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

• littlebhawk
In summary, the conversation involves a problem of calculating the time taken for an object released from a distance of 10 Earth radii to reach the surface of Earth without considering the atmosphere. Different methods were suggested, such as using an energy approach or a differential equation, but there were discrepancies in the results. Some members also expressed their concerns about the level of expertise of certain science teachers.
littlebhawk
Ok I am new to physics forum but I am sure if I am 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 don't know if I am even close to right as I've never tried a question like this before, any help would be appreciated.

Oh, this isn't a homework problem my teachers and my friends are just really curious at how to even attempt this problem.

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.

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

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?

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

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littlebhawk said:
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.

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..

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Thanks rohanprabhu that's definately a lot 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 definately don't doubt that my teachers could solve this problem if i had given them more time though, there definately 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.

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':

http://img263.imageshack.us/img263/3528/gr8fn1cz5.jpg

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

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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?

Danger said:
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?
This one and the ball-string one are both a little disturbing to me...

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?

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dst said:
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?

The student is trying to solve for the time.

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

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.

littlebhawk said:
Ok I am new to physics forum but I am sure if I am 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 don't know if I am even close to right as I've never tried a question like this before, any help would be appreciated.

https://www.physicsforums.com/showthread.php?t=156681

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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) that's 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.

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Ajwrighter said:
With out the atmosphere you’re not going to get a simple drop. With this in mind, that little bubble we are surrounded by plays a significant factor on the rate of descent.

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.

Ajwrighter said:
I'm no expert but it looks like the rock hits the ground once and then hits it again about xx time later.

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.

Ajwrighter said:
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 .

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.

Ajwrighter said:
There is about 40,000 pieces of space junk(/satalites) that's 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.

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.

rohanprabhu said:
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.

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.

K.J.Healey said:
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.

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...

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.

Eventually it has to.

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, which is only 15% of the mean value at the surface of earth.

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.

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

or,

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

Your mistake is here. You neglected the constant of integration when you integrated the velocity. After integrating you get

$$\frac{1}{2}v^2 = \frac{GM}{r} + C$$

Since at t = 0, v = 0, r = 10R_e, C = -GM/(10R_e), and

$$\frac{1}{2}v^2 = \frac{GM}{r} -\frac{GM}{10R_E}$$

Then you integrate again (it will be trickier now, if it's even possible at all - I don't think it is. I think I tried to solve this once, and there wasn't an analytic solution. Maybe I'm wrong.).

edit - you also made an error here:

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

the \sqrt{GM} shouldn't be raised to the 3/2 power, which would affect the numerical result somewhat.

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Ajwrighter said:
Eventually it has to.

you mean eventually collide?? and you mean to say that the equation took care of that on it's own? Maths doesn't work that way.

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.

this is seriously the most oblivious statement I've ever heard. When we talk about a particular phenomena in physics, we restrict the discussion to that particular topic only. When actually working out with problems, we bring in all the variables necessary. Acceleration due to gravity is independent of the atmosphere. It is purely and purely dependent only on the mass of the Earth and the separation between.

So, if there was no atmosphere, the acceleration due to gravity would be very less just outside the exosphere [as i said, around 1.4 m/s²]. The presence of atmosphere reduces this further. Plus, the resistance offered by the atmosphere [which is a fluid] is dependent on the velocity of the falling object and hence is also time-dependent.

Mute said:
Your mistake is here. You neglected the constant of integration when you integrated the velocity. After integrating you get

$$\frac{1}{2}v^2 = \frac{GM}{r} + C$$

Since at t = 0, v = 0, r = 10R_e, C = -GM/(10R_e), and

$$\frac{1}{2}v^2 = \frac{GM}{r} -\frac{GM}{10R_E}$$

Then you integrate again (it will be trickier now, if it's even possible at all - I don't think it is. I think I tried to solve this once, and there wasn't an analytic solution. Maybe I'm wrong.).

yup.. u caught it right and i did mention it in my next-to-next post: https://www.physicsforums.com/showpost.php?p=1619217&postcount=9. The thing is that after getting the equation for v as a function of 'r'.. it is really difficult to integrate it further or do anything with it. I ran the integration process through Mathematica to get the equation and got something like this: http://img263.imageshack.us/img263/3528/gr8fn1cz5.jpg

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Here's how I solved it (I got an answer of t = 658 seconds). There's a related thread on using calc- vs. non calc-based physics, this is a good argument for calc. based...

Start with F = ma:

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

$$-GM dt = r^{2} dv$$

$$-GMt + C = r^{2} v + A$$

Boundary condition: at t = 0, v = 0: C = A

$$-GMt dt = r^{2} dr$$

$$-\frac{GMt^{2}}{2} + C = \frac{r^{3}}{3}+A$$

Boundary condition: t + 0, r = 10 : C = 10^3/3 + A

When r = 1:

$$\frac{GMt^{2}}{2} \approx \frac{10^{3}}{3}$$

Substituting in for G, M, and using the correct units for r:

t = 658 s.

Note that this assumes the particle is falling directly towards the center of the earth. No rotation.

Andy Resnick said:
$$-GM dt = r^{2} dv$$

$$-GMt + C = r^{2} v + A$$

Here is where you go wrong... you are assuming that r is constant with variation in 'v', which is not the case. Only if r(v) = const. can u assume that integral.

I vaguely remember a result from astronomy which states that the period of an elliptical orbit depends only on the long axis of the ellipse. In other words an orbit of 100% eccentricity between 0r and 10r has the same period as a circular orbit between -5r and
5r.

If this is true, then we can easily get a numerical answer for this question. For example, using Kepler's law of r-cubed-over-t-squared, and recalling that a low orbit (radius = 1r) has a period of 90 minutes, it easily follows that the period of a 5r orbit must be nearly 17 hours. We then consider the object in question to be in an orbit of the same total length (10r) but with a 100% eccentricity. Therefore the object will fall to the center of the Earth from the peak of the orbit in about half this time, 8.5 hours. There is a small correction because we want the time to intersect the surface rather than the center.

Wait, does this mean the only way to solve it is numerically! I don't think so. I'll have to do a bit of research first, and then come back and see what I can help with. Probably nothing, lol.

rohanprabhu said:
Here is where you go wrong... you are assuming that r is constant with variation in 'v', which is not the case. Only if r(v) = const. can u assume that integral.

Duh... oops.

Ok, let's stick with the full nonlinear equation:

$$\frac{-GM}{r^{2}}=\frac{d^{2}r}{dt^{2}}$$

The formal solution is $r(t)=(bt+C)^{2/3}$. By substitution, $$b=-3\sqrt{GM}$$ and from the initial condition t = 0, r = 10 gives C = 10^3/2.

Plugging and chugging now gives t = 8235 s, or about 2 hours.

I got the solution from a magic book, BTW: "Handbook of Nonlinear Partial Differential Equations", Polyanin and Zaitsev. Polyanin has a whole book series and website with this stuff:

http://eqworld.ipmnet.ru/

a = -C/x^2 = v*dv/dt
At x = R (rad. of Earth), a=g, hence C = gR^2

v*dv/dt = - gR^2/x^2 = d/dx(0.5v^2)
0.5v^2 = gR^2/x + c
when x = 10R, v = 0, so:
c= -gR/10, and:
v^2 = 2gR(R/x - 1/10)

Let sqrt(2gR) = a

v = a.(R/x - 1/10)^(1/2)

v = dx/dt

dx/[(R/x - 1/10)^(1/2)] = a.dt

R/x - 1/10 = (10R - x)/(x - 10)

[(x - 10)/(10R - x)]^(1/2).dx = a.dt

I have no idea how you integrate that. Probably has no anti-derivative. Can do it numerically.

rohanprabhu said:
yup.. u caught it right and i did mention it in my next-to-next post: https://www.physicsforums.com/showpost.php?p=1619217&postcount=9. The thing is that after getting the equation for v as a function of 'r'.. it is really difficult to integrate it further or do anything with it. I ran the integration process through Mathematica to get the equation and got something like this: http://img263.imageshack.us/img263/3528/gr8fn1cz5.jpg

How do you input something like this into Mathematica? I tried:

DSolve[x''[t] == -GM/x[t]^2, x[t], t]

And it doesn't ever finish running. Then I tried adding boundary condition x[0] == a, x'[0] == 0, but that gives me an error with x'[0] (the x'[0]==0 is being evaluated as True for some reason).

DSolve[{x''[t] == -GM/x[t]^2, x[0] == r, x'[0] == 0}, x[t], t]

Edit: changed independent variable to x and it works now. Gives no solutions though, because inverse functions were used.

Edit2: used NDSolve and played around and figured it takes about 28336 seconds or 7.9 hours, for it to reach approximately zero. Why can't a bunch of people on a physics forum get answers that agree? ;)

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I did model this using excel accurate to the second, i came out with the correct veloctiy at moment of impact, with gravity equal to 9.81 (like it should be) at moment of impact, with a time of 27916 seconds or about 7.75 hours. Ill post the excel spreadsheet in a little bit I am trying to crunch it down so the file size isn't insanely large ( the spreadsheet comes out to almost 30000 rows!)

littlebhawk said:
I did model this using excel accurate to the second, i came out with the correct veloctiy at moment of impact, with gravity equal to 9.81 (like it should be) at moment of impact, with a time of 27916 seconds or about 7.75 hours. Ill post the excel spreadsheet in a little bit I am trying to crunch it down so the file size isn't insanely large ( the spreadsheet comes out to almost 30000 rows!)

I think our answers agree, if I were to make it end at one earth-radius.

Yeah I am sure they do, i finally uploaded the excel spreadsheet as a zip file, its VERY LARGE! and may take your computer a large amoutn of time to open, depending on comp.

http://www.youshare.com/view.php?file=VariableGravityField.zip

You can enter the initial values in the first row and itll work out everything else automatically.

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awvvu said:
How do you input something like this into Mathematica? I tried:

DSolve[x''[t] == -GM/x[t]^2, x[t], t]

I don't use DSolve generally.. it doesn't solve many problems. See.. we got this:

$$v = \sqrt{2} \sqrt{\frac{GM + rY}{r}}$$

$$\frac{dr}{dt} = \sqrt{2} \sqrt{\frac{r}{GM + rY}}$$

$$\sqrt{\frac{r}{GM + rY}}dr = \sqrt{2} dt$$

This is what I integrated in Matematica using:

Code:
Integrate[Sqrt[(r)/(G M + r Y)], r]

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After a lot of battling with equations and a few mountain dews later.. I've arrived at a solution.. which i believe is quite correct. So.. i request you to please take a look and point out flaws [if any]:

Assumptions for this problem: Take for any arbitrary distance 'r' from the center, let the center be the origin.
Also, instead of taking the starting position as $10R_E$, I'm assuming an arbitrary position 's'.

The acceleration of the object is given by:

$$a = -\frac{GM}{r^2}$$

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

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

$$vdv = -\frac{GM}{r^2}dr$$

$$\frac{v^2}{2} = \frac{GM}{r} + C$$

also, at $r = s$, $v = 0$ Hence,

$$GM + sC = 0$$

$$C = -\frac{GM}{s}$$

So, we have:

$$v = \sqrt{2GM} \sqrt{\frac{1}{r} - \frac{1}{s}}$$

Let, us just take out the $\sqrt{2GM} = \Phi$. We'll call it the 'planetary mass factor' [yup.. it's my turn now to use fancy terms :P], So, we have the 'planetary mass factor' as:

$$\Phi = \sqrt{2GM}$$

so..

$$\frac{dr}{dt} = \Phi \sqrt{\frac{1}{r} - \frac{1}{s}}$$

let the above equation be eqn. 1.

$$\sqrt{\frac{sr}{s - r}}dr = \Phi dt$$

Let,
$$\frac{sr}{s - r} = x$$

$$\frac{s^2}{(s - r)^2}dr = dx$$

Putting the values in eqn. 1, we get:

$$\frac{(s - r)^\frac{3}{2}}{s^2}dx = \Phi dt$$

Also, from the first assumption,

$$r = \frac{sx}{s + x}$$

So,

$$\frac{dx}{s + x} = \Phi dt$$

$$log_e|s + x| = \Phi t + J$$

J is again, a constant of integration. Now, substituting the value of 'x' in terms of r, we get:

$$\frac{s^2}{s - r} = e^{\Phi t + J}$$

$$\frac{s^2}{s - r} = e^J e^{\Phi t}$$

$$\frac{s^2}{s - r} = ke^{\Phi t}$$

On simplifying it a bit,

$$r(t) = s\left(1 - \frac{s}{ke^{\Phi t}}\right)$$

Now, r(0) = s . Solving for 'k',

$$k = 2s$$

Putting it in the equation, we get the answer as:

$$r(t) = s\left(1 - \frac{1}{2e^{\Phi t}}\right) = s\left(1 - \frac{1}{2e^{\sqrt{2GM} t}}\right)$$

This is the equation of motion I got. :D

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