How to calculate earth´s schwarzschild radius?

Question

Anyone who could write out the following equation? The equation itself is Rs (the s is a subscript s) = 2GM/c^2, and the earths schwarzschild radius is 9mm, how do you get to that answer? I found the formula on Wikipedia, but it doesn´t explain it...

Answer 1

The poor man's black hole:


Almost immediately after Newton formulated his theory of gravity, there was speculation about the possibility of "dark stars".



(There were a couple of really *old* papers by Mitchell, who you have never heard of, and Laplace, who you should have heard of.)



Remember that in Newtonian gravity the escape velocity from a compact body of mass M and radius R is given by:



(1/2) (v_{escape})^2 = G M/ R



This formula is derived simply by conservation of energy, trading off (Newtonian) kinetic energy for (Newtonian) gravitational potential energy.



Now define



R_{dark_star} = 2 G M/ c^2



Then if R is less than R_{dark_star}, we have v_{escape} greater than c, so that light cannot escape.



Of course, this logic is completely Newtonian and it is pretty much a *miracle* that exactly the same result holds in full-fledged general relativity (Einstein gravity), up to and including the factor of 2.



In general relativity we just change the name, and call it the Schwarzschild radius:



R_{Schwarzschild} = 2 G M/ c^2,



and we say that objects that are smaller than their Schwarzschild radius are to be called "black holes".



[1] Calculate the Schwarzschild radii for Moon, Earth, Jupiter, and Sun. I want the answer in centimetres/metres/kilometres, and I want at least 3 significant figures.



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A:

From the particle data book, and the CRC handbook of physics:


M_{moon} = 0.0123 M_{earth} 7.348 x 10^{22} kg

M_{earth} = 5.974(9) x 10^{24} kg

M_{jupiter} = 318 M_{earth} = 1.900 x 10^{27} kg

M_{sun} = 1.988 9(30) x 10^{30} kg

G_{newton} = 6.673(10) x 10^{-11} m^3 kg^{-1} s^{-2}

c = 299 792 458 m s^{-1}

The notation xxxxx(yy) means the last two digits of the xxxxx are
uncertain by an amount yy [one standard deviation].

============================================
Calculate:


R_S(moon) = 0.05455 mm

R_S(earth) = 4.435 028 11 mm

R_S(jupiter) = 1.410 m

R_S(sun) = 2.953 250 08 km

PS: Note that we know the Schwarzschild radius to better accuracy
than the mass... This is because from astronomical observation it is
easy to measure the product GM for the individual planets, but it is a
lot more difficult to measure G and M separately. So we know the
Schwarzschild radius of the earth and sun to 9 significant figures,
even though we cam measure G_{newton} only to 2 significant figures...
And if i had an astronomical alamanck handy, it would be possible to
deduce the Schwarzschild radius of moon and jupiter to 9 significant
figures; I am limited here by the precision of the CRC Tables for the
mass of moon and jupiter.

Answer 2

Schwarzschild Radius Formula

Answer 3

Schwarzschild Radius Equation

Answer 4

This Site Might Help You.

RE:
How to calculate earth´s schwarzschild radius?
Anyone who could write out the following equation? The equation itself is Rs (the s is a subscript s) = 2GM/c^2, and the earths schwarzschild radius is 9mm, how do you get to that answer? I found the formula on Wikipedia, but it doesn´t explain it...

Answer 5

Actually in that article on Wikipedia it simplifies the equation, and tells you what every variable stands for.

All you have to do is plug in the mass of the object and you have it.