathbf{E} \cdot d\mathbf{A} = 4 \pi G M$$

where $\mathbf{E}$ is the gravitostatic field and $M$ is the mass contained in the surface integral. the force on a small test mass $m$ in that field would be

$$\mathbf{F} = m \mathbf{E}$$

wikipedia has a link http://en.wikipedia.org/wiki/Gauss's_Law .

the reason the field get lower (proportional to $r$ ) when you're inside the planet is that the amount of stuff contained in the surface integral (the smaller sphere) is smaller.

r b-j

Danger
Gold Member