rest-mass-energy equivalent of GPE ?

[Widdekind]

Gravitational Potential Energy (GPE) ~ GM²/R

 

equivalent rest-mass energy = mc²

 

GM²/R = mc²

 

^{m/M ~ R_S/R}

 

^{(actually, 1/5 x R_S/R)}

 

^{So, order-of-magnitude, the rest-mass equivalent, of some self-gravitating body, is a similar fraction of the total actual mass, as that body's Schwarzschild radius is, to its actual physical radius.}

[xyzt]

Gravitational Potential Energy (GPE) ~ GM²/R

 

equivalent rest-mass energy = mc²

 

GM²/R = mc²

 

^{m/M ~ R_S/R}

 

^{(actually, 1/5 x R_S/R)}

 

^{So, order-of-magnitude, the rest-mass equivalent, of some self-gravitating body, is a similar fraction of the total actual mass, as that body's Schwarzschild radius is, to its actual physical radius.}

I assume that in the above "M" is the gravitational mass while "m" is the inertial mass. If so, your conjecture is many orders of magnitude off since it is well known (google "Eotvos experiment", "Einstein equivalence principle") that the two are equal to a very high order of precision.

Edited June 16, 2013 by xyzt

[Widdekind]

no -- the amount of anti-matter required to obliterate a gravity-bound ball-shaped world/planet

[xyzt]

no -- the amount of anti-matter required to obliterate a gravity-bound ball-shaped world/planet

In this case you would have indeed:

 

[math]m=M \frac{r_s}{R}[/math]

 

For example, for Earth, the ratio [math]\frac{r_s}{R}=\frac{10^{-2}}{6.4*10^6} =\frac{1}{6.4*10^8}[/math]

 

Thus, [math]m=10^{16} kg[/math]

Edited July 3, 2013 by xyzt

[Widdekind]

very vaguely, about a billionth of an earth-mass of AM => "Aldebaran"

 

such seemed readily rememberable, "mass ratio (AM:M) ~ radius ratio (R_S:R)", and seemingly works well(-est) for "classical" or "sub-relativistic" ball-shaped bodies

 

 

In this case you would have indeed:

 

[math]m=M \frac{r_s}{R}[/math]

 

For example, for Earth, the ratio [math]\frac{r_s}{R}=\frac{10^{-2}}{6.4*10^6} =\frac{1}{6.4*10^8}[/math]

 

Thus, [math]m=10^{16} kg[/math]