# Conceptually, why is acceleration due to gravity always negative?

As the title states, why is acceleration due to gravity always (-). Say you assign "up" as the positive direction. If an projectile is thrown at a 24 degree angle above the horizontal, I get that acceleration due to gravity before the vertex is negative. However, why is it not positive after the vertex? If acceleration due to gravity is negative and we assign downwards as negative, wouldn't that make acceleration positive?

What I think is that acceleration due to gravity is always towards the ground. Even if a projectile is going downwards, and we assign downwards as (-), the acceleration due to gravity is still (-), because the object still accelerating downwards. Despite it going upwards or downwards, the net acceleration of the object is downwards.

• Yes, it s always downwards, you can depend on that, always towards the center of the Earth – user146020 Mar 1 '17 at 3:31
• It's not always negative. If your +ve axis points down then it's +ve. – tfb Mar 1 '17 at 3:45
• Yes, actually its toward centre of Earth, and tge line joining that particle and centre is perpendicular to tge ground,so its downward at that point.If you are not under the particle at that instant then the acceleration isnt exactly downward(wrt to the tangent at the place you are standing), but I am saying this because I just wanted to add extra detail. The slight change in angle has no relevance here. – Red Floyd Mar 1 '17 at 3:47
• -1. Unclear what you are asking. Your 2nd paragraph is correct. Your 1st paragraph does not make sense. Replace acceleration by velocity : if velocity is -ve and downwards is -ve direction this means velocity is downwards, it does not mean velocity is upwards. – sammy gerbil Mar 1 '17 at 23:29

However, why is it not positive after the vertex? If acceleration due to gravity is negative and we assign downwards as negative, wouldn't that make acceleration positive?

It seems your misunderstanding is in understanding the concept of frame of reference. When we do calculations in physics we do this with respect to a coordinate system/frame of reference which you can chose freely (but preferably conveniently). All quantities such as position, velocity, acceleration are measured/calculated with respect to this coordinate system.

Your questions suggest that you want to consider acceleration with respect to the direction of the velocity (which does change direction itself). Your proposal is like starting with a coordinate system and once the object reaches the vertex you flip/mirror/reverse the axes of your coordinate system.

Taking your example of throwing/shooting a projectile up vertically. Let's chose the coordinates such that positive $x$ direction is up. Then, by definition the velocity at any time is $$v=\frac{dx}{dt}$$ and the acceleration is $$a=\frac{dv}{dt}=\frac{d^2x}{dt^2}$$

## Before reaching the vertex, going up

Obviously $v>0$ since the position $x$ is increasing ($dx>0$). Since the projectile is decelerating $dv<0$ and therefore $a<0$.

## After reaching the vertex, falling down

$v<0$, the projectile is going down ($dx<0$). The projectile is accelerating, i.e. $|v(t+dt)|-|v(t)| > 0$, but since the velocity is negative this can be written as $v(t)-v(t+dt)>0$. Therefor $a=\lim_{dt\to 0}(v(t+dt)-v(t))/dt<0$

Of course you can do the same reasoning in a different coordinate system where the $x$-axis is pointing down.

The negative sign is just a definition, and it doesn't mean anything except for signifying the direction of gravity. So, the gravitational acceleration is really, $a=g\hat y$. Towards the center of the Earth, the gravity is defined to the be negative. At all points on the trajectory of a projectile, the gravitational acceleration points in the same direction, which is downwards toward the center of the Earth. So, the sign for the gravitational acceleration is always negative. However, it doesn't matter too much how you define the gravity and in any direction.

It is about the perspective we are using about the forces. When you put a +Q ball near to another +Q ball you increase the potential energy of the system against the Electrical forces. When you leave the system electrical forces will then push both balls a far from each other. So when we make work against a system we consider it + because we increase the potential energy of the system but when natural forces make work just the opposite way we did we say that it's - because it works "against" what we did. If we were to be electrical forces we would say that we did + work and the others make - work.

On your case it's really the same. We work against gravity to increase the height of a thing and when we stop gravity takes it back.

I think the OP is confusing acceleration and direction of motion. Acceleration does not depend on the direction of the motion. According to the Second law of motion, (for constant mass) $\vec F=m\vec a$, acceleration is in the sense of the resultant force acting on the particle. So, it does not matter that the particle is moving in which direction; as long as the resultant force acting on it does not change, the acceleration won't change.

I don't have enough reputation to make a comment, so I'm adding this as a separate answer.

The other answers are not wrong, but I feel it is worth pointing out that this is related to the fact that, as far we know, there is only type of gravitational charge (i.e. mass), which we by convention call positive. If I recall correctly there are planned/ongoing experiments to see whether antimatter might have a negative gravitational charge. I don't believe many people expect that to be the case, but it is evidently something considered worth the effort to check.

Related questions:

Negative Mass and gravitation

Electrical force vs gravitational force

Acceleration due to gravity in itself is not negative but it is directed toward center of earth (downward) and we take Downward direction as negative by convention. And as force of gravity is pointing to same direction at every point on the trajectory hence acceleration due to gravity is same for before or after the vertex.

## protected by Qmechanic♦Mar 1 '17 at 16:33

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