Is the value of g negative or positive?

In summary, g is a vector pointing to the center of the Earth that has a positive magnitude when used in the upwards direction and a negative magnitude when used in the downwards direction.
  • #1
Ali Hamaiz
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I know that in formus there are many answers about this question .I just want the value of g (will it be positive or negative) in this example which I am going to give , I threw a ball vertically upward the value of g at its max height (vi=0) ,what will be the value of g +10 or -10 m/s^2. I think it can be both if you consider the cartisean plane's cordinate , it depends upon refrence.
 
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  • #2
g is a vector (pointing to the center of the Earth).
 
  • #3
Ali Hamaiz said:
I threw a ball vertically upward the value of g at its max height (vi=0) ,what will be the value of g +10 or -10 m/s^2. I think it can be both if you consider the cartisean plane's cordinate , it depends upon refrence.

What sign did you use for the initial velocity? ditto the displacement/height?
 
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  • #4
This means with the sign of Vi you decide the value of g to be positive or negative.
 
  • #5
Correct. It depends which direction (up or down) you define as positive.
 
  • #6
In all the textbooks that I remember using (there may be exceptions elsewhere!), g is defined as a positive number, the magnitude of the (vector) acceleration of a freely-falling object near the Earth's surface. It is used with a + or - sign depending on the coordinate system.

If position, velocity, and acceleration are defined to be positive in the upwards direction and negative in the downwards direction, then the (y-component of) acceleration of a freely-falling object is ##a = -g## (more precisely ##a_y = -g##) and its position as a function of time is ##y = y_0 + v_{0y} t + \frac 1 2 a_y t^2 = y_0 + v_{0y} t - \frac 1 2 gt^2##.

If position, velocity, and acceleration are defined to be positive in the downwards direction and negative in the upwards direction, then the (y-component of) acceleration of a freely-falling object is ##a = +g## (more precisely ##a_y = +g##) and its position as a function of time is ##y = y_0 + v_{0y} t + \frac 1 2 a_y t^2 = y_0 + v_{0y} t + \frac 1 2 gt^2##.
 
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  • #7
You start with a diagram. You put xy axes on it. That defines the sign of your y (vertical) position. If you choose to call 'up' positive, then g will be negative. If you choose 'down' to be positive (say you are dropping things off a bridge) then your g will be positive. It doesn't matter what you choose as long as you stick with the signs throughout . :smile:
Soon your diagram can be in your head.
 
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  • #8
I got it thanks .
 
  • #9
The direction of the ball's motion does not determine whether g is positive or negative.
If you call the directional sense of gravitational force negative, then g will also be negative.
 
  • #10
I always tell my students to double check that the real effect of their equations is a downward acceleration.

Best practice is to define your coordinate system on the paper with y clearly pointing up. But if you point y down, gravitational acceleration is going to be positive.
 
  • #11
Svein said:
g is a vector (pointing to the center of the Earth).

jtbell said:
g is defined as a positive number, the magnitude of the (vector)

It basically boils down to what you mean by "g".
 
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  • #12
My students often hated me for forcing them to do this, but in at least 3 or 4 instances, I make them solve a problem by placing the origin in another location, and changing the direction that they call positive. I do this for projectile motion and simple Coulomb's law problems.

The whole exercise was meant to show that the location on where we define to be "zero" and the orientation of our coordinate axis are really arbitrary, and that no matter how we do this, it should not affect the final answers. Mother Nature doesn't give a hoot on which direction we call "positive" and "negative", and thus, your answer shouldn't either.

Zz.
 
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  • #13
ZapperZ said:
The whole exercise was meant to show that the location on where we define to be "zero" and the orientation of our coordinate axis are really arbitrary, and that no matter how we do this, it should not affect the final answers. Mother Nature doesn't give a hoot on which direction we call "positive" and "negative", and thus, your answer shouldn't either.

...as long as they are consistent in their definition.
 

FAQ: Is the value of g negative or positive?

1. Is the value of g always negative?

No, the value of g can be either positive or negative depending on the location on Earth. At the equator, g is positive, while at the poles, it is negative. In most other locations, g is positive but can vary slightly.

2. How is the value of g measured?

The value of g is measured using a device called a gravimeter, which measures the gravitational acceleration of an object. This can also be calculated using the formula g = Gm/r^2, where G is the universal gravitational constant, m is the mass of the Earth, and r is the distance from the center of the Earth to the object.

3. What factors affect the value of g?

The value of g can be affected by factors such as altitude, latitude, and local geology. It can also vary slightly due to the rotation of the Earth and the gravitational pull of other celestial bodies.

4. Can the value of g change over time?

Yes, the value of g can change over time due to factors such as tectonic movements and changes in the Earth's mass distribution, but these changes are typically very small and difficult to measure.

5. Why is it important to know the value of g?

Knowing the value of g is important for a variety of scientific and practical reasons. It helps us understand the structure and composition of the Earth, and it is also crucial for many engineering and navigation applications.

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