Finding Gravitational Acceleration

[Martyn59]

Hi if I understand it correctly, this is the process to find freefall acceleration of a falling body.

S=ut+1/2at^2 Initial time and displacement is zero so,
S=1/2at^2 find acceleration
a=2(S/t^2)
graph for change in displacement over time squared
a= 2(yf-yi/xf-xi) (f=final i= initial)

I almost understand why that works,

But I thought that acceleration was found by change in velocity over change in time. a=v-u/t
But when I use this formula on the data from my experiments I get a wrong answer.any help would be appreciated.

Thank you.

 

[PeroK]

Hi if I understand it correctly, this is the process to find freefall acceleration of a falling body.

S=ut+1/2at^2 Initial time and displacement is zero so,
S=1/2at^2 find acceleration
a=2(S/t^2)
graph for change in displacement over time squared
a= 2(yf-yi/xf-xi) (f=final i= initial)

I almost understand why that works,

But I thought that acceleration was found by change in velocity over change in time. a=v-u/t
But when I use this formula on the data from my experiments I get a wrong answer.any help would be appreciated.

Thank you.

That formula should work - but only if the object starts at rest. I expect you meant $$a = 2\frac{y_f - y_i}{(t_f - t_i)^2}$$

 

[Dale]

But I thought that acceleration was found by change in velocity over change in time. a=v-u/t
But when I use this formula on the data from my experiments I get a wrong answer.

This will give you the average acceleration. Since the acceleration is constant the average acceleration is the same as the acceleration. This formula is correct, but your measurements above are measuring position, not velocity. So although the formula is correct, it is not particularly relevant or helpful in this situation.

 

[kuruman]

If you have a collection of positions $y_i$ and associated times $t_i$ for an object moving under constant acceleration, you are correct in saying that the equation $y_i=v_0 t_i -\frac{1}{2}gt^2$ is the one to use. You do not explain the procedure you used, specifically whether (a) the object was released from rest many times and you measured its time of flight $t_i$ for varying heights $h_i$ or (b) you released the object once and you measured the times $t_i$ it took to reach a known distance $y_i$. In either case, the appropriate method for analyzing the data is to take the ratio $y_i/t_i$ and make a plot of this on the ordinate (vertical axis) versus $t_i$ on the abscissa (horizontal axis). The result should be a straight line of the form $Y=mX+b$. I leave it up to you to figure out the significance of the slope $m$ (it's not the mass) and the intercept $b$.

 

[italicus]

@Dale

are you sure that g is constant? What if , for example, a geostationary satellite stops its revolution around the Earth, and falls radially, in an adeguate reference frame?
Or the Earth falls radially towards the Sun? It takes around 64 days for the Earth to reach our star in free fall: apply conservation of energy and Kepler.
For a geostationary satellite, the distance from the centre of the Earth is about R _S = 42371 km; applying the Newton law for universal gravitation, and assuming g_T = 9.81 m/s² ( on the Earth surface, R_T= 6371 km) you have that:

g_S/g_T = (R_T/R_S)² = 0.0226

that means : g_S = 0.2218 m/s²

this is the value of the acceleration of gravity at a distance from Earth equal to the orbital radius of the geostationary satellite. The free fall takes place with an increasing acceleration, isn’t a uniformly accelerated motion.the value of g_S can be also evaluated from the value of its speed and the orbital radius , because :

v = √(g_SR_S)

this formula can be found equating the centripetal force to the gravitational force, and doesn’t depend on the mass of the satellite ; so :

v²/R_S = g_S

the speed is easily calculated knowing that the satellite is geostationary.

sorry, I’m not able to use Latex.

 

[berkeman]

are you sure that g is constant? What if , for example, a geostationary satellite stops its revolution around the Earth, and falls radially, in an adeguate reference frame?

Most likely he is doing table-top experiments in his lab, so assuming a constant value for "g" is probably fine:

But when I use this formula on the data from my experiments I get a wrong answer.

And there is an easy fix for this:

sorry, I’m not able to use Latex.

Just click on the "LaTeX Guide" link below the Edit window to learn the basics.

 

[DaveC426913]

Just click on the "LaTeX Guide" link below the Edit window to learn the basics.

You know what drives me crazy about these guides?

They never ever say what the tags are to bracket latex (since that's site specific).

Is it [ latex ]? [ LATEX ]? [ tex ]? [ TEX ]? < TEX >, etc.?

This trips me up every time and is by far the most common reason why I fail to use latex. (It's different on every site, so it's not like once I figure it out I'll never forget).

 

[berkeman]

You know what drives me crazy about these guides?

They never ever say what the tags are to bracket latex (since that's site specific).

Is it [ latex ]? [ LATEX ]? [ tex ]? [ TEX ]? < TEX >, etc.?

This trips me up every time and is by far the most common reason why I fail to use latex. (It's different on every site, so it's not like once I figure it out I'll never forget).

Well, at least at PF it is straightforward (I agree that it seems different on different websites):

 

[Dale]

are you sure that g is constant?

Given the formulas that the OP is using and the level of their post I assumed that they are a student doing exercises for a for a class in an introductory physics class, not a NASA scientist working on satellite deorbiting maneuvers. This is indeed an assumption on my part, but unless the OP clarifies otherwise it seems a safe assumption.

 

[italicus]

Given the formulas that the OP is using and the level of their post I assumed that they are a student doing exercises for a for a class in an introductory physics class, not a NASA scientist working on satellite deorbiting maneuvers. This is indeed an assumption on my part, but unless the OP clarifies otherwise it seems a safe assumption.

How exaggerate are you! NASA scientists, for a simple deduction of g being variable , from a great distance toward the Earth? You want to joke! We are not talking of Hohmann orbit transfer!

Anyway, in an Earth limited laboratory of height H, assuming g = constant is fine, in Newtonian mechanics.

 

[kuruman]

Given the formulas that the OP is using and the level of their post I assumed that they are a student doing exercises for a for a class in an introductory physics class, not a NASA scientist working on satellite deorbiting maneuvers. This is indeed an assumption on my part, but unless the OP clarifies otherwise it seems a safe assumption.

I started adding my own reply to the above, but then I thought to myself that this thread does not need another tangential post. Perhaps the mentors following this thread might consider instituting some kind of guideline that limits the number of non-OP posts before the OP has had a chance to reply. This would be only appropriate to homework help threads which this one appears to be. Too many helpers often spoil the thread in the sense that the OP loses sight of the goal.

I do not wish to start another tangent within this thread. If my suggestion merits further discussion, @berkeman and/or @Dale please let me know. In that case, this post will self-destruct and a new thread will start in Advisor Lounge.

 

[Martyn59]

That formula should work - but only if the object starts at rest. I expect you meant $$a = 2\frac{y_f - y_i}{(t_f - t_i)^2}$$

Yes, darn it, when I wrote it out I forgot to square the time on the x axis.

I've been giving this some thought
I think I've figured out why the other formula doesn't work. (change in velocity over change in time)
I was dropping a weight from 1900mm and getting 59 hundredths of a second and calling that 3.22 m/s
dropping it again from 500mm and getting 30 hundredths of a second and calling that 5.6m/s
a= v-u/▲t so a=3.22-1.67/.29 so a=5.34 m/s obviously wrong. I figure the problem is that what I'm calling initial velocity, and final velocity are actually periods of acceleration.

so I have to use the other one.. ( s=ut+1/2at^2)
When I use that formula I get an answer which is close. but not right.
I've read that when doing the experiment (measuring elapsed time of a falling mass) I should do multiple drops and use the shortest time obtained for each height. I can see the reason for that. so i'll redo the experiment and hopefully I'll get closer to 9.8m/s^2

Getting the right answer is important, but I also really want to properly understand why the math works.
Thanks for your help I appreciate it.

 

[Martyn59]

This will give you the average acceleration. Since the acceleration is constant the average acceleration is the same as the acceleration. This formula is correct, but your measurements above are measuring position, not velocity. So although the formula is correct, it is not particularly relevant or helpful in this situation.

Thank you, yes, I've been thinking a bit more deeply about the whole thing, and I see that now. Thanks for your help.

 

[Martyn59]

If you have a collection of positions $y_i$ and associated times $t_i$ for an object moving under constant acceleration, you are correct in saying that the equation $y_i=v_0 t_i -\frac{1}{2}gt^2$ is the one to use. You do not explain the procedure you used, specifically whether (a) the object was released from rest many times and you measured its time of flight $t_i$ for varying heights $h_i$ or (b) you released the object once and you measured the times $t_i$ it took to reach a known distance $y_i$. In either case, the appropriate method for analyzing the data is to take the ratio $y_i/t_i$ and make a plot of this on the ordinate (vertical axis) versus $t_i$ on the abscissa (horizontal axis). The result should be a straight line of the form $Y=mX+b$. I leave it up to you to figure out the significance of the slope $m$ (it's not the mass) and the intercept $b$.

Okay, I think I've almost got this M= change in height over change in time squared. No idea about intercept b though. I'll look it up. Thanks for your help. I appreciate it.