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Doing a physics experiment where we are required to calculate the acceleration from gravity, for a simple high school assignment. And as part of an extended discussion, I would like to take into account air resistance aswell. But looking into the formula, I notice that there are a lot of variables included in it. Such as drag coefficient, rho, velocity, and area.

Which are the best values to assume for these variables if I want to get an average value for the effect of air resistance on gravity?

Context: Dropping a 1.5cm diameter marble from a ceiling onto floor, in a room of about 20 degrees Celcius. And the main aim is to calculate the gravity of earth using the constant acceleration formula. All I need this air resistance value for is to manipulate the equation of acceleration to get it closer to the value of 10m/s^2, constant of gravity. Formula I'm talking about: Cd * .5 * rho * V^2 * A

Also what does the velocity part in the formula mean? The initial velocity, the final velocity of the marble?

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  • $\begingroup$ The velocity means the velocity, i.e. at every point between initial and final the drag is different. You could take the average velocity over distance to get average resistance. (as work is force*distance) The drag coefficient of a sphere is about .47 en.wikipedia.org/wiki/Drag_coefficient. Really it may need to be calibrated for any unusual shape. $\endgroup$ – JMLCarter Mar 19 '17 at 11:25
  • $\begingroup$ Mine is a nice 1.5 diameter spherical marble, so not many issues there. So it doesn't depend on the SIZE of the marble at all? And how can I find the average velocity? is it the same as adding initial/final velocities and dividing by 2? $\endgroup$ – user352935 Mar 19 '17 at 14:12
  • $\begingroup$ Sure there is a term for area, I think it is cross sectional area. I'm wondering what equation you are looking at... $\endgroup$ – JMLCarter Mar 19 '17 at 14:13
  • $\begingroup$ Yes, added it to the question. So this cross sectional area would be a circle, because the object is a sphere? $\endgroup$ – user352935 Mar 19 '17 at 14:17
  • $\begingroup$ Of course, you didn't need to ask that I suspect. $\endgroup$ – JMLCarter Mar 19 '17 at 14:24
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Since this was a "simple school assignment", taking air resistance into account is probably not appropriate. If the drop height was only a few meters, the effect of air resistance on a marble is unlikely to make a significant impact on your results. (If the object was a ping pong ball, air resistance would be significant over a few metres.) Other errors in your method are likely to be much larger. There is no point making a correction for it. It might appear smart, but it is only smart when the correction is significant.

It is much more useful to investigate the real errors in your experiment, which are likely to be due to timing.

The article When does air resistance become significant in free fall (The Physics Teacher, AAPT 2011) concludes :

... Blaming air resistance for errors in free‐fall experiments in introductory physics laboratories is almost never justified.

Rather than try to apply a correction (which is not straightforward - see below) you could do a calculation to show that air resistance is much smaller than the gravity force on the marble. You have the correct formula. You can measure cross-sectional area $A=\pi r^2$. You can estimate the final velocity $v=\sqrt{2gh}$ when air resistance is greatest. And you can look up values for air density $\rho$ at 20C and drag coefficient for a sphere.


The velocity used in the formula is instantaneous velocity. In your experiment the velocity is continuously increasing so the air resistance is continuously increasing hence also the deceleration it causes. Worse still, if air resistance is significant then you cannot know what the velocity will be until you have taken air resistance into account - a kind of Catch 22 situation. You would need to solve a differential equation - perhaps numerically - in order to find what the fall time would be with air resistance. However, you can get an over-estimate of the effect by using the final velocity.

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  • $\begingroup$ The point of the experiment is not to blame the air resistance for the incorrect result, in fact doing so would get me further away from the answer. The point is that by taking into account air resistance, no matter how small it may be, shows insight and clear understanding of the experiment we are doing. It is not a must, but something that is recommended. $\endgroup$ – user352935 Mar 20 '17 at 12:26
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    $\begingroup$ By taking air resistance into account you are saying that you think it will make a significant improvement in your result. This is not true. The author says that in almost all cases (ie except for very accurate work, or a very long fall time) the effect is insignificant. ... Yes, it shows insight to be aware that air resistance (and also the buoyancy of the air, the rotation of the earth, the presence of the moon and sun, etc) could affect the result of a very accurate measurement. However, trying to correct for such effects while not correcting much larger timing errors is not smart. $\endgroup$ – sammy gerbil Mar 20 '17 at 16:27
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    $\begingroup$ Who recommended that you calculate a correction for air resistance? I doubt that it was either the lab script or your teacher. $\endgroup$ – sammy gerbil Mar 20 '17 at 16:30
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    $\begingroup$ @user352935 I just noticed something in the question that fits into the theme of this answer. For example, you said you're using $10 \frac {m}{s^2}$ for gravity. That's already $1.9$ % off from the more commonly referenced $9.81 \frac {m}{/s^2} $ as acceleration due to gravity. Your results would be more accurate by using $9.81$ instead of $10$. This difference is probably more signifigant than air resistance (and both are probably going to make no difference compared to how well you measure). Any effect of drag should be far overshadowed by experimental error. $\endgroup$ – JMac Mar 20 '17 at 19:07
  • $\begingroup$ I DON'T think that it should make a significant improvement in my result, by including air resistance, the calculated value of gravity only deviates further from the standard one from textbooks. Like I said, it's just to show understanding. I have not been asked to use the formula, just merely acknowledge it, but I am taking it a step further by actually calculating it, no matter how rough the final answer may be. Getting the closest answer to gravity was NOT the point of the experiment. $\endgroup$ – user352935 Mar 21 '17 at 13:17

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