Falling Meteor

Classical Mechanics Level 5

A meteor of initial mass $m_{0}$ enters the earths atmosphere and starts falling from rest under the influence of gravity. Assume that the meteor gains mass from the cloud at a rate proportional to the momentum of the meteor, $\frac { d{ m }_{ 0 } }{ dt } =k{ m }_{ 0 }v$ , where $v$ instantaneous velocity of the meteor. If $k$ is equal to $5 m^{-1}$ , what will be the terminal velocity of the meteor?

Take into consideration the buoyant force of the cloud

Details and assumption

You may neglect air resistance.

Take gravitational acceleration $9.8m/{ s }^{ 2 }$

The answer is 1.4.

1 solution

Anish Puthuraya
Mar 22, 2014

A good problem, but I think it would have been better if $\displaystyle g$ wasn't assumed to be constant.

Writing Newton's 2nd Law,

$F = \frac{dp}{dt}$

$mg = \frac{d(mv)}{dt} = \frac{dm}{dt}v+\frac{dv}{dt}m$

Now,
It is given that,

$\frac{dm}{dt} = kmv$

Substituting this value back in the equation,

$mg = kmv^2+m\frac{dv}{dt}$

$g = kv^2+\frac{dv}{dt}$

Note that, when the terminal velocity is achieved, the value of the velocity will not change further (it is obvious why). In other words,

$\frac{dv}{dt} = 0$

Hence,

$g = kv^2$

$v_{term} = \sqrt{\frac{g}{k}} = \boxed{1.4m/s}$

You don't consider buoyant force!

jatin yadav - 7 years, 2 months ago

When I wrote this solution, it was not mentioned in the question to take buoyant force into consideration. But there is no density given, so how do I consider it?

Anish Puthuraya - 7 years, 2 months ago

Well, I considered it like this(assuming the meteor was initially small) :

$m-m_{0} = \rho V$

$\Rightarrow ,F_{b} = (m-m_{0}) g$

$\Rightarrow F_{net} = mg - F_{b} = m_{0} g$ , which turns out to be constant. If you differentiate $m_{0}g = v \frac{dm}{dt} = kmv^2$ and put $\frac{dv}{dt}=0$ you get $v_{steady} = 0$ .

jatin yadav - 7 years, 2 months ago

@Jatin Yadav – @jatin yadav correct me if i am wrong but i think the solution is correct if we take buoyant force into consideration. We know that in a cloud buoyant force and gravitational force are balanced. Lets take the mass of the cloud gained by the rock to be $m_{c}$ .During time of falling there is an external gravitational force acting on the system and a buoyant force in the cloud that is keeping the mass element $m_{c}$ from falling. Since the total force on the mass element $m_{c}$ is zero, the total external force acting on the system is just the gravitational force acting on the meteor. Even though the solution doesn't show that

Beakal Tiliksew - 7 years, 2 months ago

@Beakal Tiliksew – That was what I was saying. The gravitational and buoyant force on $m_{c}$ is balanced. Hence, the net force on system is the gravitational force on $m_{0}$ that is $m_{0} g$ . This was what I got. You will apply newton's law on system. Hence, my equations are correct!

jatin yadav - 7 years, 2 months ago

@Jatin Yadav – I have made a modification of the problem to clear things up.

${ m }_{ 0 }g={ m }_{ 0 }\frac { dv }{ dt } +v\frac { d{ m }_{ 0 } }{ dt } \quad \rightarrow \quad \frac { d{ m }_{ 0 } }{ dt } =k{ m }_{ 0 }v$ ,

$\frac { dv }{ dt } =0,\Longrightarrow g=k{ v }^{ 2 }\quad$

My apologies for any confusion that i may have caused, thanks for the follow up

Beakal Tiliksew - 7 years, 2 months ago

@Beakal Tiliksew – why both force is balancing??????

Naman Negi - 7 years, 1 month ago

@Naman Negi – A cloud stays aflot because the the buoyant force balance gravitational force

Beakal Tiliksew - 7 years, 1 month ago

ps taking buoyant force into consideration was stated from the beginning;)

Beakal Tiliksew - 7 years, 2 months ago

Good one!

Shikhar Jaiswal - 7 years, 2 months ago

I probably should have done that>

Beakal Tiliksew - 7 years, 2 months ago

Did the exact same way, but wrote lesser steps. Just two lines. Yeah non constant g varying with height would have been a good problem. I directly used the formula F= Fext+ vrel(dm/dt), well using basic ideas is also good, as always I try less to mug up equations in physics but at the end some I forget the original derivation but the formula is still there. Good refresher.

Bala Tweakbytes - 7 years, 2 months ago

Exactly what I did.

Amish Naidu - 7 years, 1 month ago

Falling Meteor

The answer is 1.4.

1 solution

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