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Falling objects and Einstein’s equivalence-principle?


Hello2
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Hello, I would like to know more about falling objects and Einsteins equivalence, and I have a question:

 

The acceleration of falling objects with different weights  differ on Earth when it is done in vacuum or not;

Is that also the case with ‘falling objects’ as in Einstein’s rocket that is accelerating?

Edited by Hello2
Forgot first sentence;
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21 minutes ago, Hello2 said:

The acceleration of falling objects with different weights  differ on Earth when it is done in vacuum or not;

Er, no. The acceleration of falling objects is the same regardless of their weight (mass). This was demonstrated by Galileo about 400 years ago.

 

Here is an overview: http://physics.ucr.edu/~wudka/Physics7/Notes_www/node49.html

It is very easy to prove this is the case by using a couple of Newton's equations.

The force on an object with mass m is given by: f = G * m * M / r2, where M is the mass of the Earth, G is the gravitational constant and r is the distance between the centre of the Earth and the object.

In other words, the force is proportional to the mass of the object: f = g * m (where 'g' is just a constant factor that includes G, M and r).

But the relationship between force and acceleration of an object is given by: f = m * a

Rearranging, we get a = f / m

Substituting f from above: a = g * m / m

So, the acceleration a = g,

g, from above, is G * M / r2, which on Earth is about 9.8m/s2

https://www.wolframalpha.com/input/?i=G+*+(mass+of+Earth)+%2F+(radius+of+Earth)^2

 

Edited by Strange
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10 minutes ago, Strange said:

Er, no. The acceleration of falling objects is the same regardless of their weight (mass). This was demonstrated by Galileo about 400 years ago.

 

And also ,I presume under any form of acceleration (in a vacuum)?

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2 minutes ago, geordief said:

And also ,I presume under any form of acceleration (in a vacuum)?

Not necessarily. They key thing is that the force due to gravity is proportional to mass and the acceleration is inversely proportional to mass.

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1 hour ago, Strange said:

Not necessarily. They key thing is that the force due to gravity is proportional to mass and the acceleration is inversely proportional to mass.

So if you are on a spaceship and hit the after burners some  of the cargo loose in the hold would move to the rear faster than other items?

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15 minutes ago, geordief said:

So if you are on a spaceship and hit the after burners some  of the cargo loose in the hold would move to the rear faster than other items?

No, because F = ma, and a = F/m

The more massive ones feel a stronger force, but have more resistance to acceleration. The two effects cancel, as they have to.

In addition, gravitational mass and inertial mass are the same, so you can't tell the difference between ma and mg in a uniform field.

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24 minutes ago, swansont said:

No, because F = ma, and a = F/m

The more massive ones feel a stronger force, but have more resistance to acceleration. The two effects cancel, as they have to.

In addition, gravitational mass and inertial mass are the same, so you can't tell the difference between ma and mg in a uniform field.

That's what I imagined. I was trying to make sense of Strange's "not necessarily" which seemed to me to imply the opposite of my suggestion just two posts up (I misinterpreted ,no doubt). 

Edited by geordief
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(Do you know next professor:

Julius Sumner Miller Lesson 6: Concerning Falling Bodies & Projectiles

https://www.youtube.com/watch?v=EV9wIJF6PaE

??)

On 31-8-2018 at 4:12 PM, Hello2 said:

---

The acceleration of falling objects with different weights  differ on Earth when it is done in vacuum or not;

Is that also the case with ‘falling objects’ as in Einstein’s rocket that is accelerating?

 

I thought I had seen video’s saying: There is no difference between the gravity on Earth and “falling objects” because of acceleration in that Einstein’s rocket.

What I meant to ask was: On Earth there is a difference when falling objects fall in a vacuum or fall in not a vacuum.

Is there also that difference in Einstein’s rocket when it is accelerating?

 

 

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2 hours ago, Hello2 said:

Is there also that difference in Einstein’s rocket when it is accelerating?

Yes.

I should note here though that the equivalence is between uniform acceleration, and a uniform gravitational field. This means the equivalence holds only locally, in a small enough region. On larger scales, a gravitational field due to a mass distribution (such as a planet) is not uniform.

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