Welcome to NRICH.

 
Satellite falling to Earth
[Editor: The following question isn't entirely resolved, but the discussion of factors involved in determining the descent of a satellite towards Earth are preserved here.]
By Arun Iyer on Thursday, September 06, 2001 - 07:39 pm:

There is say a satellite revolving around the earth at a quite a good height h. Say now it suddenly stops and falls down to earth.In what time will it come down?
(This is just my own made up question so the ending expression for time may include some terms i have not introduced here)

I have tried a lot though if you see i got stuck on just one complication,
we know that g at a height h is->
gh=g[R/(R+h)]2
Since g here is varying i am not able to put it in any of the kinematic equations.

love arun

By Jim Oldfield on Thursday, September 06, 2001 - 07:56 pm:

Glancing quickly at your equation it looks unfamiliar, though it could well be right. I wouldn't use it though, when we know:
F=Gm1m2/r2 (Newton's law of gravitation).
F=ma (Newton's second law, constant mass)
Þ a=Gm/r2 (where m is the mass of the Earth here)

\ d2r/dt2 = Gm/r2
and now solve this differential equation to relate r and t (you will need to set boundary conditions).

By Arun Iyer on Friday, September 07, 2001 - 08:18 pm:

Jim,
i think you are overlooking some facts.Practically speaking we have here another force to reckon with that is drag force of the air. Due to this force,the body achieves certain maximum velocity called the terminal velocity.After achieving this velocity,the body does not accelerate.

There are again much more complications regarding terminal velocity (which should vary quite significantly in this particular problem).All of this is too long to write down right now.

I hope you get the feel of my problem.
love arun

By Andrew Hodges on Saturday, September 08, 2001 - 09:50 pm:

Well we know that the drag force is crudely proportional to the velocity squared, so we can simply take account of that fact in the differential equation...

By Andrew Hodges on Saturday, September 08, 2001 - 09:56 pm:

I seem to end up with the differential equation:

r'' + k(r')2 = Gm/r2

where acceleration, a = r'' and velocity v = r'

This looks like a killer equation to solve so the best thing to do would be to define the initial conditions and solve it numerically.

By Jim Oldfield on Sunday, September 09, 2001 - 01:29 am:

Ah, the bane of all physicists, I did indeed forget air resistence. But even if we assume it proportional to speed squared (this is true at very high speeds, at low speeds it is proportional to speed, and in practice it is somewhere in between! :-P) we are still forgeting something. As you might expect, air resistance is dependant upon (in fact proportional to if I remember correctly) the density of the fluid it is travelling through. But Arun said "a satellite revolving around the earth at a quite a good height," so we are talking about starting in a low (or even zero) density atmosphere, and this varies as we come closer to the earth. I will look on the internet for details of the variation, but in all honesty I'm not optimistic.

Jim

By Jim Oldfield on Sunday, September 09, 2001 - 02:21 am:

I have just read that atmospheric pressure varies exponentially with height. I leave it to you to solve the resulting differential equation (!)

differential equation

Jim

By Jim Oldfield on Sunday, September 09, 2001 - 02:25 am:

Oops, that first minus should be a plus (since (negative) acceleration due to air resistence and acceleration due to gravity act in opposite directions).

Jim

By Arun Iyer on Sunday, September 09, 2001 - 09:39 am:

It does not end here my friends.

Point No.1->
the air resistance not only drags the body up but also must be setting up various types of frictional forces which must be taking place at atomic level.These forces tend to deform or even sometimes burn the object.(this observation is purely my opinion so it is open for criticism)

Point No. 2->
What about terminal velocity?
In my opinion,while solving this problem,we should actually divide the path of falling satellite into 3 different zones---1>low pressure 2>moderate pressure 3>high pressure.According to me,the body must attain three different terminal velocities while traversing through its path.As i already said above,the body doesn't accelerate after it has reached terminal velocity.Hence,the body travels with zero acceleration during three parts of its path.

If i am wrong at some points,please do correct me.Actually i am trying to solve this problem for nearly a year or so and the points i have mentioned above were certain things i could not rectify.

love arun

By Jim Oldfield on Sunday, September 09, 2001 - 05:58 pm:

1. You are right, if the air resistance deforms the body it will effect the value of the air resistance (and also on the mass, affecting other parts of the equation). If there are big solar panels or something sticking out this could have quite an effect, but if the body is quite centralised hopefully there will be little difference.

2. I disagree (IMHO). Of course a body will never QUITE reach terminal velocity, but will come closer and closer to reaching it. Since this is a natural reflection of the physics involved, I think the equation when solved (not that I can do it) will account for that.

Does anyone have any idea how to solve the equation (a, b and m are constants)? I don't even know how to do it numerically. The plus and minuses have changed again, it should definitely be right this time J.

Corrected differential equation

Jim

By Brad Rodgers on Sunday, September 09, 2001 - 08:33 pm:

I think I have an idea of how to solve it, but I don't think that it will have a closed form solution. First, write the equation as

d2r/dt2+a×ebr(dr/dt)2=c/r 2

not really caring at the moment what a,b, and c are.

Note that this is equivalent to

d/dt(exp(aebr/b)dr/dt)=c×exp(aebr/b)/r 2

putting exp(aebr/b)dr/dt=x, we are given

dx/dt=(dx/dr)×(dr/dt)=c×exp(aebr/b)/r2

multiplying by exp(aebr/b), we get

x×dx=(c×exp(2aebr/b)/r2)dr

we can now integrate both sides, however, I'm not sure the the RHS has an algebraic integral.

Brad

By Arun Iyer on Sunday, September 09, 2001 - 09:02 pm:

This is the same integral you have posted under the topic COMPLEX INTEGRAL, isn't it?

I don't know why but i have feeling that RHS might involve the term Ei(x) and g [Euler's constant]

If it does involve such terms,we will have then to reevaluate the entire thing.
love arun

By Arun Iyer on Monday, September 17, 2001 - 08:03 pm:

Well,
anybody lucky to have got the solution to that differential equation?
love arun