Free fall

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in free fall without air resistance what is the force that causes the constant acc? Is it the object weight ?
And in free fall with air resistance what is the applied force ? Is it the weight of the object and the resistive force is the air resistance? And the net force is the one that causes the acc?
 

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haruspex
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in free fall without air resistance what is the force that causes the constant acc? Is it the object weight ?
Yes. Weight is the force exerted by a gravitational field on a mass. But note that the acceleration is never actually constant. As the object gets closer to the source of the gravitational field, the force, and therefore the acceleration, increase.
And in free fall with air resistance what is the applied force ? Is it the weight of the object and the resistive force is the air resistance? And the net force is the one that causes the acc?
Yes.
 
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Yes. Weight is the force exerted by a gravitational field on a mass. But note that the acceleration is never actually constant. As the object gets closer to the source of the gravitational field, the force, and therefore the acceleration, increase.
Huh? But g is 10m/s !
 
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Huh? But g is 10m/s !
As you move further away from the Earth's surface, the value of g decreases proportional to 1/distance2. Only at sea level is g ~10m/s2. You do have to be very high though until you notice the difference, for the astronauts in the international space station g is still > 9m/s2.
 
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Chronos
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If you fell all the way to the center of the earth, your acceleration would be zero - although you would have a very respectable velocity.
 
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If you fell all the way to the center of the earth, your acceleration would be zero - although you would have a very respectable velocity.
Ummm ok, i take it that your point is about tha fact that gravitational pull is exerted from the centre of the earth?
 
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Just to double check if what I assume is correct, for free fall without air resistance, the applied force is the weight, the acceleration is g-10m/s , because for resultant force f=ma and in this case weight is the resultant force as there is no air resistance present, so weight divided by the mass of the object gives us the acceleration due to gravity for free fall , right ?
 
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Just to double check if what I assume is correct, for free fall without air resistance, the applied force is the weight, the acceleration is g-10m/s , because for resultant force f=ma and in this case weight is the resultant force as there is no air resistance present, so weight divided by the mass of the object gives us the acceleration due to gravity for free fall , right ?
Right. However, note that acceleration is measured in meters per second squared, not meters per second. The value on earth would therefore be 9.8 m/s2.
 
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i take it that your point is about tha fact that gravitational pull is exerted from the centre of the earth?
no...only when you are outside the radius of the earth.

In F = ma, when a = g, F = W......so W = mg
 
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Gravitational pull is definetly exerted from the centre of the Earth. The distances in the equations are measured from the centre of the eARTH...R NOT H !!!
Inside the Earth, falling down a mine shaft, the pull is still towards the centre but may not be 9.8m/s^2

"We know much, we understand little."
 
  • #11
haruspex
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Gravitational pull is definetly exerted from the centre of the Earth.
This is not quite true, and can lead to misunderstandings.
The gravitational pull of the earth is exerted independently by every little speck of matter in it.
If you are outside a uniform hollow shell, it just happens that the net pull of that shell is exactly the same as if the entire mass were concentrated at the shell's centre. This only happens in 3 dimensions. If you are inside the shell, all those individual pulls just happen to cancel out, leaving no net field.
From this, it follows (treating the earth as uniform and spherical, neither of which is true) that the pull from above earth's surface at distance r from its centre is proportional to 1/r2, whereas if r is less than the radius of the earth it is proportional to r.
In practice, the earth is an oblate spheroid. You are further from the centre when at the equator than when at the poles, but because it's not a sphere, the field does not act as though all the mass were at the centre; the reduction in gravity is less than you might expect based purely on the extra distance from the centre. OTOH, the earth's spin makes apparent gravity about .3% less.
 

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