What is the acceleration of gravity on a strange planet?

[clockworks204]

1. A rock is dropped from rest from a height above a strange planet, and a strobe-light photograph is taken. The image is damaged in transmission to Earth so that an unknown part of the top of the picture is lost. However, five successive images of the falling rock can be seen. The spacing between the remaining images corresponds to 2.3360 m, 2.4000 m, 2.4640 m, and 2.5280 m, and the flash rate is 10.0 flashes per second. Calculate the acceleration of gravity on that planet ( m/s2).



2. V0(change in time) + 1/2(accel.)(change in time squared)
S1 + S2= V0(2)(change in time) + 1/2(accel.)(change in time squared)



3. I'm stuck... I know that change in time is 1/8 sec, S1=1.79m, S2=1.85m. I'm not quite sure how to proceed after that!

 

[heterotic_]

1. A rock is dropped from rest from a height above a strange planet, and a strobe-light photograph is taken. The image is damaged in transmission to Earth so that an unknown part of the top of the picture is lost. However, five successive images of the falling rock can be seen. The spacing between the remaining images corresponds to 2.3360 m, 2.4000 m, 2.4640 m, and 2.5280 m, and the flash rate is 10.0 flashes per second. Calculate the acceleration of gravity on that planet ( m/s2).
2. V0(change in time) + 1/2(accel.)(change in time squared)
S1 + S2= V0(2)(change in time) + 1/2(accel.)(change in time squared)
3. I'm stuck... I know that change in time is 1/8 sec, S1=1.79m, S2=1.85m. I'm not quite sure how to proceed after that!

If there are 10 flashes per 1 second, in perfect intervals, each of the 5 remaining images are 1/10th of a second apart, making the change in time for each image of the rock .1 seconds.

If the change in distances are 2.3360 m, 2.4000 m, 2.4640 m and 2.5280 m, and v = d/t, then we have velocities of

v1 = 2.3360 m/.1 s = 23.36 m/s
v2 = 2.4000 m/.1 s = 24 m/s
v3 = 2.4640 m/.1 s = 24.64 m/s
v4 = 2.5280 m/.1 s = 25.28 m/s

There are a couple of ways you could determine the acceleration due to gravity. Where g = the acceleration due to gravity;

g = Δv/Δt
g = 2d/t^2
g = v^2/2d

According to Newton's Law of Gravitation, the acceleration of gravity increases as you approach the center of mass. The acceleration due to gravity we look for then, is going to be an average over a relatively small distance over the surface of the planet with the information you were given.

If it were me, I'd find the accelerations at each point, average them out and give that as my answer. I'd use the first equation, simply because I think it will be the most accurate, though the difference would be negligible. There would only be three equations, though, Δv is the change in velocity.

 

[Bartek]

If the change in distances are 2.3360 m, 2.4000 m, 2.4640 m and 2.5280 m, and v = d/t,

You can not use formula v=d/t because it is accelerated motion (with - unknown - initial velocity).

Appropriate formula is d=Vinit*t + 0.5*a*t^2. There are two unknown, so you need two equations - needed data are on the photo.

regards

 

[heterotic_]

You can not use formula v=d/t because it is accelerated motion (with - unknown - initial velocity).

Appropriate formula is d=Vinit*t + 0.5*a*t^2. There are two unknown, so you need two equations - needed data are on the photo.

regards

We had a similar problem in class and this was the solution we used to solve it, we assumed there was no drag due to an atmosphere. The initial velocity IS known. The problem states the rock is dropped from a rest (initial velocity of 0), the unknowns are the initial height, the time in transit and the force of gravity exerted on the rock(and in turn its acceleration & velocity), the masses and sizes of the rock and the planet, respectively.

The only think we know for certain is that the camera takes 10 photos per second and the distances the rock fell in 1/10th of a second times 4.

 

[Bartek]

The initial velocity IS known. The problem states the rock is dropped from a rest (initial velocity of 0),

But "unknown part of the top of the picture is lost". We don't know how many flashes are lost. First saved point represent rock falling with unknown velocity from unknown height at unknown time.

We can use only distance between first and second saved point (=2.3360 m), first and third (=2.3360 m + 2.4000 m)... etc. And of course time between snapshots.

Of course when we calculate acceleration we can find initial height and time of first flash.

regards

 

[heterotic_]

But "unknown part of the top of the picture is lost". We don't know how many flashes are lost. First saved point represent rock falling with unknown velocity from unknown height at unknown time.

We can use only distance between first and second saved point (=2.3360 m), first and third (=2.3360 m + 2.4000 m)... etc. And of course time between snapshots.

Of course when we calculate acceleration we can find initial height and time of first flash.

regards

d = 1/2(at^2)
and
d + 2.3360 = 1/2(a(t+0.1)^2)
and
d + 2.3360 + 2.4000 = 1/2(a(t+0.2)^2)

right?

2.336 = 1/2(a(t+0.1)^2) - 1/2(at^2)
2.336 = 0.1a(t+0.05)
a = (t + 0.05)/.2336

and

4.736 = 1/2(a(t+0.2)^2) - 1/2(at^2)
4.736 = 0.2a(t+0.1)
a = (t + 0.1)/.9472

so

(t + 0.05)/.2336 = (t + 0.1)/.9472
4.28082t+0.214041 = 1.05574t+0.105574
t = .034 s

2.336 = 1/2(a(.44)^2) - 1/2(a(.34)^2)
2.336 = 1/2(a(.1936)) - 1/2(a(.1156))
2.336 = .0968a - .0578a
2.336 = 0.039a
a = 59.89 m/s^2
Doing it the way we did in class:
v(initial) = 23.36 m/s
v(average)1 = (23.36+24)/2 = 23.68 m/s
v(average)2 = (23.36+24+24.64)/3 = 24 m/s
v(average)3 = (23.36+24+24.64+25.28)/4 = 24.32 m/s

v(total average) = 23.84

IMO, one just takes longer.

 

[Bartek]

d = 1/2(at^2)
and
d + 2.3360 = 1/2(a(t+0.1)^2)
and
d + 2.3360 + 2.4000 = 1/2(a(t+0.2)^2)

right?

Absolutely right (I assume, that d is "lost" height and t is time from start to first saved point).

Its equivalent to pair of equations:

2.3360=Vinit*0.1 + 1/2*a*0.1^2
2.3360+2.4000=Vinit*0.2 * 1/2*a*0.2^2

where Vinit is velocity in 1st saved point. IMHO it is easier way :-)

Then you wrote:

2.336 = 1/2(a(t+0.1)^2) - 1/2(at^2) - right
2.336 = 0.1a(t+0.05) - right
a = (t + 0.05)/.2336 - I'm afraid wrong! a=23.36/(t+0.05)

regards
Bartek
ps
Its beter to consider 1st, 3rd and 4th point in equations instead 1st, 2nd and 3rd - to reduce measurement errors.

 

[heterotic_]

Absolutely right (I assume, that d is "lost" height and t is time from start to first saved point).

Its equivalent to pair of equations:

2.3360=Vinit*0.1 + 1/2*a*0.1^2
2.3360+2.4000=Vinit*0.2 * 1/2*a*0.2^2

where Vinit is velocity in 1st saved point. IMHO it is easier way :-)

Then you wrote:

2.336 = 1/2(a(t+0.1)^2) - 1/2(at^2) - right
2.336 = 0.1a(t+0.05) - right
a = (t + 0.05)/.2336 - I'm afraid wrong! a=23.36/(t+0.05)

regards
Bartek
ps
Its beter to consider 1st, 3rd and 4th point in equations instead 1st, 2nd and 3rd - to reduce measurement errors.

Sorry about the mistaken calculation above. I was multitasking doing my own physics homework, then got logged and had retype everything and mistakenly multiplied instead of dividing. Anyway, I wasn't asking if I was right or wrong. Why do people say "I'm afraid" when pointing out an error? Are you really afraid that I'm wrong?

2.336 = 1/2(a(t+0.1)^2) - 1/2(at^2)
2.336 = 0.1a(t+0.05)
a = 23.36/(t+0.05)

4.736 = 1/2(a(t+0.2)^2) - 1/2(at^2)
4.736 = 0.2a(t+0.1)
a = 23.68/(t+0.1)

23.36/(t+0.05) = 23.68/(t+0.1)
t = 3.6

2.336 = 1/2(a(3.7)^2) - 1/2(a(3.6)^2)
2.336 = 0.365a
a = 6.4 m/s^2

Either way, it's an average acceleration, no matter which points you use, it's an average. There's not enough data about the planet or the rock to use the gravitational constant to get a more accurate answer. You're making this problem more complex than it needs to be.

You're not really saying why the other is incorrect either - it's not! Both come to the same answer. And honestly, when I typed out the velocities the first time, it was fairly obvious by the constant displacement by .64 every tenth of a second what the acceleration was.

a = Δv/Δt
a = .64/.1
a = 6.4 m/s^2

Here are some other familiar numbers without using the equal equations.

v(first point to second point) = 23.36 m/s
v(average)1 = (23.36+24)/2 = 23.68 m/s

In this problem, you do not need to use equivalent equations. We are assuming a vacuum and we have enough data to use a less complicated system, I was correct in my original statement, and even I overcomplicated it a bit considering the numbers were very much constant to one acceleration (unlike gravity would ordinarily behave).

My point to begin with is that there is more than one way to get to the right answer. As they say, the more you complicate something simple, the more prone you are to errors.

 

[Bartek]

Ok

Anyway, I wasn't asking if I was right or wrong. Why do people say "I'm afraid" when pointing out an error? Are you really afraid that I'm wrong?

I do not fight against you. I wrote you are right and your equations can give an aswer...

I think, it is better way I showed abowe. For many reason. I'm metrologist... "broken picture" is raw data of experiment. In "my" method I do not have to assumed antything. I have four points (x,t) and have to calculate acceleration. It is more clear for me.

And I think it's easier to solve.

But of course you can calculate other way.

regards
Bartek
ps
I don't know why do people say "I'm afraid" when pointing out an error. As you can see my english is not good (I'm affraid ). Maybe it's the reason? Or maybe tradition?

 

[heterotic_]

I do not fight against you. I wrote you are right and your equations can give an aswer...

I think, it is better way I showed abowe. For many reason. I'm metrologist... "broken picture" is raw data of experiment. In "my" method I do not have to assumed antything. I have four points (x,t) and have to calculate acceleration. It is more clear for me.

And I think it's easier to solve.

But of course you can calculate other way.

regards
Bartek
ps
I don't know why do people say "I'm afraid" when pointing out an error. As you can see my english is not good (I'm affraid ). Maybe it's the reason? Or maybe tradition?

I'm getting dual degrees in atmospheric science and astronomy right now.

Your first reply said that "you cannot use v = d/t". Doesn't your equation also assume it is a vacuum? If we assume there is an atmosphere, we'd have to use another Method, right? Aren't we also assuming that there also isn't any liquid during the fall? If there were, we'd have apply Archimedes' principle, right?

 

[Bartek]

I'm getting dual degrees in atmospheric science and astronomy right now.

Congratulations.

Bartek
ps
"you cannot use v=d/t" not because of atmosphere or vacuum. You cannot use it because 2.3360m < 2.4000m < 2.4640m < 2.5280m. Velocity is not constant. Increase of velocity is constans. Vacuum is conclusion - not prediction.

 

[heterotic_]

Congratulations.

Bartek
ps
"you cannot use v=d/t" not because of atmosphere or vacuum. You cannot use it because 2.3360m < 2.4000m < 2.4640m < 2.5280m. Velocity is not constant. Increase of velocity is constans. Vacuum is conclusion - not prediction.

You can determine instantaneous velocity at each point, because at any instant in time, velocity will be constant. Since acceleration is the rate of change in velocity, given the instantaneous velocities and time of change, that gives you an accurate acceleration, as I already proved above.