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Gravitation Constant

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Tags: constant, gravitation
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User Name
#1
Sep5-06, 09:15 PM
P: 22
I know it is 9.8 m/s^2. However, we're covering Kinematics, and I'm just wondering how you know whether you use -9.8 or +9.8.

At first, I thought that if the object was traveling up, you would use +9.8 and if it was going down, you would use -9.8. But I just went over a problem that used +9.8 when an object was dropped.

So now I'm really confused. ~_~
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tony873004
#2
Sep5-06, 09:41 PM
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Since up is usually considered positive, it's common to use -9.8. But it depends on how your equations are written.

With this formula, you'd use -9.8
[tex]
y = y_0 + v_0 t + \frac{1}{2}gt^2
[/tex]

With this formula, you'd use +9.8
[tex]
y = y_0 + v_0 t - \frac{1}{2}gt^2
[/tex]

It will not change depending on whether the objects is going up or down. Gravity always pulls down. It pulls down on objects that are moving down, making them go faster. It pulls down on objects that are moving up, slowing their upward velocity.
MeJennifer
#3
Sep5-06, 09:46 PM
P: 2,043
Perhaps it is my error, but do schools teach students that the gravitational constant is -9.8?
My understanding of the gravitational constant is something entirely different.

tony873004
#4
Sep5-06, 09:50 PM
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Gravitation Constant

Quote Quote by MeJennifer
Perhaps it is my error, but do schools teach students that the gravitational constant is -9.8?
My understanding of the gravitational constant is something entirely different.
You're right, and I didn't catch that. I assume the OP meant the acceleration due to gravity on the surface of Earth which is approximated as a constant.

The gravitational constant, G, is 6.67x10-11Nm2/kg2
MeJennifer
#5
Sep5-06, 09:57 PM
P: 2,043
Quote Quote by tony873004
The gravitational constant, G, is 6.67x10-11Nm2/kg2
Except for a typo:
6.67x10-11Nm2/kg-2
borisleprof
#6
Sep5-06, 10:05 PM
P: 36
Hello everybody,
gravitationnal acceleration is g = 9.8 m/s^2.
It is when you analyse a situation let's say that an object is moving upward, you know that its weight is downward, so when you have a situation in which there are two opposite directions, you choose one that is positive (usually the upward direction)and the other one as negative.

If an object is falling on the floor, it is useless to speak of a negative accelerationduring its fall because you have to consider only one direction (here downward)
tony873004
#7
Sep5-06, 10:09 PM
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Quote Quote by MeJennifer
Except for a typo:
6.67x10-11Nm2/kg-2
No typo. You'd only use -2 if you pulled the kg up to the numerator. Both are correct:

6.67x10-11Nm2/kg2

6.67x10-11Nm2kg-2
MeJennifer
#8
Sep5-06, 10:11 PM
P: 2,043
Quote Quote by tony873004
No typo. You'd only use -2 if you pulled the kg up to the numerator. Both are correct:

6.67x10-11Nm2/kg2

6.67x10-11Nm2kg-2
Yes you are right!
borisleprof
#9
Sep5-06, 10:14 PM
P: 36
The gravitationnal constant is realy 6.67x10^-11 Nm^2/kg^2
tony873004
#10
Sep5-06, 10:16 PM
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Quote Quote by MeJennifer
Yes you are right!
But just for the record, I do make my fair share of typos. Thanks for checking.
User Name
#11
Sep6-06, 09:05 PM
P: 22
Yes, my teacher taught us that the gravitational constant is -9.8 m/s^2.

Is he teaching it wrong? It doesn't seem wrong to me when he does example problems on the board. It's just that when I do bookwork, I have to use +9.8 to get the right answer..
Andrew Mason
#12
Sep6-06, 11:43 PM
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Quote Quote by User Name
Yes, my teacher taught us that the gravitational constant is -9.8 m/s^2.

Is he teaching it wrong? It doesn't seem wrong to me when he does example problems on the board. It's just that when I do bookwork, I have to use +9.8 to get the right answer..
If she actually calls it the gravitational constant you should correct her. It is g, the acceleration at an average place on the earth's surface due to earth gravity. It is not the gravitational constant a) because it is not constant (it varies as 1/r^2 where r is the distance from the earth centre and b) the gravitational constant is G, as explained above.

AM
User Name
#13
Sep6-06, 11:52 PM
P: 22
Oh, that is my mistake. He doesn't call it the gravitational constant. It's the acceleration due to earth's gravity.

Sorry.
tony873004
#14
Sep7-06, 02:04 AM
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Quote Quote by User Name
Oh, that is my mistake. He doesn't call it the gravitational constant. It's the acceleration due to earth's gravity.

Sorry.
I'm relieved your teacher isn't that confused. You'll encounter G, the real gravitational constant, in later chapters.

As to whether or not you use +9.8 or -9.8, it depends on what formula you use. My physics teacher liked to use the positive number, so he gave us the formula and warned us to only use it for freefall:
[tex]
v_f = v_i - gt
[/tex]
In this case, you get to plug in "g" as a positive number 9.8. The fact that gravity pulls down is taken care of by the formula, so you don't need the negative sign with the number.

But this formula is usually written as
[tex]
v_f = v_i + at
[/tex]
"a" is a more generic term than "g". Forces other than gravity can cause an object to accelerate. In this case, you need to recognize that gravity pulls down, which in most coordinate system is negative. Here you'd use -9.8 for a.
tony873004
#15
Sep7-06, 02:13 AM
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Interesting fact:
"g" (that's little g, not big G), is not simply acceleration due to gravity. It's net downward acceleration which is affected slightly by your latitude on the globe. At the equator, the tangental velocity of Earth's rotation is higher and the centrifical force makes you ligher. So "g" is acceleration due to gravity - acceleration caused by centrifical force. (Centrifical force not really a force, but in certain reference frames, such as this, it makes sense). This is the reason that you'll never see "g" expressed by more than 3 significant digits (ie. you don't see 9.812734755372). It varies over the surface of the Earth for this and other reasons. So to express it more precisely than 9.8 or 9.81 doesn't make sense.
HallsofIvy
#16
Sep7-06, 04:39 AM
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One of the things you have to do in any kinematics problem is "set up a coordinate system". Not necessarily a 2 dimensional graph, but you do need to decide where "x= 0" will be, which directions are positive and negative, etc.

It is fairly standard to choose "up" positive and "down" negative. Given that, then g is negative, since gravitational force is downward, whether something is being dropped or thrown upward. If the object is thrown downward, its initial velocity is negative (downward), if thrown upward, its initial velocity is positive. An object that is "dropped" has 0 initial velocity. But in either case the acceleration is downward and is negative.

It is less common, but can be done to choose downward to be positive. If, for example, a problem asks the distance downward and object will fall, since the "downward" is assumed and the distance will be a positive (really, unsigned) number, it might make sense to choose downward to be the positive direction. Then acceleration, since it is in that direction, will be positive. If the object is initially thrown upward, its initial velocity would be negative.

There are no "coordinate systems" in nature- you are free to set one up in whatever way makes the problem simplest. And it is a very good idea to state clearly how it is set up.
potatoman177
#17
Jan7-10, 10:19 PM
P: 1
what does the m^2 mean in the gravitational constant
6.67^-11Nm^2/kg^2


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