# Mass falling in Earth's Gravity help

• kraigandrews
In summary, a particle falling in Earth's gravity experiences a force of air resistance given by F = −αv v, where v is the velocity, v is the speed, and α is a constant. The magnitude of the force is αv^2, and the direction is opposite to the velocity. To determine the constant α, the terminal speed of the particle (20.2 m/s) and its mass (2.6 kg) are given. Using the equation F=mdv/dt, the solution is found to be α = mg/v^2, which is equal to 1.263 with units of kg/s. There may be some confusion about the notation used for the force, as it is written as F
kraigandrews

## Homework Statement

A particle falls in Earth's gravity. The force of air resistance is F = −αv v where v is the velocity, v is the speed, and α is a constant. (The direction of the force is opposite to the velocity, and the magnitude of the force is αv^2.) If the terminal speed is 20.2 m/s and the mass is 2.6 kg, determine the constant α.

F=mdv/dt

## The Attempt at a Solution

m(dv/dt)=mg-aplhav
thus v(t)=(mg/aplha)(1-e^(-(alpha/m)t))
so aplha if I am correct should be (mg/Vterminal)=1.263, however I am not sure of the units for this. My guess would be kg/s however my homework programs says that is incorrect, and also I feel I may have neglected the [/B]v[/B] and the magnitude of the opposing force causing a wrong answer.
If someone could shed some light on this it is greatly appreciated.

From your formula it seems that you need to use

F=-αv2 so you'd have

ma = mg - αv2

If that's the case then what does the F=-αv mean? To me, the question is pretty confusing as far as what goes where.

At terminal velocity the velocity is constant. So the acceleration must be nil. What can you say about the forces acting in that case?

it would be dv/dt=a=0 thus the net forces are balanced, I guess my confusion is what is the question asking for as the resistive force F=-αv or F=-αv^2?

kraigandrews said:
it would be dv/dt=a=0 thus the net forces are balanced, I guess my confusion is what is the question asking for as the resistive force F=-αv or F=-αv^2?

Well, I'd go with the magnitude of the resistive force and make the appropriate assumption about its direction for a body falling straight down

I do find the notation vv a bit odd. I mean, why not v with a unit vector for direction? Maybe it's just me...

The magnitude of F = −αv v, where the second v is a vector is -av^2. And it points in a direction opposite to the vector v if a is a positive constant.

gneill said:
Well, I'd go with the magnitude of the resistive force and make the appropriate assumption about its direction for a body falling straight down

I do find the notation vv a bit odd. I mean, why not v with a unit vector for direction? Maybe it's just me...

I think the problem statement says the magnitude is av^2. So I don't think they mean the vector v to be a unit. As you said.

Last edited:
Dick said:
I think the problem statement says the magnitude is av^2. So I don't think they mean the vector v to be a unit.

Hi Dick, That was my point; it wasn't a (magnitude)*(unit-vector) construct, but rather a magnitude*vector (vv as opposed to v2u).

On reflection, I can see the utility of doing it this way when the magnitude you want is the square of the vector. vv rather than v2*(v/v).

gneill said:
Hi Dick, That was my point; it wasn't a (magnitude)*(unit-vector) construct, but rather a magnitude*vector (vv as opposed to v2u).

On reflection, I can see the utility of doing it this way when the magnitude you want is the square of the vector. vv rather than v2*(v/v).

Hi gneill, yeah, I realized what you meant after I posted. Nothing wrong with the unit vector notation either. At least it makes the velocity dependence clearer than it was to kraigandrews.

## 1. How does mass affect the rate of falling in Earth's gravity?

Mass does not affect the rate of falling in Earth's gravity. According to the law of universal gravitation, all objects fall at the same rate regardless of their mass. This means that a feather and a brick, if dropped from the same height, will hit the ground at the same time.

## 2. Does the shape of an object affect its rate of falling in Earth's gravity?

No, the shape of an object does not affect its rate of falling in Earth's gravity. As mentioned before, all objects fall at the same rate regardless of their shape or mass. The only factors that affect the rate of falling are air resistance and the strength of the gravitational force.

## 3. What is the acceleration due to gravity on Earth?

The acceleration due to gravity on Earth is approximately 9.8 meters per second squared (m/s²). This means that for every second an object falls, it increases its speed by 9.8 m/s. This value can vary slightly depending on the location on Earth, as well as the height and mass of the object.

## 4. Can an object ever stop accelerating in Earth's gravity?

No, an object cannot stop accelerating in Earth's gravity unless it encounters an opposing force that is equal to or greater than the force of gravity. This is because gravity is a constant force that acts on all objects, causing them to continuously accelerate towards the center of the Earth.

## 5. How is the rate of falling affected on other planets?

The rate of falling on other planets is affected by the strength of their gravitational force. For example, on a planet with a stronger gravitational force than Earth, objects will fall faster. On a planet with a weaker gravitational force, objects will fall slower. However, the rate of falling will still be the same for all objects regardless of their mass or shape.

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