How Do You Model an Apple Falling into a Black Hole Using Integral Calculus?

[Rapidrain]

Just for fun I'd like to find the equation r(t) which defines the position of an apple as a function of time falling from a distance Ro (radius of the earth) down to a basketball with the mass of the earth--probably a black basketball hole.

All websites show the equations for a distance where the gravitational acceleration of the Earth (g = 9.8 m³/s²) is, for all practical purposes, constant--Pisa tower, etc. I am curious as to whether I can show the position in an equation from Ro all the way to the event horizon of my black basketball hole.

I got this far : v(t) = sqrt ( 2*g / r(t) - 2*g/Ro ) //&1

That second member I set to -2*g/Ro because I am dropping the apple from Ro away from the black basketball hole and the initial velocity is 0.

a little algebra gets me : dt = 1/(sqrt(2*g/r(t) - 2*g/Ro)) dr

now I integrate that guy and end up with :

t = (1/sqrt(2*g))*( -r*Ro*sqrt(1/r - 1/Ro) - sqrt(Ro³)(arctan(Ro/r - 1)) //&3

and it doesn't work when using 200, 400, 600 meters
It yields a time which is 55,000,000-fold that of the values calculated using the tower of Pisa-scale formula r = 0.5*g*t²

Is my v(t) formula wrong at &1 ?

Or is my integration wrong in &3?

I've been trying to solve this for two months now, can anybody see an error in my logic?

 

[Mindscrape]

On first look, though let me think about it, I don't see how this should be any different than a basketball actually falling to Earth under the general law of gravitation. You know that Fg=GmM/r^2 (since you can find g from Gauss's Law for gravitation, and then Fg=mg), so you just have to solve for the differential equation until the edge of the apple hits the basketball. When this happens things get a bit more difficult.

 

[Rapidrain]

What I believe you are describing is the situation all teenagers know from High School physics class. At the surface of the Earth the distance to the center (r) is effectively constant since you are observing something fall for less that 1% of r, and there it makes the whole gravitation mathematics quite straightforward. This is Mickey Mouse.

What I have is an apple falling towards an object with the mass of the Earth from an initial distance of Ro (6.371*10^6 meters). I want to describe a situation where r changes dramatically and where the 9.822 m/s^2 of the high school textbooks changes from its surface of the Earth value as r changes.

As you say, the rest is differential calculus. And as you say also, it gets difficult. This is why I am asking this question.

 

[D H]

(g = 9.8 m³/s²)

I got this far : v(t) = sqrt ( 2*g / r(t) - 2*g/Ro )

Look at your units! g is 9.80665 m/s², not 9.8 m³/s².

So let's back up a step or two. Assume the Earth has a spherical mass distribution (density is a function of distance from the center of the Earth). For an object released with zero initial velocity at time t=0 and distance r[/sub]0[/sub] from the center of the Earth, conservation of energy requires that

\frac 1 2 v(t)^2 - \frac{GM_e}{r(t)} = - \frac{GM_e}{r_0}

or

v(t)^2 = 2GM_e\left(\frac 1{r(t)} - \frac 1 {r_0}\right)

The quantity GM_e is well-known. The product GM_e=398,600.442 km³/s² (note well: not 9.8 m³/s²), is known to much better precision than either the universal gravitational constant G or the mass of the Earth M_e.

Note that there is no assumption of a uniform gravity field here. This in fact assumes a 1/r² gravity field. Note also that the sign of the velocity is ambiguous here. You are concerned with the time interval t=0 and the time at which the object hits the Earth. During this interval the velocity vector is downward. dr/dt is negative. Thus you should be using

v(t) = -\sqrt{2GM_e\left(\frac 1{r(t)} - \frac 1 {r_0}\right)}

You can solve this to yield the time t(r) at which the object reaches some distance r from the center of the Earth. Inverting this to yield r(t) is going to be a bit more difficult.

 

[Rapidrain]

>>Look at your units! g is 9.80665 m/s2, not 9.8 m3/s2.

<br /> v(t) = -\sqrt{2GM_e\left(\frac 1{r(t)} - \frac 1 {r_0}\right)}<br />

In that guy up there I agree with you.

As you correctly point out my value of g is not 9.8. This confusion comes because I am not using g = G*Me/r^2 where r = the surface of the earth, instead I'm using g = G*Me which has a value of 3.9868*10^14 m3/s2.

I had this equation one month ago. Getting it down to t(r) is my goal now. My calculus is really rusty. And that is what drove me to this forum.

The issue of field uniformity is irrelevant at this point. I've got the mass of the Earth in a basketball and hope to avoid all extraneous factors, keeping it as plain as possible.

 

[Mindscrape]

What I believe you are describing is the situation all teenagers know from High School physics class. At the surface of the Earth the distance to the center (r) is effectively constant since you are observing something fall for less that 1% of r, and there it makes the whole gravitation mathematics quite straightforward. This is Mickey Mouse.

What I have is an apple falling towards an object with the mass of the Earth from an initial distance of Ro (6.371*10^6 meters). I want to describe a situation where r changes dramatically and where the 9.822 m/s^2 of the high school textbooks changes from its surface of the Earth value as r changes.

As you say, the rest is differential calculus. And as you say also, it gets difficult. This is why I am asking this question.

No, actually what I am referring to is
\nabla \cdot \mathbf{g} = - 4 \pi G \rho
Not at all assuming that g is constant (in fact it depends on 1/r^2 as the equation states), nothing too "Mickey Mouse." I did, however, assume a constant density, and that we are using a Gaussian surface outside the radius of the effective Earth, which gives you the inverse square law you are familiar with.

You could also use the same "Mickey Mouse" type potential from high school with conservation of energy as DH has suggested.

 

[Rapidrain]

Keeping it simple, I am not referring to gaussian surfaces nor anything more complex than a point of gravitational force, my black hole basketball and an apple falling towards that source of gravity. Just these two entities. No earth, no other mitigating factors.

Simply the problem of solving the equation below into t(r) :

<br /> v(t) = -\sqrt{2GM_e\left(\frac 1{r(t)} - \frac 1 {r_0}\right)}<br />

 

[D H]

Do you want a nudge in the right direction or the answer?I'll assume you want a nudge for now. Rewrite the expression for velocity, dr/dt, as

\frac{dr}{dt} = -\sqrt{\frac{2\mu_e}{r_0}}\sqrt{\frac{r_0}{r}-1}

where μ_e is the Earth's standard gravitational parameter, μ_e=GM_e. The first radical on the right hand side is just a constant. The second radical suggests a trig substitution. See how far you can get using the substitution

\frac{r}{r_0} = cos^2\theta

and assume theta is between 0 and pi/2 inclusive, making both sine and cosine zero or positive.

 

[Rapidrain]

Now that looks like fun. I'll get on it in the next 24 hrs.
Thank you muchly DH !

 

[Rapidrain]

this came out really sweet.
I used the cos^2 x = r/ro substitution as you suggested and it all came out to be :

dt = (cos x)² dx

and that gave me :

t = BFC*[ x/2 + (1/4)*(sin 2x) + C ]

where BFC is a big fat constant. 2*(r_e**3/2)/sqrt(2GM_e)

It took a half hour to figure in a noisy cafe and another 20 minutes at home to confirm with XL. What a journey.

Many Thanks DH.

 

[D H]

Now you're getting started. Good!

Note that I said "getting started". You might want to express t as a function of r rather than as a function of theta (or x in your case). You might also want to evaluate that integration constant. You know that the object started at a distance r=r₀ at time t=0. That makes this an initial value problem. There is no unknown constant of integration here.