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I'm a mathematics student currently tackling differential equations, but I'm not very good at interpreting problems so I hope you can help me with this one. I don't really know if the problem is actually a "real simplified" physical model but I thought this was the place to ask.

Problem: On the interior of the earth, the gravitational force is proportional to the distance from the center of the earth. Assume you make a hole from the surface all the way down to the center and let a rock fall freely down there. What velocity would the rock have when it reaches the center?

My attempt: Let $r(t)$ be the distance between the rock and the center of the earth at time $t\geq 0$. Let $R$ be the radius of the earth. So $r(0)= R$, and we want to find the velocity $v(t^*),$ for the time $t^*$ such that $r(t^*)=0.$ Now, if the distance traveled by the rock at time $t$ is given by $x(t)$, then it seems to me that

$$x(t) = R - r(t), \text{ when } \, t\in[0,t^*], \text{ and if so, then}$$ $$\frac{dx}{dt}=-\frac{dr}{dt}, \text{ and} \, \frac{d^2x}{dt^2}=\frac{dv}{dt}=-\frac{d^2r}{dt^2}$$

Now, I interpret the fact that the gravitational force is proportional to the distance from the center of the earth as:

$$g(r) = K r(t), \text{ for some constant} \, K.$$

Also, I don't know if I'm allowed to use this, but I think there is a Newton law that states that $F=ma=m \frac{dv}{dt}$. Assuming the mass of the rock is $m=1$ and since the rock is free falling (so the only force acting on it is the gravitational force), we get

$$Kr(t)= g(r) = \frac{dv}{dt} = -\frac{d^2r}{dt^2},$$

so, we have the differential equation

$$\frac{d^2r}{dt^2} = -Kr(t), r(0)=R.$$

Now my questions:

  1. Is this reasoning correct? Can I deduce $r'(0)$ from this data? If not I would get a family of solutions so I guess I can but I'm not sure.

  2. If $K<0$ then $r(t)$ is a sum of exponential functions, but it seems to me that, if this is an actual physical model and my intuition is not wrong, that $r(t)$ would oscillate between poles [that is, between $(0,R)$ and $(0, -R)$] , so we can assume $K>0$ right?

  3. If this is an actual physical model, what does the constant $K$ stand for? Something related to the distribution of mass of the earth or something?

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You can think it like this: the earth-rock system is very similar to a spring-ball system if you ignore any pertubations i.e. wind resistance and Coriolis effect, and the centre of the earth corresponds to the origin of spring.

The force of gravity is $$F=G\frac{Mm}{r^2}$$ and earth mass $M$ is $$M=\frac{4}{3}\pi \rho r^3$$ So, the force actually becomes $$F=\frac{4}{3}\pi\rho Gmr$$ For your case, just 1D case is considred. But, you know the force always has an opposite direction compared to displacement vector. So, strictly, the force is $F=-\frac{4}{3}\pi\rho Gmx=-kx$ with $k>0$. Finally, the motion equation can be expressed as $$m\ddot{x}+kx=0$$ Clearly, it has an ossicilation solution,i.e. $x(t)=x_0cos(\omega t)$ with $x_0$ the initial position at $t=0$, and the angular frequency is $\omega=\sqrt{\frac{4}{3}\pi\rho G}$

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  • $\begingroup$ Welcome to Physics SE! Please note that we don't answer homework or worked example type questions. Please see this Meta post on asking homework/exercise questions and this Meta post for "check my work" problems. That being said, you shouldn't provide complete answers to a homework question. $\endgroup$ – Yashas Mar 19 '17 at 5:04
  • $\begingroup$ @Feynman thank you for your answer. I have some questions if you don't mind: Your solution seem to imply that $x'(0)=0$ right? how did you arrive at that result? and what do you mean by angular frequency? I'm no familiar with physics at all so in my mind there is no angle involved in a 1-dimentional motion. Thanks! $\endgroup$ – Luis Vera Mar 19 '17 at 5:17
  • $\begingroup$ Yes, I assume $x^\prime(0)=0$. The exact solution for the motion equation is $x(t)=Acos(\omega t)+Bsin(\omega t)$, and the constant $A$ and $B$ will be detemined uniquely by initial conditions.Ahh, What's the $\omega$ or angluar frequency? You just substitute the solution into the motion equation and drops the duplicate. By definition, Angluar frequency $\omega=2\pi f$ with $f$ the frequency, e.g. 100Hz $\endgroup$ – Feynman Mar 19 '17 at 5:32
  • $\begingroup$ @Feynman thank you. My argument to understand why $x'(0)=0$ would be that, since no force is being applied to the rock, $v(0)=0$, and for some tiny quantity $0 + \Delta t,$ the gravitational force of the earth starts to act on it, so it starts to fall. Now, for the angular frequency $\omega$, seeing in wikipedia that $\omega = \frac{2\pi}{T}$, where $T$ is the time it takes for the particle to return to it's initial position, it seems to me that the angular frequency is a ratio that relates the period of the cosine function ($2\pi$) relative to the period of the oscillatory motion involved. $\endgroup$ – Luis Vera Mar 19 '17 at 5:49
  • $\begingroup$ You fully misunderstand why $x^\prime(0)=0$. At the beginning, just thinking of a spring, the rock has a maximum displacement wrt the center of earth, so, earth will exert a maximum force to the rock, but it is still static. In the next step, or some tiny time, the force will reduce and the velocity of rock will increase, and in the center of earth the velocity will reach to maximum and meanwhile the force is zero. $\endgroup$ – Feynman Mar 19 '17 at 6:03
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I think you are worrying too much it is in simple hormonic motion (SHM) hence the velocity is maximum and acceleration is zero as the equation of gravity at depth of $d$ is given by $g'=g(1-\frac{d}{R})$ where $g'$ is new gravity and R is radius of the Earth hence at center the gravity is zero but you have derived it under constant gravity condition hence you have to think better.

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