# Crazy physics problem! (big debate in my high school)

1. Feb 21, 2008

### littlebhawk

Ok im new to physics forum but im sure if im going to get help ill probably get it here, this problem has gone UNSOLVED by everyone who has tried it to include my physics and calculus teacher.
Here is the problem:
There is an object 10 earth radii from the CENTER of earth, it is released with zero initial velocity, how long will it take to hit the surface of earth assuming no atmosphere?

Obviously gravity increases as the object moves closer to earth, i found the integral of force of gravity versus distance from one earth radius from center, to 10 earth radii away to get the work done on the object, changed it to the kinetic energy at moment of impact, and got my final velocity. My final velocity was 10608.1 m/sec, I plugged this into Vf^2=Vi^2 + 2ad to solve for what i thought would be sort of an "average" acceleration, i got a= 0.9799 m/sec^2, plugged this into Vf=Vi+at, solved for T and got 10825.6 seconds, or about 3 hours.

I dont know if im even close to right as i've never tried a question like this before, any help would be appreciated.

2. Feb 21, 2008

### littlebhawk

Oh, this isnt a hw problem my teachers and my friends are just really curious at how to even attempt this problem.

3. Feb 21, 2008

### mgb_phys

Without calculus you can work out an estimate by doing it in steps.
Calculate the force at a distance of 10radii, and so the acceleration and the time taken to move say 0.1radii.
Then repeat at 9.9radii and so on ( a spreadsheet might help!)
Remember the solution ends at 1radii not zero.

You can also write a differential equation for the force and acceleartion as a function of distance - but this approach will at least give you an answer to check against.

4. Feb 21, 2008

### nicksauce

Use an energy approach, with Gravitational PE + Kinetic E = Constant. Solve the resulting ODE and get the equation of motion.

5. Feb 21, 2008

### Danger

Welcome to PF, Littlebhawk.
I don't mean to sound facetious or insulting here; it's an honest question. Why is this guy a science teacher if he doesn't know science?
I had a chemistry teacher in grade 11 who knew only the stuff in the textbook; if you went a hair off of the page, he hadn't a clue. Shouldn't someone be an expert in the field that he teaches?

6. Feb 21, 2008

### rohanprabhu

Take for any arbitrary distance 'r' from the center, let the center be the origin, the gravitational force acting on the body:

$$F_g = -\frac{GM_em}{r^2}$$

Hence, the acceleration at that point:

$$g = -\frac{GM_e}{r^2}$$

So, you have:

$$\frac{dv}{dt} = -\frac{GM_e}{r^2}$$

or,

$$\frac{dv}{dr} \frac{dr}{dt} = \frac{GM_e}{r^2}$$

here, 'v' is the instantaneous velocity with which the object traverses the radial distance. so.. we have,

$$v \frac{dv}{dr} = -\frac{GM_e}{r^2}$$

or,

$$v = -\sqrt{\frac{2GM_e}{r}}$$

hence,

$$dt = -dr \sqrt{\frac{r}{2GM_e}}$$

on solving the equation,

$$r(t) = -\left(\frac{3t}{2\sqrt{GM_e}}\right)^\frac{3}{2} + C$$

for us, we take that at t = 0, $r = 10R_e$. Hence,

$$r(t) = 10R_e -\left(\frac{3t}{2\sqrt{GM_e}}\right)^\frac{3}{2}$$

For, $r = R_e$, we get:

$$9R_e = \left(\frac{3t}{2\sqrt{2GM_e}}\right)^\frac{3}{2}$$

solve that to get:

$$t = \frac{2 \sqrt{2GM_e} (9R_e)^\frac{2}{3}}{3} \approx 3.59396\sqrt{GM_e} (R_e)^\frac{3}{2}$$

using values for G and $M_e$,

$$t = 2.79921 \times 10^{12} s$$

which... i think is definately wrong... because it is too long a time period. [it's around 88,000 years]. I may have made a mistake somewhere..

Last edited: Feb 21, 2008
7. Feb 21, 2008

### rohanprabhu

The answer you got will actually be a lot different than the actual answer.. [but not in the explosive proportions as mine].. because the acceleration is time dependent.. 'average acceleration' doesn't really count for anything.

EDIT:
I ran the equations again.. and Mathematica gave me 2 solutions for the equation, one being the previous 88,000 years.. and the other one as $t = 16293242.127s$ which is around 188 days. It still seems like quite long though..

Last edited: Feb 21, 2008
8. Feb 21, 2008

### littlebhawk

Thanks rohanprabhu thats definetly alot further in calculus then i couldve gone, ill be taking calculus next year so all the calculus i really knew was how to get the integral and derivatives of simple equations and how to use the TI-83 to find the solution.

Oh and Danger, your right it is a perfectly valid question, i definetly dont doubt that my teachers could solve this problem if i had given them more time though, there definetly capable of it. My science teacher helped me last year enough to get a 5 on the AP chem test, even though i had never taken chemistry before in my life. Also, he has gone many years teaching physics with algebra and questions like these only come up when imaginative students have a brainstorm.

9. Feb 21, 2008

### rohanprabhu

well.. i did a few things wrong.. basically with my first integration.. when I get the constant 'C', i have to take it inside the square root and stuff.. so i just ran a few equations in Mathematica, which led me to this correct equation of t as a function of 'r':

and it looks quite horrible. I'm too sleepy now.. i'll see if there's anything that i can do tomorrow...

10. Feb 21, 2008

### TVP45

You can do a simple "back of the envelope" calculation to find an upper limit on the time. You know gravitational force varies as the inverse of distance squared. You're ten times farther out, so that's one-hundredth. So, just take the worst case and let a be .098 m/s^2 all the way down. What do you get for time?

11. Feb 21, 2008

### Staff: Mentor

This one and the ball-string one are both a little disturbing to me...

12. Feb 21, 2008

### dst

I don't understand why you can't just apply simple energy conservation to this? I.e. -

O x ---------------------------X

Big X = full gravitational potential, small x = GPE completely converted to KE. Then you could assume a constant velocity and hence find the time taken. What's wrong with this approach? No need for calculus, am I missing something?

Last edited: Feb 21, 2008
13. Feb 21, 2008

### kmarinas86

The student is trying to solve for the time.

14. Feb 21, 2008

### TVP45

Yeah, even if you don't know how to set up the equations for the exact solution, you should know how to throw this in Excel with 100 steps, as mgb suggested, and get an approximate answer. I share Danger's misgivings, not because I think it's terrible to not know, but because this sort of thing tends to turn people off from science.

15. Feb 21, 2008

### kmarinas86

Last edited: Feb 21, 2008
16. Feb 21, 2008

### Ajwrighter

I'm no expert but it looks like the rock hits the ground once and then hits it again about xx time later. With out the atmosphere you’re not going to get a simple drop. Remember, even though the gravity inside the atmosphere is 9.8m/s^2 and out side the atmosphere its still 9.8m/s^2 in fact you could go a thousand miles past the moon and the gravity will probably of changed to about 9.75m/s^2 . With this in mind, that little bubble we are surrounded by plays a significant factor on the rate of descent. There is about 40,000 pieces of space junk(/satalites) thats falling towards earth, each with the same amount of gravity pulling them down. And I promise you, when you wake up tomorow that same junk will still be up there falling.

Last edited: Feb 21, 2008
17. Feb 21, 2008

### rohanprabhu

Agreed. We'll have to consider the Fluid resistance of air (Drag Force). We'll also have to consider the wind and many other meteorological factors, which is pretty difficult.

How does it hit it xx time later? You mean it rebounds from the surface and stuff. We never computed the equation considering collision dynamics.

No it is not. 'g' which is acceleration due to gravity and not gravity is different even on the surface of the earth because of the earth not being perfectly spherical. The value 'g' is inversely proportional to r², so there is no way that 'g' outside the atmosphere will be close to what's inside. Outside the atmosphere, it will be way different. Considering the Exosphere, which is 10,000 km from the earth's center, $g \approx 1.487~ms^{-2}$ which is only 15% of the mean value at the surface of earth.

that was quite philosophical... and again it's not 'Gravity', it's 'acceleration due to gravity'. Also, when u say 'pulling it down', you are reffering to a force. The force is different on objects of different mass. Also, it's not 40,000... we have around 6216 pieces of junk.

18. Feb 21, 2008

### K.J.Healey

Probably passes through the earth and comes all the way back and hits the same point. Can't allow R to go negative, that should solve it.

19. Feb 21, 2008

### rohanprabhu

You mean in an SHM [not exactly SHM.. but a harmonic motion nevertheless] kind of way... well.. that certainly is a possibility and I guess that's the correct explanation of the two answers...

20. Feb 22, 2008

### Ajwrighter

Eventually it has to.

Actually. Our atmosphere acts as a boundry in which the acceleration due to earth's gravity becomes much smaller. Because of the amount of distance from Earth's atmosphere, the rate of an object's desent is much smaller. However the acceleration due to earth's gravity is still very closely the same. The reason why an object no longer falls as rapidly is the neglect of mass. The amount of force from the center of the earth vrs the point in which the escape velocity is achieved is significantly diffrent.

Last edited: Feb 22, 2008