Hey guys so I was looking over my class notes and I just wanted some clarification. I was wondering if air resistance is taken into account, would that have any bearing or would it possibly change gravity? I was just looking through my notes and just came across this and I noted that it would change gravity but for some reason I just can't seem to conceptualize this. Could someone please clarify this? Thanks a lot!
Effect of Air Resistance on Gravity?
- Thread starter dmplz707
- Start date
- Joined
- May 25, 2007
- Messages
- 4,119
- Reaction score
- 962
Change gravity? no.
But it will create a force that opposes the velocity, so for an object in free fall the drag force will point upward, opposite of gravity. This will cause a decrease in the net force down, so a is reduced. If the object falls fast enough (and has a shape that catches air as it travels), then the drag force can become significant enough to perfectly offset gravity, resulting in a constant speed of descent (terminal velocity).
But it will create a force that opposes the velocity, so for an object in free fall the drag force will point upward, opposite of gravity. This will cause a decrease in the net force down, so a is reduced. If the object falls fast enough (and has a shape that catches air as it travels), then the drag force can become significant enough to perfectly offset gravity, resulting in a constant speed of descent (terminal velocity).
- Joined
- Jun 10, 2011
- Messages
- 60
- Reaction score
- 1
It doesn't change gravity so to speak, though it does oppose gravity in free-falling objects.
It's analogous to if friction were opposing you push a box across the floor. The force you put into pushing the box is the same, but friction is a separate and opposite force that reduces net force. Similarly, the force of gravity will be in the downward direction with a magnitude m*g, but the force of air resistance will oppose the direction of motion, and will act as a separate force that does not reduces gravitational force, but net force.
It's analogous to if friction were opposing you push a box across the floor. The force you put into pushing the box is the same, but friction is a separate and opposite force that reduces net force. Similarly, the force of gravity will be in the downward direction with a magnitude m*g, but the force of air resistance will oppose the direction of motion, and will act as a separate force that does not reduces gravitational force, but net force.
- Joined
- Jun 26, 2010
- Messages
- 412
- Reaction score
- 12
Recall where the acceleration due to gravity, g, came from in the first place. Newton's law of gravitation states that:
If we let r be the distance from the center of the earth, m1 and m2, be the masses of the Earth and your falling body respectively, we could calculate the gravitation force between two objects like we would with any other force. However, this is annoying for lots of applications, so when we're near the surface of the Earth, we factor out the mass of the object and consider all the rest to be constants and thus it falls out as g. So, the acceleration due to gravity is ultimately a consequence of Newton's law of gravitation and we accordingly use F = mg instead of the more complicated, albeit complete, version.
I should point out that we also play the same game with potential energy and approximate gravitational potential energy as U = mgh, which is incorrect when we're far away from the surface of the Earth.
That help?
