Miniature Black Holes: Calculating Distance from Person

[Gold3nlily]

Homework Statement

This problem is awesome! I like this chapter; Its really interesting. I think I just get a little impatient when I can't figure out the answer right away... I appreciate any help.

Miniature black holes: Left over from the big-bang beginning of the universe, tiny black holes might still wander through the universe. If one with a mass of 3 × 10¹¹ kg (and a radius of only 4 × 10^-16 m) reached Earth, at what distance from your head would its gravitational pull on you match that of Earth's? Assume free-fall acceleration ag=9.83 m/s2.

Homework Equations

F = (GMm)/r²

G = 6.67x10^-11m³/Kgs²
rep = distance to Earth as seen by person
rbp = distance to black hole as seen by person
Me = mass of Earth = 5.96 x 10²⁴ kg
Mp = mass person = not given (probably cancel out somehow)
Mb = mass black hole = 3 × 10¹¹ kg
Rb= radius black hole = 4 × 10^-16 m
Re = radius Earth = 6.38 x 10⁶ m

The Attempt at a Solution

want to find distance when forces are equal...
Fep = (G*Me*mp)/rep²
Fbp = (G*Mb*mp)/rbp²
Fep = Fbp

(G*Me*mp)/rep² = (G*Mb*mp)/rbp²
divide both sides by G and mp
(Me)/rep² = (Mb)/rbp²

but I am stuck. I think:
rep = distance between Earth and person + Re
and
rbp = distance between person and B.H. + Rb

and I want to solve for "distance between person and B.H." but I don't know "distance between Earth and person". Surely it would be small, but can't be zero... would it just be the radius of the earth? (When I used this in the equation I only got the mass of the black hole as my answer, so I think the equation might be flawed.)

Any suggestions?

 

[phinds]

This won't be any help, but my distinct impression is that such a black hole would evaporate very quickly (some small fraction of a second) so for one to be still hanging around from the BB is unthinkable.

 

[Gold3nlily]

This won't be any help, but my distinct impression is that such a black hole would evaporate very quickly (some small fraction of a second) so for one to be still hanging around from the BB is unthinkable.

Hmm, that is interesting. Thanks for your input.

I finally figured it out. It was WAY simpler than I thought!

I used F=(G*M)/(r + d)²
set this equation equal to the force of Earth (ag--> 9.83)
Then solved for d (d = variable we are looking for)
Yay!

 

[gneill]

This won't be any help, but my distinct impression is that such a black hole would evaporate very quickly (some small fraction of a second) so for one to be still hanging around from the BB is unthinkable.

Evaporation time goes up very quickly with mass. A good approximation for the lifetime of a black hole of mass M is:

T = 8.40716 \times 10^{-17} \left( \frac{M}{kg} \right)^3 sec

For a 3 x 10¹¹ kg mini black hole, that amounts to about 72 billion years!

 

[Gold3nlily]

Evaporation time goes up very quickly with mass. A good approximation for the lifetime of a black hole of mass M is:

T = 8.40716 \times 10^{-17} \left( \frac{M}{kg} \right)^3 sec

For a 3 x 10¹¹ kg mini black hole, that amounts to about 72 billion years!

Neat!