Fall Duration Calculator: Objects of Different Mass from 10m to 2m

[Luc B]

Hello.

Sorry for my English...
I have made some calculations. By Iterative calculation. With Excel.
The time that need two objects to fall from 10 meters to 2 meters.
The mass of the objects i choose are:
1 and 10 kilos.
2 and 10 kilos
5 and 10 kilos.
And.
1 and 9 kilos
3 and 7 kilos.
5 and 5 kilos.

What i found as duration surprise me.
More time for 1 and 10 kilos than for 5 and 10 kilos.
And the same time for 1 and 9 kilos than for 5 and 5 kilos.

Can someone redo my calculations?

Luc B.

 

[phinds]

Can someone redo my calculations?

WHAT calculations? I don't see any calculations. If I understand your problem correctly, you obviously have the wrong answer so what we need is to see YOUR calculations to see where you went wrong

 

[Luc B]

Thanks for your reply.
How do i make the calculation?
First, i calculate the distance an object travels. x.m1 = y . m2 and x+y = 10 - 2 meters.
I calculate the cinetic energy. 1/2 . m . v² at different places along the travels.
And i calculate the average speed for each portion of the travels.
And i have the duration of the travel.
With Excel, it is possible to divide the travels in thousands parts.

What are the results?

1 and 10 kilos. 14.4 days
2 and 10 kilos. 13.8 days
5 and 10 kilos. 12.3 days
And.
1 and 9 kilos. 15.1 days
3 and 7 kilos. 15.1 days
5 and 5 kilos. 15.1 days.

Luc B.

 

[phinds]

DAYS ?

 

[jbriggs444]

x.m1 = y . m2 and x+y = 10 - 2 meters.

What is x? What is m1? What is y? What is m2? What is the relevance of x+y?

Like @phinds, I still see no calculations.

 

[Luc B]

Hello.

Phinds. Yes, days. But if you want, give me other mass and distance, and Excel make the calculation in a few micro secondes...
Jbriggs444. x is the distance that travels one object. y the distance that travels the other object.
And the total distance is x + y. here 8 meters (10 - 2)

I have already post this in french forum. My first language.
A lot of people say me that i am wrong.
Noone make the calculation by himself....

Luc B.

 

[phinds]

Phinds. Yes, days.

OK, SOMETHING is weird here. You problem statement says you want the time for an object to fall from 10 meters to 2 meters. This would be a few seconds at most, yet you say DAYS. ?

 

[Luc B]

Hello phinds.

It is not object that fall on the earth.
It is an object of 1 kg that fall towards an object of 10 kg.
And sorry, it takes days.....
But perhaps, my calculation are wrong.
When an astronaut goes out the international station. He doesn't fall on its in a few seconds. Does he?

Luc B.

It is very late in my country. Sorry but i have to sleep. See you tomorrow...

 

[Nidum]

You have two objects in free space and the only forces acting on the objects are mutual gravitational attraction . You are attempting to work out how long it takes for the objects to move towards each other by a specified distance . Initial separation distance is known and initial relative velocity between objects is zero .

That's not too difficult a calculation if the objects are relatively small compared to the distances involved ?

 

[phinds]

It is not object that fall on the earth.
It is an object of 1 kg that fall towards an object of 10 kg.

Ah. OK, I misunderstood the problem.

 

[berkeman]

Ah. OK, I misunderstood the problem.

Not your fault. The original question was lacking clarity in the extreme; we all were scratching our heads.

 

[pixel]

You have two objects in free space and the only forces acting on the objects are mutual gravitational attraction . You are attempting to work out how long it takes for the objects to move towards each other by a specified distance . Initial separation distance is known and initial relative velocity between objects is zero .

That's not too difficult a calculation if the objects are spherical and relatively small compared to the distances involved ?

But maybe not very simple as the acceleration is not constant. I found this thread that has some suggested solutions (which I can't vouch for):
https://physics.stackexchange.com/q...-masses-will-collide-due-to-Newtonian-gravity

 

[Janus]

Here they give an equation to find the fall time for two masses from a distance of r to a distance of x:
https://en.wikipedia.org/wiki/Equations_for_a_falling_body

Note that if x and r remains unchanged, then the relative time only depends on
\frac{1}{\sqrt{2u}}

where u = G(M+m)
with M and m being the masses involved.
1kg+10kg = 10kg and 5kg+10kg =15kg, and 1/10 is larger than 1/15, which agrees with your results with the 1kg-10kg pair and the 5kg-10kg pair.

1kg+9kg, 3kg+7kg, and 5kg+5kg all equal 10kg, so you should get the same fall time for all three combinations of masses.

 

[Luc B]

Hello.
And thanks you very much.

That means:
1/ Aristote was rigth. Heavy objects fall faster ...
2/ Galilée was rigth. All the objects of the Earth fall in the same way on earth.
3/ Einstein was wrong. And time is not relative.
That's all for now

Luc B.

 

[Dale]

That means:
1/ Aristote was rigth. Heavy objects fall faster ...
2/ Galilée was rigth. All the objects of the Earth fall in the same way on earth.
3/ Einstein was wrong. And time is not relative.

Or
4/ You have misunderstood what one or more of the above said

You may want to read about the reduced mass
https://en.m.wikipedia.org/wiki/Reduced_mass

 

[Luc B]

Hello Dale.

I will come back here in a few time.
I am sure that i am wrong somewhere.
Perhaps in the calculation.
I am sure someone will redo them. And i will find my mistake.
See you here later.

Luc B.

 

[Dale]

Did you even bother to read the link? What is the reduced mass of the combinations you mentioned? What is the gravitational force? How does it depend on the product and on the sum of the masses?

The conundrum you mention here does not need numerical calculations. It can be investigated analytically very easily using the concept of the reduced mass. Which is the reason that I pointed it out to you

 

[Luc B]

Hello Dale.
Thanks for your answer.
I know about reduced mass. I have read a lot but in french. I promise i will read in English. Very good exercice for me.
But i think (not sure) it is an approximation. But i am not sure.
I have an other question. I am not sure i use the good words.
The liberation speed of the Earth is 11,2 km/s. I know how we can calculate its. How we find its.
We take only the attractive force of the planet on object that goes away.

But

If we have a small planet of 1000 kg. And a rocket of 1000 kg.
What is the liberation speed of the rocket about this planet.

Sorry. But i do not find its.

Luc B.

 

[Dale]

But i think (not sure) it is an approximation.

No, it is exact.

 

[Luc B]

Hello Dale

Ok ok. I will read it.
Did you make any analitic calculation?
To confirm or not my numbers?
I read it. In English...

Luc B.

 

[Luc B]

Hello

Yes. You are rigth. It is exact.
But they do not speak about time.
The same that in the french forum.
A lot of people say "hey, it is knowed".
But no one make a CALCULATION.

So

Aristote was rigth.
Galilée was rigth...

Luc B.

 

[Janus]

Hello Dale.
Thanks for your answer.
I know about reduced mass. I have read a lot but in french. I promise i will read in English. Very good exercice for me.
But i think (not sure) it is an approximation. But i am not sure.
I have an other question. I am not sure i use the good words.
The liberation speed of the Earth is 11,2 km/s. I know how we can calculate its. How we find its.
We take only the attractive force of the planet on object that goes away.

But

If we have a small planet of 1000 kg. And a rocket of 1000 kg.
What is the liberation speed of the rocket about this planet.

Sorry. But i do not find its.

Luc B.

In both cases to get an exact answer you need to take the sum of the masses into account. With the case of a 1000 kg rocket and the Earth, this works out to be 5.972 x 10^24kg +1000kg, which for all practical purposes equal 5.972 x 10^24 kg, as the difference in the answer you'd get by using just the Earth's mass and the sum of the masses is too small to reasonably measure (in fact, the uncertainty in the value for the mass of the Earth is much larger than 1000 kg).

With the 1000 kg planet and 1000 kg rocket, using the sum of the masses vs.just the mass of the planet makes a noticeable difference in the answer. Whether or not you use the sum of the masses, or just that of one depends on the relative masses and just how accurate you need the answer to be.

 

[Luc B]

Hello.

Ok. Ok.
But
To calculate the liberation's speed (are these words the good words?) of a rocket from the earth, we take care about the force that the Earth exerts on the rocket. And the acceleration (deceleratin...) of the rocket. But we do not take care of the force that the rocket exerts on the earth. And the acceleration of the earth.
If the rocket is the same mass than the planet that it leaves, they is no liberation's speed.
I think. I am not sure.

Luc B.

 

[jbriggs444]

liberation's speed (are these words the good words?)

Are you talking about "escape velocity" -- the minimum speed which an object would need to have in order for a free fall trajectory to take it infinitely far away from a gravitating object?

If so, the concept of "reduced mass" makes it immediately clear that an escape velocity exists and is well defined, even in the case where the mass of the escaping object is not negligible.

 

[Luc B]

Ok Jbriqqs444.
Thank you very much. In french we say "vitesse de libération". And i learn in English "escape velocity".
And what is the value for a rocket of 1000 kg that escapes a planet of 1000 kg.
You must say in whish Galilean reference system you are working i think.

Luc B.

 

[jbriggs444]

And what is the value for a rocket of 1000 kg that escapes planet of 1000 kg.

At what initial separation? Escape velocity depends on initial separation as well as mass. Since you began this thread by speaking of an eight meter fall (from 10 m down to 2 m), let us work to calculate the escape velocity from a separation of 2 m. This is a radius of 1 m from the combined center of mass.

The Physics Forums way is not to spit out answers. Our way is to guide you to calculate the answers for yourself.

https://en.wikipedia.org/wiki/Reduced_mass

What is the reduced mass for the case of two 1000 kg masses?

 

[CWatters]

A lot of people say "hey, it is knowed".
But no one make a CALCULATION.

You have to be careful about relying on calculations. You can't use Newton's laws to prove Newton is correct. What you really need is data. Newton was correct until someone noticed a problem with the orbit of Mercury.

 

[Luc B]

Hello
Jbriqqs444. The reduce mass is 500 kg. In case of two mass of 1000 kg...
Cwaters. People has made calculations about solar system before we see Mercury problem...

Thank for your answers.

Luc B

 

[jbriggs444]

People has made calculations about solar system before we see Mercury problem...

What question are you asking?

The reduce mass is 500 kg. In case of two mass of 1000 kg...

Agreed, the reduced mass is 500 kg.

The idea is that this can be used as the if it were the inertial mass of the object in describing its movement with respect to the gravitating primary. The object's gravitational mass remains unchanged.

What is the gravitational potential energy of one 1000 kg object in the field of the other?

 

[Janus]

Ok Jbriqqs444.
Thank you very much. In french we say "vitesse de libération". And i learn in English "escape velocity".
And what is the value for a rocket of 1000 kg that escapes a planet of 1000 kg.
You must say in whish Galilean reference system you are working i think.

Luc B.

The escape velocity would be found by:
v_e = \sqrt{\frac{2G(M+m)}{r}}
G is the gravitational constant,
M and m are the masses of the planet and rocket.
r is the radius of the planet (or more properly, the distance between the center of the planet and the rocket for the point you wish to calculate the escape velocity for.)
In this case the escape velocity is the relative velocity the two must have with respect to each other to prevent their mutual gravity from ever pulling them back together. It doesn't matter whether you are measuring the rocket's velocity relative to the planet or the planet's velocity relative to the rocket.

 

[Luc B]

Hello Science Advisor Jbriqqs444

My question was: Are my calculations correct? About time of objects falling one to an other.
And i haven't still any answer.

About escape velocity, before making any calculation about speed (velocity), how many time need an object to go to the infinity.
An infinite time i think. And it will never be there.
So, is there any speed (velocity) whish send its to there?
I think not.
Am i rigth?

Thanks if you answer.

LucB.

 

[PeroK]

Hello Science Advisor Jbriqqs444

My question was: Are my calculations correct? About time of objects falling one to an other.
And i haven't still any answer.

About escape velocity, before making any calculation about speed (velocity), how many time need an object to go to the infinity.
An infinite time i think. And it will never be there.
So, is there any speed (velocity) whish send its to there?
I think not.
Am i rigth?

Thanks if you answer.

LucB.

If you fire something away from the Earth at a given speed, then either:

a) It eventually falls back to Earth

b) It doesn't (ever fall back).

The lowest speed at which it never comes back is the escape velocity.

 

[Luc B]

Hello Perok.
Thanks for this information...

Fire. You means "tirer" in french.
Do not forget the third low of Newton. action and reaction.
Specially if you are NOT on the earth. But on a very small planet.
And your gun is very big.

Luc B.

 

[PeroK]

Hello Perok.
Thanks for this information...

Fire. You means "tirer" in french.
Do not forget the third low of Newton. action and reaction.
Specially if you are NOT on the earth. But on a very small planet.
And your gun is very big.

Luc B.

Yes, "tirer". I would never forget Newton's third law!

The escape velocity is the relative velocity between the rocket and the planet. A normal planet hardly moves at all. For an asteroid, say, then it's definitely the relative velocity that you need. Not the velocity of the rocket in the original rest frame.

In any case, there is an escape velocity.

In fact, for a 1000 kg asteroid, you could easily jump off it and you would never come back to it. The escape velocity would be very low.

 

[Luc B]

Hello Perok.

I agree with you.
But what is the relative velocity between an asteroid of 1000 kg and a bullet of 1000 kg.
I know this forum is not to give answer but to help people.
My answer is: there is no velocity possible. The object fall always back. OR. Never reach the infinite.
Why. Because of the time, the infinite duration. Whish not exist.

Mathematic is a language. It is not physic.
Take an apple. Cut its in a infitive pieces.
What have you? An infinite of small, very small piece of apple.
Put them together. You have a cuted apple.Divide 3 by infiinity. You have nothing. And you don't respect the first low of thermodynamic.
Am i rigth?

Luc B.

 

[PeroK]

Hello Perok.

I agree with you.
But what is the relative velocity between an asteroid of 1000 kg and a bullet of 1000 kg.
I know this forum is not to give answer but to help people.
My answer is: there is no velocity possible. The object fall always back. OR. Never reach the infinite.
Why. Because of the time, the infinite duration. Whish not exist.

Mathematic is a language. It is not physic.
Take an apple. Cut its in a infitive pieces.
What have you? An infinite of small, very small piece of apple.
Put them together. You have a cuted apple.Divide 3 by infiinity. You have nothing. And you don't respect the first low of thermodynamic.
Am i rigth?

Luc B.

No, you are wrong.

If the bullet comes back, then it must come back at some (finite) time. Try to calculate that time and you'll find that there is no such time. Therefore, the bullet never comes back.

There is no need to take about "infinity".

 

[Luc B]

Hello gold member... Perok

Ok ok. I am wrong.

I think the bullet comes always back..
So in some (finite) time.
And the escape velocity doesn't exist.

But you say that escape velocity exist.
And your " colleague" Janus has given me how calculate its.
My computer said: 0.00036524 meter per second.
Is that rigth, according to you?

Luc B.

 

[PeroK]

Hello gold member... Perok

Ok ok. I am wrong.

I think the bullet comes always back..
So in some (finite) time.
And the escape velocity doesn't exist.

But you say that escape velocity exist.
And your " colleague" Janus has given me how calculate its.
My computer said: 0.00036524 meter per second.
Is that rigth, according to you?

Luc B.

Yes, that's right.

If you say the bullet comes back, then you have to give a time when it comes back. That's what it means to "come back". So, if you take a velocity of 0.000364 m/s, then you can say what time it comes back.

But, if you have a velocity of 0.000366 m/s, then the equation for the time that the bullet comes back has no solution.

And, if an equation has no solution, then it's not good maths or physics to say: eh bien, I don't believe that - it must have a solution! If there is no solution, there is no solution, and the bullet never comes back.

 

[PeroK]

@Luc B

Here's another way to look at it. Suppose you fire a rocket from Earth at 20,000 m/s (which is geater than the esacpe velocity). One thing you can calculate easily is how much the rocket slows down. In this case, the rocket never goes slower than 16,500 m/s.

So, not only does it not come back, but it is always traveling away from the Earth at more than 16,500 m/s.

 

[Luc B]

Hello Perok.

I think you are rigth.
I will make my own calculation tomorow.
It's sleeping time in my country.
I have had a "good" discussion with your "friend" Janus. About Eddington photographics en 1919 and a little experiment of Fresnel (+/- 1800).

{request for personal information deleted by a moderator}

See you tomorow.

Luc B.

 

[CWatters]

Cwaters. People has made calculations about solar system before we see Mercury problem..

You missunderstand what I was saying.

If you use an equation invented by someone you will get the same answer as that person. That does not mean they are 'correct'.
My point is that you cannot use

 

[Luc B]

Ok ok. CWatters

There are a lot of writers here. You and some others.
The same than in a french forum. The physics forum of the Liège University. And Sentinelle is a very important writer.
And all of you say that i am wrong.
Ok. I agree. I am wrong.
Are you happy? Is that rigth for you?
It is for me.
Time to sleep in my country.
See you tomiorrow, CWatters. Happy to know you.
Happy to have found someone who will say me "What to do".

Luc B.

 

[Luc B]

Ok ok Nidium

Do the calculation. If it is not so difficult.
And Thanks a lot.

Luc B.

 

[Dale]

Hello Dale

Ok ok. I will read it.
Did you make any analitic calculation?
To confirm or not my numbers?
I read it. In English...

Luc B.

The numbers are not the important thing. The relationship you found is the important thing:

What i found as duration surprise me.
More time for 1 and 10 kilos than for 5 and 10 kilos.
And the same time for 1 and 9 kilos than for 5 and 5 kilos.

This relationship can be obtained analytically using the reduced mass, Newton’s law of gravitation, and Newton’s 2nd law. It is super easy, you can probably do it in your head.

 

[Luc B]

Hello Dale

What i found is correct. Don't it?
And i come back with my conclusions. Whish are also correct. Following to me.
And may be, i am wrong. May be.

If we have the Earth. And nothing change about its. And there is an object of 100 kg that falls on it. It take more time than a object of 500 kg. I must be specific. Same height, Exactly same Earth. The same conditions. And the objects didn't come from the Earth. Aristote was rigth in - 350...
We have the Earth. And we take something from the Earth. And let its fall. Always from the same height. No matter the mass we take from the Earth to let its fall. It takes always the same time. And in 1600, Galilée was rigth in 1600.

But.

Now people think that all the objects fall on the Earth in the same way. And this is not correct.

Luc B.

 

[ZapperZ]

Hello Dale

What i found is correct. Don't it?
And i come back with my conclusions. Whish are also correct. Following to me.
And may be, i am wrong. May be.

If we have the Earth. And nothing change about its. And there is an object of 100 kg that falls on it. It take more time than a object of 500 kg. I must be specific. Same height, Exactly same Earth. The same conditions. And the objects didn't come from the Earth. Aristote was rigth in - 350...
We have the Earth. And we take something from the Earth. And let its fall. Always from the same height. No matter the mass we take from the Earth to let its fall. It takes always the same time. And in 1600, Galilée was rigth in 1600.

But.

Now people think that all the objects fall on the Earth in the same way. And this is not correct.

Luc B.

First of all, there is a language issue here, and it is a factor in deciphering what you are saying, and what you understood of what has been written to you. But be VERY careful in trying to push your own agenda. If you did not clearly understand the PF rules that you had read when you joined this forum, it might do you go to go back and re-read and UNDERSTAND those rules. They are strictly enforced.

Next, your claim is both correct and not correct, but it depends on the situation.

https://www.physicsforums.com/insights/why-is-acceleration-due-to-gravity-a-constant/

IF you are dealing with "g" being a constant, then ALL objects will fall at the same time. The mathematics describing the physics will prove this. However, once either the object is comparable to the size of the earth, or if one removes the constancy of "g", then one has to redo the calculation. I do not believe you have done the latter IF you are still under the assumption that "g" is a constant.

The infamous experiment of dropping objects of different weights and then hitting the ground at the same time are done on a terrestrial scale and with the explicit assumption that "g" doesn't change, which is a perfectly valid approximation. Classical mechanics in itself (i.e. your "calculation") is an approximation.

Zz.

 

[Dale]

What i found is correct. Don't it?

Yes

And i come back with my conclusions. Whish are also correct.

I don’t think so. You state your conclusion as a contest between people. As though they were wrestling or something and there can only be one winner.

Your conclusions are not only poorly stated as a personal contest, but they are also not logically connected to the calculation. A correct analysis would be to start with the same experimental scenario, identify the laws used by each person, determine the prediction for each, and compare the predictions to the actual results of the experiment.

The classical result is reached by using both Newton’s 2nd law and Newton’s law of gravity, so the conclusion clearly agrees with Newton. As far as I know, Galileo also used Newton’s laws, but I don’t know what laws Aristotle used.

 

[Luc B]

Hello
To Mr ZapperZ.
First. I have written: "And may be, i am wrong. May be". But i have seen no other number than mine. I am waiting for that. To compare.
Then, i have not made the calculation with g constant. I have made the calculation following the laws of Newton. Yes it is an approximation. Because i make my calculations by iteration. But it is very easy to control the precision. You can make the calculation with the maximum speed for each portion. And the minimum speed... And with excel, it is easy to divide the distance in 2 or 3000 portions...
But my calculations can be wrong...

To Mr Dale
Thank you for your reply.
BUT
Aristote lived round -350 (ante christum natum...). He was a thinker. He has said that our Earth was in the center of the universe. Motionless. And he had only his inner ear (oreille interne) to say that. And he had look to falling objects. And a stone goes faster than a leaf (feuille d'arbre). Aristote didn't use any law. Only his mind...
2000 years later, we are round 1600, Galilée began to teach young people in Pise, italy i think. And he was teaching Aristotle's ideas.....
And it is also said that he dropped stones from the top of the pisa tower. (tour de PIse, Italia). He had also no law for the gravitation effects.
And Newton came round 1700. And has written the tree laws. About gravitation, force and movement..............

F = G.m1.m2/d² F= m1 . a1 and F12 = F21. That all.

Perhaps, i am not sure. Your conclusions are NOT correct.

It is clearly impossible that Galileo has used the law of Newton...

Luc B.

 

[ZapperZ]

Hello
To Mr ZapperZ.
First. I have written: "And may be, i am wrong. May be". But i have seen no other number than mine. I am waiting for that. To compare.
Then, i have not made the calculation with g constant. I have made the calculation following the laws of Newton. Yes it is an approximation. Because i make my calculations by iteration. But it is very easy to control the precision. You can make the calculation with the maximum speed for each portion. And the minimum speed... And with excel, it is easy to divide the distance in 2 or 3 portions...
But my calculations can be wrong...

Please read the link that I have given you. Many of what you've stated in your posts have been addressed.

Unless you are solving a differential equation that produces a non-analytical solution, there is no extra benefit in solving this ".. by iteration..". My point still stands. If M_earth>>m_object on earth, then what is the big deal here? The physics shows that "g" can be accurately modeled as a constant. Do you think people who build houses consider the variation in "g" when they construct those buildings?

Zz.

 

[Luc B]

Ok. Ok. Mr ZapperZ

You are rigth. For the falling of object on Earth. From distance that human people use currently, there is not significant difference to use the Newton's Laws or a g constant.
And when Galileo said :"All object fall the same way on Earth". If we stay with objects on the human scale that fall from height also on a human scale, It's the truth.
I have made the calculations between objects of some kilo.
WHY?
Because i have NO computer that can make a différence between an object of 1 kg whish fall on Earth. And an object of 10 kg. I hope you understand why.
It is because Earth mass + 1 of 10 kg is always Earth mass for a computer.
With your raisonning, Jupitert falling on the Earth is the same than 1 kg falling on the Earth.
Sorry to shake your truths

Luc B.