You are Here: Home >< Physics

# Acceleration Due to Free Fall Watch

1. Question 1 (c)

The acceleration due to free fall will be :

the resultant force on mass/mass of object.

Now we have the centripetal force acting towards the centre of the planet and the gravitational force of attraction that the planet exerts on the mass... again towards the centre of the planet.

Thus, shouldn't it be the sum of the centripetal force on the mass and the gravitational force and not the difference ?
2. bump
3. (Original post by Ari Ben Canaan)
Question 1 (c)

The acceleration due to free fall will be :

the resultant force on mass/mass of object.

Now we have the centripetal force acting towards the centre of the planet and the gravitational force of attraction that the planet exerts on the mass... again towards the centre of the planet.

Thus, shouldn't it be the sum of the centripetal force on the mass and the gravitational force and not the difference ?
No. For 2 reasons.
If your argument was correct, then as the Earth spins faster the acceleration of free fall would get greater. The opposite is the case. If the Earth span fast enough such that the centripetal acceleration was equal to g, then objects on the surface would not appear to fall at all, giving a measured value of zero for g. So clearly from a common sense perspective, it cannot be the sum of the two, but rather the difference.
Why?
There is only one real physical force acting on the object, that is the gravitational attraction of the Earth. F. This is constant.
When the Earth is not spinning, letting the object fall produces an acceleration towards the centre g, due to gravity where F = mg
If the object is also moving in a circle it has a centripetal acceleration (towards the centre) a.
This acceleration also has to be provided by the force F (This is the only place it can come from). Both accelerations are towards the centre and add.
So now F = m(g' +a) where g' is the new acceleration measured due to gravity.
F=mg' + ma
So mg' = F - ma
4. (Original post by Stonebridge)
x
I realise its been a month since this was posted but I think I've finally understood this.

Just check if my explanation correct.

Centripetal force can be thought of as the resultant acting on an object moving in a circle.

Gravitational Force - Normal Force = centripetal force

Gravitational Force - mg = centripetal force

Leading to Gravitational Force - Centripetal force = g (m=1 because acc. due to free fall is resultant force per UNIT mass).

I realise you've said essentially the same thing but this is just me re-encapsulating it in my own words.
5. Also, apparent weightlessness would be when Gravitational force = centripetal force, right ?
6. (Original post by Ari Ben Canaan)
I realise its been a month since this was posted but I think I've finally understood this.

Just check if my explanation correct.

Centripetal force can be thought of as the resultant acting on an object moving in a circle.

Gravitational Force - Normal Force = centripetal force

Gravitational Force - mg = centripetal force

Leading to Gravitational Force - Centripetal force = g (m=1 because acc. due to free fall is resultant force per UNIT mass).

I realise you've said essentially the same thing but this is just me re-encapsulating it in my own words.
Yes if by g here you mean the "measured" value of g. (What I called g')

(Original post by Ari Ben Canaan)
Also, apparent weightlessness would be when Gravitational force = centripetal force, right ?
Yes.
"Weightlessness" can be a problem because of the different definitions people have of it. In particular how they define "weight" in space.
Weightlessness is certainly a "feeling". (Or lack of!)
If you are in circular orbit around the Earth you will certainly feel "weightless" and the centripetal force on you will be equal to the gravitational force.
7. (Original post by Stonebridge)
Yes if by g here you mean the "measured" value of g. (What I called g')

Yes.
"Weightlessness" can be a problem because of the different definitions people have of it. In particular how they define "weight" in space.
Weightlessness is certainly a "feeling". (Or lack of!)
If you are in circular orbit around the Earth you will certainly feel "weightless" and the centripetal force on you will be equal to the gravitational force.
So basically the cosmonauts/astronauts in the ISS are experiencing weightlessness because they are at a particular distance from the earth such the gravitational force = centripetal force.....
8. i think it's better explained due to the velocity rather than the distance at which they are travelling, mv^2/r = g. f.
9. (Original post by Ari Ben Canaan)
So basically the cosmonauts/astronauts in the ISS are experiencing weightlessness because they are at a particular distance from the earth such the gravitational force = centripetal force.....
Yes. They are in orbit and the centripetal force on them is equal to the gravitational force on them.

Updated: April 7, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### HELP!

Mum wants to come to Freshers

### How posh are you?

Poll
Study resources

The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.