Rehash: speed of falling object at Earth's centre

[DaveC426913]

On another forum has appeared the ol' "drop a ball down a tunnel that passes through the Earth's centre" discussion.

Someone has naively calculated the speed of the object at a ridiculous number - a fraction of the speed of light!

I'm pretty sure their error is that they are not taking into account the drop off in g as your distance from Earth's centre decreases. (i.e. they don't realize it, but they're actually calculating the speed of a ball falling towards an Earth-mass black hole with zero radius).

So, what is the velocity of the ball as it passes through the centre of the Earth (ignoring all the usual factors: vacuum, Earth's rotation, etc.)?

Thx.

 

[berkeman]

Sounds like a fun puzzle. I don't have time to do the integral at the moment (pesky work), but I'll try to do it tomorrow if I have time. What's a good number to use for the radius and total mass of the Earth?

 

[berkeman]

Doh. That was easier than I thought. Just equate final KE to initial PE for the object on the surface. I get about 8000m/s, but I'm not at all sure of my starting numbers. I'm in a meeting and can't go look stuff up, but I swag'ed these numbers:

R = 2000 miles = 3.2^6m approx.
I just used g = 9.8m/s^2 instead of bothering with M of the Earth

Is this what you get, Dave? Nothing close to c obviously.

 

[tony873004]

I'm not sure if this is right (I'm going to try berkeman's method in a minute), but:
I remember reading once that the period for such an object to do one complete circuit and end up back at the hole where it was dropped would be the same as the period for circular orbital velocity at Earth's surface. So:

$$P = 2\pi \sqrt {\frac{{a^3 }}{{GM}}} = 2\pi \sqrt {\frac{{\left( {6378000m} \right)^3 }}{{{\rm{6}}{\rm{.673}} \times {\rm{10}}^{{\rm{ - 11}}} \left( {kg \cdot m/s^2 } \right)m^2 /kg^2 \times 5.97 \times 10^{24} kg}}} = 5071s$$
so it will take 2535 seconds to get to the center of Earth which is 6378 km away.

$$v_{avg} = \frac{{\Delta d}}{t} = \frac{{6378}}{{2535}} = 2.5km/s$$

It begins with a velocity of 0.

$$v_{avg} = \frac{{v_f - v_i }}{2} \Rightarrow v_f = 2v_{avg} + v_i = 2 \times 2.5km/s + 0 = 5km/s$$

 

[tony873004]

Radius of Earth = 6378000 m (~4000 miles)
Mass of Earth = 5.97E24

$$\frac{{ - GMm}}{r} = \frac{1}{2}mv^2 \Rightarrow mv^2 r = - 2GMm \Rightarrow v = \sqrt {\frac{{ - 2GM}}{r}}$$
This yields the escape velocity formula (after getting rid of the minus sign).

PE KE method 2:
$$\begin{array}{l} mgh = \frac{1}{2}mv^2 \Rightarrow v = \sqrt {\frac{{mgh}}{{\frac{1}{2}m}}} \\ \\ v = \sqrt {\frac{{9.81m/s^2 6378000m}}{{\frac{1}{2}}}} = {\rm{11186}}m/s \\ \end{array}$$
This yields escape velocity as well. That can't be right, can it?

 

[DaveC426913]

Are you guys factoring in the drop-off in g as you plunge into the Earth?

At 2000 miles from the centre, you're under the influence of much less than 1/2g. It drops to 0 by the centre.

 

[Doc Al]

I agree with berkeman's answer. Realize that the force varies linearly with distance from the Earth's center: from 0 at the center to mg at the surface. The force equation is: F = -(mg/R)r. Thus the motion is simple harmonic, just like a mass on a spring. The max speed is given by sqrt(gR).

 

[franznietzsche]

I agree with berkeman's answer. Realize that the force varies linearly with distance from the Earth's center: from 0 at the center to mg at the surface. The force equation is: F = -(mg/R)r. Thus the motion is simple harmonic, just like a mass on a spring. The max speed is given by sqrt(gR).

Might want to think about that again. Your peak potential energy in an SHM is
$$U = \frac{1}{2} (\frac{m g}{R}) r^2$$
$$\qquad = \frac{1}{2} m g R$$

So you're off by a factor of $$\sqrt{2}$$ in the velocity. Further

$$F = -\frac{G M_r m}{R^2}$$
In order for the force to scale linearly with r (i.e. have a SHM), you require that
$$F = -k r$$
$$k = \frac{G M_r m}{R^3}$$
$$\frac{M_r}{R^3} = a$$
where a is constant. From there:
$$\frac{ d M}{dR} = 4 \pi R^2 \rho$$
$$\qquad = 3 a R^2$$
$$\rho = \frac{3 a}{4 \pi}$$

i.e., you have assumed constant density, which is not correct (this isn't the source of the problem though), and unnecessary.

Much better to use conservation of energy than force (always better to do so when forces vary in potentially unknown ways over the path, but you know the potential energy at the starting and ending points of the path):

$$U = -\frac{GMm}{R}$$
$$K_f - \Delta U = 0$$
(it is the change in potential energy that matters, not its actual value)
$$\Delta U = \frac{GM_{R_{surf}}m}{R_{surf}}$$
$$\frac{1}{2} m v^2 = \frac{GM_{R_{surf}}m}{R_{surf}}$$
$$v = \sqrt{\frac{2GM_{R_{surf}}}{R_{surf}}$$

or about 11,680 m/s.

 

[Janus]

If you take the difference between the Gravitational Potential at the surface of the Earth:

$$U = - \frac{GMm}{R}$$

With R being the RAdius of the Earth.

and the potential at the center of the Earth (Not zero!):

$$U = - \frac{3GMm}{2R}$$

you get :

$$\Delta U = \frac{3GMm}{2R} - \frac{GMm}{R}$$

$$\Delta U = \frac{GMm}{R}\left ( \frac{3}{2}-1\right )$$

$$\Delta U = \frac{GMm}{2R}$$

$$\frac{v^2m}{2} = \frac{GMm}{2R}$$

$$v = \sqrt{\frac{GM}{R}}$$

v= ~7.9 Km/sec or the same as the orbital speed at the surface of the Earth.

 

[franznietzsche]

If you take the difference between the Gravitational Potential at the surface of the Earth:

$$U = - \frac{GMm}{R}$$

With R being the RAdius of the Earth.

and the potential at the center of the Earth (Not zero!):

$$U = - \frac{3GMm}{2R}$$

I'm trying to get this, and I can't seem. How do you get this result?

 

[tony873004]

...Realize that the force varies linearly with distance from the Earth's center: from 0 at the center to mg at the surface...

Assuming this to be true, I computed it numerically with 1 second time steps. It does turn out to be equal to circular velocity for a satellite orbiting at Earth's surface (~7.9 km/s) (ignore air resistance).


Graph of data : http://orbitsimulator.com/PF/drop.GIF
(Why no image link in this subforum? Other subforums in PF have image links.)

Private Sub Drop()
Dim a As Double, v As Double, x As Double, aMax As Double, aMin As Double
Const G As Double = 0.000000000066725, M As Double = 5.97E+24, R As Double = 6378000

aMax = G * M / R ^ 2
v = 0
x = R
Open "c:/drop.txt" For Output As #1
    For k = 1 To 6000
        a = -x / R * aMax
        v = v + a
        x = x + v
        Print #1, Str$(x/1000) + "," + Str$(v)
    Next k
Close #1
End Sub

 

[George Jones]

I'm trying to get this, and I can't seem. How do you get this result?

Typically, the potential energy for a harmonic oscillator is written as

$$U = \frac{1}{2} k r^2,$$

but any other potential that differs from this by a constant will also work, so actually

$$U = \frac{1}{2} k r^{2} + c.$$

Usually, $c$ is chosen to be zero so that the potential is zero at the equilibrium point of the oscillator. In this case, however, the zero for gravitational potential is chosen at infinity. Matching (at the surface of the Earth) this choice of potential outside the Earth to the harmonic oscillator potential inside the Earth forces $c$ to be non-zero.

Setting the two potentials equal at the surface of the Earth gives

$$- \frac{GmM}{R} = \frac{1}{2} \frac{GmM}{R^3} R^2 + c,$$

which results in

$$c = - \frac{3}{2} \frac{GmM}{R}.$$

 

[George Jones]

Pull together what everyone has written into one coherent mass.

Let $M$ be the mass of the Earth and $R$ be the radius of the Earth, which is assumed to be spherical, and to have constant density

$$\rho =\frac{M}{V}=\frac{M}{\frac{4}{3}\pi R^3}.$$

As the body of mass $m$ falls through the Earth, only the part of the Earth that is closer to the Earth's centre than it influences its motion. So, if the body is a distance $r$ from the centre of the Earth, only a sphere of radius $r$ need be considered. The mass of this sphere is

$${}{}{\RD{\CELL{M_{r} &=&\rho V_{r}}}{}{}{}\RD{\CELL{ &=&\rho \frac{4}{3}\pi r^{3}}}{}{}{}\RD{\CELL{ &=&\frac{M}{\frac{4}{3}\pi R^{3}}\frac{4}{3}\pi r^{3}}}{}{}{}\RD{\CELL{ &=&\frac{r^{3}}{R^{3}}M.}}{}{}{}}$$

The acceleration of the body is then

$${}{}{\RD{\CELL{m\ddot{r}=-\frac{GmM_{r}}{r^{2}}=-\frac{GM}{R^{3}}r,}}{}{}{}}$$

so

$${}{}{\RD{\CELL{\ddot{r}+\omega ^{2}r=0,}}{}{}{}}$$

which is the equation for simple harmonic motion with (angular) frequency

$${}{}{\RD{\CELL{\omega =\sqrt{\frac{GM}{R^{3}}}.}}{}{}{}}$$

This has immediate solution

$${}{}{\RD{\CELL{r &=&R\cos \omega t}}{}$$

$${}{}\RD{\CELL{v &=&-\omega R\sin wt=-\sqrt{\frac{GM}{R}}\sin \omega t\text{,}}}{}{}{}}$$

so (as Janus has already posted), the maximum speed is

$${}{}{\RD{\CELL{\sqrt{\frac{GM}{R}}.}}{}{}{}}$$

The speed also can be written as

$${}{}{\RD{\CELL{|v| &=&\sqrt{\frac{GM}{R}}\sqrt{1-\cos ^{2}\omega t}}}{}{}{}\RD{\CELL{ &=&\sqrt{\frac{GM}{R}\left( 1-\frac{r^{2}}{R^{2}}\right) },}}{}{}{}}$$

which also can be derived easily from the conserved energy of a harmonic oscillator.

 

[Doc Al]

Might want to think about that again. Your peak potential energy in an SHM is
$$U = \frac{1}{2} (\frac{m g}{R}) r^2$$
$$\qquad = \frac{1}{2} m g R$$

So far, so good.

So you're off by a factor of $$\sqrt{2}$$ in the velocity.

Why do you say that? Just set the peak kinetic energy ($1/2 m v^2$) equal to that peak potential energy and you'll get:

$$v_{max} = \sqrt{gR} = \sqrt{GM/R}$$

Further

$$F = -\frac{G M_r m}{R^2}$$
In order for the force to scale linearly with r (i.e. have a SHM), you require that
$$F = -k r$$
$$k = \frac{G M_r m}{R^3}$$
$$\frac{M_r}{R^3} = a$$
where a is constant. From there:
$$\frac{ d M}{dR} = 4 \pi R^2 \rho$$
$$\qquad = 3 a R^2$$
$$\rho = \frac{3 a}{4 \pi}$$

i.e., you have assumed constant density, which is not correct (this isn't the source of the problem though), and unnecessary.

I admit that I assumed a constant density, for simplicity. But if you make any other assumption, you'll get a different answer.

Much better to use conservation of energy than force (always better to do so when forces vary in potentially unknown ways over the path, but you know the potential energy at the starting and ending points of the path):

$$U = -\frac{GMm}{R}$$
$$K_f - \Delta U = 0$$
(it is the change in potential energy that matters, not its actual value)
$$\Delta U = \frac{GM_{R_{surf}}m}{R_{surf}}$$
$$\frac{1}{2} m v^2 = \frac{GM_{R_{surf}}m}{R_{surf}}$$
$$v = \sqrt{\frac{2GM_{R_{surf}}}{R_{surf}}$$

or about 11,680 m/s.

For a correct treatment using change in gravitational PE, see the responses by Janus and George.

 

[DaveC426913]

"...but you know the potential energy at the starting and ending points of the path..."

The math is way too complex for me, but are these formulae taking into account the drop-off in Fg along the path? It seems to me this is an awfully special case - completely different from any other orbital mechanics problem.

 

[George Jones]

"...but you know the potential energy at the starting and ending points of the path..."

The math is way too complex for me, but are these formulae taking into account the drop-off in Fg along the path? It seems to me this is an awfully special case - completely different from any other orbital mechanics problem.

Yes, the drop-off in the force of gravity is taken into account.

The speed of the object when it is a distance $r$ from the centre of the Earth is

$$|v| = \sqrt{\frac{GM}{R}\left( 1-\frac{r^{2}}{R^{2}}\right) },$$

where $R$ is the radius of the Earth and $M$ is the mass of the Earth. At the centre, $r = 0$ gives the result that Janus gave at the bottom of post #9.

 

[Janus]

Also,

Starting with

$$M_r = \frac{Mr^3}{R^3}$$

For the mass effecting the falling object at any given radius r inside the Earth (As per George Jones' post).

We can take a different route by first finding the force of gravity at any radius r:

$$F_g = \frac{GmM_r}{r^2}$$

$$F_g = \frac{Gm\frac{Mr^3}{R^3}}{r^2}$$

$$F_g = \frac{GmMr}{R^3}$$

By integrating this, we get the gravitational potential of the object wrt to the center for any r :

$$U = \frac{GmMr^2}{2R^3}+C$$

In this case we will use the Earth's center as our reference point, so we set the potential energy to zero at the center. (this allows us to get rid of the constant (C).

At the surface r=R, so at the surface:

$$U = \frac{GmMR^2}{2R^3}$$

$$U = \frac{GmM}{2R}$$

(Note this is not the same as the normal expression for gravitational potential, but remember, that expression sets the zero potential at infinity, and we are stting it to zero at the center of the Earth.)

Since our expression directly gives the potential difference between surface and center, we can equate it to kinetic energy as we did before, and we get the same answer as we did before.