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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)

[tex]F_{g, Earth}=\frac{GM_{Earth}m_{me}}{R_{Earth}^2}[/tex]

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

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

[tex]F_{g, Earth}=F_{g, Black Hole, my head}-F_{g, Black Hole, my head}[/tex]

[tex] \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}[/tex]

The G's and [tex]m_{me}[/tex]'s cancel

[tex] \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}[/tex]

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

[tex] \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}[/tex]

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.

[tex] x=\frac{M}{d^2}-\frac{M}{(d-2m)^2}[/tex]

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

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

[tex] x=\frac{M(d-2m)^2} {d^2(d-2m)^2} -\frac{Md^2} {d^2(d-2m)^2} [/tex]

[tex] x=\frac{M(d-2m)^2 - Md^2} {d^2(d-2m)^2} [/tex]

[tex] x=\frac{M(d-2m)^2 - Md^2} {d^2 (d-2m) (d-2m)} [/tex]

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