Non constant acceleration in gravitation field

[philipp2020]

To calculate the exact fall time of a mass, constant g is often used for a short distance
traveled as for example in this paper:

http://scitation.aip.org/getpdf/servlet/GetPDFServlet?filetype=pdf&id=AJPIAS000044000009000855000001&idtype=cvips&doi=10.1119/1.10277&prog=normal

In this paper they come to a average time of 0.24470 for an object droped from 1 feet.

Now tried to calculate the exact fall time value including non consistent acceleration, but I couldn't come quiet close to the same values as with constant acceleration. My result shows it should take 0.8483364538s for the object to fall from one feet. As for mass of the Earth I was using 5.974*10^24kg and the radius of the Earth 6371000m. I don't think that the unprecise numbers for mass and radius are the reason for the different result, so there must be something wrong in my formula. Maybe someone could help me find the problem with my equations?

To calculate the falling time I used the following formula for non consistent acceleration and then solved for t at the end:

<br /> a = \frac{-GM}{r^2}t

v =\int_0^t \! \int_{0637100,348}<br /> ^{6371000} \frac {-GM}{r^2}t\,dt\,dr = -3.42\times \frac {1}{2}t^2

S = \int v \,dt = \int_0^t <br /> \frac{1}{2}t^23.42\,dt =\frac{1}{6}t^3\times-3.42

Ps: Sorry for double post, forgot title in first one...

 

[rcgldr]

In this case, acceleration is a function of position, not time. (You'd need to go through a lot of math to determine acceleration as a function of time). Posts #11 and #19 in this older thread show the math for the time it takes for two point objects at some distance apart to collide due to gravity.

https://www.physicsforums.com/showthread.php?t=635188

 

[philipp2020]

Thanks very much for the answer!

As I wanted to calculate what difference it makes to include both masses A and B or only the mass A I am a bit confused right now. :)

Getting to step one in the other thread the following is used:

v= dr/dt

a = dv/dt

multiply by dv/dt by dr/dr:

a = (dr dv)/(dt dr) = v dv/dr

This gets you to the first step:

v dv/dr = -G (m1 + m2) / r2

v dv = -G (m1 + m2) dr / r2Here m2 is also included as the mass of the dropped object. But for all experiments so far starting with galileo, the mass of m2 doesn't matter. For example a feather in a vacuum arrives at the exact same time on the floor as a lead ball. So why is it possible to use the 2nd mass in this step?

Or is allowed due to special relativity, that the time for a faster (more accelerated through gravity) object goes more slowly and so the effects cancels each other out and the time until the objects meet is the same?

 

[rcgldr]

...

v dv = -G (m1 + m2) dr / r2

Here m2 is also included as the mass of the dropped object. But for all experiments so far starting with galileo, the mass of m2 doesn't matter.

If m1 is much greater than m2, such as m1 = Earth's mass = 5.9736 x 10^24 kg, then unless m2 is similarly huge, it won't significantly affect the answer. If m2 was 6 kg, it would only affect the result by about 1 / 10^24.

 

[philipp2020]

If m2 was 6 kg, it would only affect the result by about 1 / 10^24.

That is what I tried to calculate. How much the time difference would be for including small masses in the calculation for m2.

So if the difference is only 10^24, what is the meaning of experiments which only can measure time differences until 10^-4 or something. For example like dropping a feather and lead ball in a vacuum.

Also the upcoming ESA Microscope experiment only measures until 10^-15:

http://sci.esa.int/science-e/www/object/index.cfm?fobjectid=36013
http://eotvos.dm.unipi.it/opendiscussion/microscope-ao_final.pdf

 

[TurtleMeister]

That is what I tried to calculate. How much the time difference would be for including small masses in the calculation for m2.

So if the difference is only 10^24, what is the meaning of experiments which only can measure time differences until 10^-4 or something. For example like dropping a feather and lead ball in a vacuum.

Also the upcoming ESA Microscope experiment only measures until 10^-15:

http://sci.esa.int/science-e/www/object/index.cfm?fobjectid=36013
http://eotvos.dm.unipi.it/opendiscussion/microscope-ao_final.pdf

That's a good question. It's one that I've been asking for many years, in various different ways. Newton was the first person to do this type of experiment. He used a pendulum, and experimented with using bobs made of different materials. Ever since then, experimenters have been repeating the experiment with ever increasing sensitivities. But as you've discovered with your equations, all of them will fall way short. All of them will give a null result. Even if there were a small gravitational composition dependence, these experiments would not be able to detect it. Here's a modern example:

http://www.npl.washington.edu/eotwash/eotwashwhat

After reading "what is a torsion balance", click the link "experiments" and then "equivalence principle".

One way to greatly reduce the sensitivity problem is to control the composition of the large mass, not the small one. This seems like a very simple solution. But to the best of my knowledge, no one is doing it. So for the past few years, I've just assumed that there is something wrong with my line of thinking.

 

[philipp2020]

After reading "what is a torsion balance", click the link "experiments" and then "equivalence principle".

One way to greatly reduce the sensitivity problem is to control the composition of the large mass, not the small one. This seems like a very simple solution. But to the best of my knowledge, no one is doing it. So for the past few years, I've just assumed that there is something wrong with my line of thinking.

I am not 100% sure, but I think 10^-15 with ESA's Microscope Mission in 2015 will be the most precise measurement so far concerning gravitational forces. Somebody may correct me please.

To control the composition of the large mass, wouldn't it then not be better to make the torsion balance experiment somewhere at the lagrange points and not take Earth as a huge reference for the large mass?

 

[TurtleMeister]

Yes, the Microscope Mission will be the most precise so far. STEP, which is still in development, will reach 10^-18. The Eot-Wash torsion balance experiments have reached the level of 10^-13.

In the torsion balance experiment, the Earth is not used at all. It is of course impossible to control the composition of the earth. The main purpose of a torsion balance is to cancel out the effects of the Earth's gravitational field. The large mass in a torsion balance experiment is called the "attractor" and the small mass is called the "test mass". Both of these masses are man made, so their compositions can be easily controlled.

In all of these experiments (from Newtons pendulum experiments, to STEP), they are comparing the passive gravitational mass of two relatively small test masses of different composition. These are the small masses in your equations, and the ones linked by rcgldr. Obviously if there were a gravitational dependence on the composition of mass it would be a very small one, otherwise we would have detected it by now. So how can we expect these experiments to give anything other than a null result when it is the composition of the small mass that is being controlled? Your question and my question are basically the same philipp.

 

[TurtleMeister]

Just wanted to add that according to http://www.onera.fr/news/2011-1212-microscope-mission-conference.php link, the Microscope Mission will launch in 2017. There is also an Italian mission called "Galileo Galilei", or GG, but I have been unable (and too lazy) to find very much info on it.

Also philipp, don't expect a satisfying answer to your question. Or, if you do get one then please let me know about it. Sorry if I spoiled your thread, as I have noticed that other PF members seem to "run for the hills" when they see me posting on this topic. :)