# Algebra problem in a force question

1. May 9, 2005

### tony873004

I've got this boiled down to an algebra problem.

At what distance from a 1 Solar mass black hole must you be so that the tidal force experienced by my body is equal to the gravitational force at Earth's surface.

Assume I am 2 meters, and 70 kg (the 70 kg's will drop out later, but they ask for force, so I'll include them for now)

$$F_{g, Earth}=\frac{GM_{Earth}m_{me}}{R_{Earth}^2}$$

$$F_{g, Black Hole,my feet}=\frac{GM_{Black Hole}m_{me}}{d_{black hole,my feet}^2}$$

$$F_{g, Black Hole, my head}=\frac{GM_{Black Hole}m_{me}}{d_{black hole,my head}^2}$$

$$F_{g, Earth}=F_{g, Black Hole, my head}-F_{g, Black Hole, my head}$$

$$\frac{GM_{Earth}m_{me}}{R_{Earth}^2}= \frac{GM_{Black Hole}m_{me}}{d_{black hole,my feet}^2}-\frac{GM_{Black Hole}m_{me}}{d_{black hole,my head}^2}$$

The G's and $$m_{me}$$'s cancel

$$\frac{M_{Earth}}{R_{Earth}^2}= \frac{M_{Black Hole}}{d_{black hole,my feet}^2}-\frac{M_{Black Hole}}{d_{black hole,my head}^2}$$

My head is 2 meters further from the black hole than my head, so:

$$\frac{M_{Earth}}{R_{Earth}^2}= \frac{M_{Black Hole}}{d_{black hole,my feet}^2}-\frac{M_{Black Hole}}{(d_{black hole,my feet}-2m)^2}$$

Clean this up a little. Mass of Earth and radius of Earth are known, so let's just call the left side of the equation X for now.

$$x=\frac{M}{d^2}-\frac{M}{(d-2m)^2}$$

Here's where I get stuck :grumpy: How do I isolate d ?

Here's my effort, but this is probably wrong:

$$x=\frac{M(d-2m)^2} {d^2(d-2m)^2} -\frac{Md^2} {d^2(d-2m)^2}$$

$$x=\frac{M(d-2m)^2 - Md^2} {d^2(d-2m)^2}$$

$$x=\frac{M(d-2m)^2 - Md^2} {d^2 (d-2m) (d-2m)}$$

I'm just shuffling numbers around and getting nowhere :zzz:

2. May 9, 2005

### Staff: Mentor

If you really want to solve it exactly, multiply both sides by $d^2(d-2m)^2$ and then get a fourth order equation in d.

But... assuming that 2m << d, use an approximation. Several ways to do it. One way is to write $(d - 2m)^2 = d^2(1 - 2m/d)^2$, then use a binomial expansion of that last term, ignoring higher powers of (2m/d), since they are really small.

An even better way is to realize that for small changes in x, $\Delta f = (df/dx) \Delta x$.

3. May 10, 2005

### tony873004

I still don't get it. I just don't remember how to do this kind of algebra
btw... the 2m can just be written as 2 because m is not a variable, it means 2 meters.

I know the answer is 3781979 meters because I did it this way with programming, which is ok to get credit on the problem, but I want to know how to do it the algebra way.

Code (Text):

Private Sub btnCompute_Click()
Dim G As Double, Mearth As Double, Mbh As Double, Mme As Double
Dim Rearth As Double, Rbh As Double, Hme As Double
Dim Fearth As Double, Fbhhead As Double, Fbhfeet As Double
Dim Fdifference As Double

sstep = Val(txtStep) 'Amount black hole’s radius will be increased with each step
Rbh = Val(txtRbh) 'starting value for the radius of the Black Hole
G = 0.00000000006673 ' Gravitational Constant
Rearth = 6371000 ' Radius of the Earth in meters
Mearth = 5.97E+24 ' Mass of the Earth in kilograms
Mbh = 1.9891E+30 ' Mass of the black hole in kilograms (1 Solar Mass)
Hme = 2 ' My height in meters
Mme = 70 ' My mass in kilograms (value is irrelavant to final answer)
Fearth = (G * Mearth * Mme / Rearth ^ 2) ' Compute Force at Earth's Surface
txtFearth = Fearth ' Display Force at Earth's surface

Do
Rbh = Rbh + sstep ' Increment the black hole's radius by 1 step size
Fbhfeet = (G * Mbh * Mme / Rbh ^ 2) ' Force of black hole at my feet
Fbhhead = (G * Mbh * Mme / (Rbh + Hme) ^ 2) ' Force of black hole at my head
Fdifference = Fbhfeet - Fbhhead ' Difference between these 2 forces
Loop Until Fdifference < Fearth ' Keep going until you get me an answer

txtDifference = Fdifference ' Display the black hole forces difference
txtRbhComputed = Rbh ' Display the distance from the black hole that I must stay
End Sub

4. May 10, 2005

### tony873004

I figured it out. But I had to keep the highest power and eliminate the rest.