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Hi everyone, I am new here. I am a Junior In high school, so I haven't taken Physics yet (next year), and I have a question. If you have two spheres that are identical in every way except one is denser, which will hit the ground first if dropped from the same height on Earth (With an atmosphere)? Basically I am asking, does density affect rate of fall in an atmosphere. This has probably been asked before, but I wasn't able to find anything on it. Thanks :)

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F=ma

[math]F_g=G\frac{m_1 m_2}{r^2}[/math]

 

We can combine them to form:

 

[math]m_1 a_g=G\frac{m_1 m_2}{r^2}[/math]

 

Which simplifies to:

 

[math]a_g=G\frac{m_2}{r^2}[/math]

 

This means that the acceleration due to gravity is independent of the mass of the bodies being accelerated.

 

Gravity accelerates all objects at 9.8m/s2, regardless of their density or weight.

 

This is only true in the special case of objects near earth sea level or an equivalent scenario.

Edited by ydoaPs
multiple post merged
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Assuming "identical in every way except one is denser" also implies that the denser one has a higher total mass (the only other possibility would be having a different diameter) the denser one will hit the ground earlier than the less dense one. That is because the motion of the denser object is less influenced by friction with the atmosphere that hinders the free fall Capn and yourdadonapogos assumed (they probably didn't catch the "in an atmosphere"-part).

Edited by timo
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Cap'n, yourdad, you guys aren't quite 100% correct.

 

The key phrasing here is "in an atmosphere". In an atmosphere, the difference in densities can be important -- there is a buoyancy force that while very, very small in air, is non-zero.

 

Drag laws are also dependent on density differences. Let me post some things in detail here.

 

First, we need to define a Reynolds number for a sphere:

 

[math]Re_p = \frac{d_p U \rho_f}{\mu_f}[/math]

 

where

[math]d_p[/math] is the diameter of the sphere,

[math]U[/math] is the velocity of the sphere,

[math]\rho_f[/math] is the density of the fluid (air in this case), and

[math]\mu_f[/math] is the viscosity of the fluid.

 

The units of all the terms are chosen to make [math]Re_p[/math] a dimensionless number.

 

The next correlation to use is to calculate the Archimedes number: This is a number that is independent of the terminal velocity

 

[math]Ar = \frac{d_p^3 \rho_f (\rho_p - \rho_f) g}{\mu^2} [/math]

 

[math]d_p[/math] is the diamter of the particle

[math]\rho_f[/math],[math]\rho_p[/math] is the density of the fluid, particle respectively. The density of air is 1.246 kg/m^3

[math]g[/math] is the acceleration due to gravity = 9.8 m/s^2

[math]\mu[/math] is the viscosity of the fluid = 1.78*10^-5 kg/(m*s)

 

There is a correlation between the Archimedes number and the Reynolds number at terminal velocity.

 

The correlation between Ar and Re at terminal velocity (Re_t) is:

 

[math] Re_t = (2.33 Ar^{0.018} - 1.53 Ar^{-0.016} )^{13.3} [/math]

 

The terminal velocity of two objects -- identical in all other respects except two different densities -- will have slightly different terminal velocities. This shows up because of the slight difference in buoyancy between the two objects, and because the density difference does appear in the most accurate drag correlations. The Archimedes number shows the influence in the density of the object. (Please see Khan and Richardson, 1989, as a good reference showing where these correlations come from. K&R also cite many good papers that detail the experiments that validated these correlations.)

 

Now, in air, is it going to be big? No, of course not. In air, the difference is going to be tiny -- tiny enough that it is probably going to be negligible. You would need hypersensitive measuring devices to detect the difference -- and a perfectly controlled volume of air (i.e. inside a well insulated building with no drafts) to minimize the effects of air currents that will occur. In fact, in actual practice, in almost all calculations involving air, the buoyancy term is dropped because it is so small and doesn't change the final result more than a few thousandths of a percent.

 

But, in all complete technical correctness (a la Futurama -- the best kind of correct!), an object's density will have an effect on it's terminal velocity in a fluid like the atmosphere. And, since the terminal velocity is different, the time is takes to fall a given distance will also be different.

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Wow, thanks for that in depth explanation! Makes perfect sense to me, you just have to think of the air like a liquid (in terms of the importance of density). Thanks :D

 

Well, both gas and liquids are fluids. And, the behavior of all fluids is remarkably similar, especially once you start considering them in terms of the dimensionless variables, like Reynolds number and Archimedes number (and Froude number and Euler number, etc.)

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  • 4 weeks later...

originally posted by cap'n refsmmatt

The answer is, they'll hit the ground at the same time. Gravity accelerates all objects at 9.8m/s2, regardless of their density or weight.

 

isn,t it in vaccum. it happens only if there is vaccum.in the normal state the object which is heavier strikes the ground first

 

i believe i am correct for i am not a genius like you

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No, the object with the largest weight to drag ratio will hit the ground first. Try dropping a piece of newspaper and a lead ball that is lighter than the newspaper. The lead ball will land first. Or better, dropping an inflated balloon and an uninflated balloon. The inflated balloon weighs more but will fall slower.

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No, the object with the largest weight to drag ratio will hit the ground first. Try dropping a piece of newspaper and a lead ball that is lighter than the newspaper. The lead ball will land first. Or better, dropping an inflated balloon and an uninflated balloon. The inflated balloon weighs more but will fall slower.

 

Yes but then the objects are no longer identical except for mass. Pradeep has it correct now.

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