Natural units in the Schwarzschild Metric

[PeteSampras]

Homework Statement

My Teacher says that in the Schwarzschild metric he uses natural units, where he writes
##g_{rr}=1-2M/R##
He says that for one neutron star ##R=5## corresponds to approx 13 KM.

Homework Equations

##1l_p=1,616 \cdot 10^{-35}m##

The Attempt at a Solution

Unfortunately he does not says me (perhaps with a litle of arrogance) if ##R=5 l_p## or not...

But, if evaluate ( from ##1-2GM/(c^2R)##)

##c^2R=(3 \cdot 10^8)^2 \cdot 5 l_p=##

##(3 \cdot 10^8 m/s)^2 \cdot 5 \cdot 1,616 \cdot 10^{-35}m=##

##7,272 \cdot 10^{-18} m^3/s^2##

But I don't know what my teacher want to say "your R=5 is similar to 13KM". More information he does not want to give me.

Help please

 

[mfb]

This is purely relativity, there is no Planck length involved.

It would be better to say R=5M. A black hole has its event horizon at R=2M (the point where g_rr gets 0), so R=5M is well outside that.

 

[vela]

Homework Statement

My Teacher says that in the Schwarzschild metric he uses natural units, where he writes
##g_{rr}=1-2M/R##
He says that for one neutron star ##R=5## corresponds to approx 13 KM.

Homework Equations

##1l_p=1,616 \cdot 10^{-35}m##

The Attempt at a Solution

Unfortunately he does not says me (perhaps with a litle of arrogance) if ##R=5 l_p## or not...

But, if evaluate ( from ##1-2GM/(c^2R)##)

##c^2R=(3 \cdot 10^8)^2 \cdot 5 l_p=##

Why is ##l_p## there and what happened to ##G##?

Try plugging in the values and simplifying the units on ##G/c^2##. And then remember that in natural units, you're setting ##G/c^2## equal to 1.