Changes between Version 1 and Version 2 of u/erica/scratch

Timestamp:

01/27/16 13:53:33 (10 years ago)

Author:

Erica Kaminski

Comment:

—

u/erica/scratch

@@ -1 @@
-Having a kinetic energy (per unit mass) at r that is higher than would have been acquired from freefall alone, i.e. solving the equation:
+Having a kinetic energy (per unit mass) at r that is higher than would have been acquired from freefall alone (which would just be GM/r), i.e. solving the equations:
 
 [[latex($ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = 1.01 (\frac{GM}{R})$)]]
 
-again where, 
+[[latex($ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = 1.3 (\frac{GM}{R})$)]]
+
+again where the specific kinetic energy at the surface of the star (R) after having fallen from a distance r away from the center of the star is given by, 
 
 [[latex($ke(R)=ke(r)-\frac{G*M}{r}+\frac{G*M}{R} $)]]
 
 
-|| '''ke(r)''' || '''Distance (pc)''' || '''Error (%)'''||
-|| 10*freefall || 6.8x10^-7^ || 30 ||
-|| 100*freefall || 7.4x10^-6^ || 30 ||
-|| 1000*freefall || 7.5x10^-5^ || 30 ||
-|| 10*freefall || .00002 || .01 ||
-|| 100*freefall || .0002 || .01 ||
-|| 1000*freefall || .002 || .01 ||
+
+|| '''ke(r)''' || '''Error (%)''' || '''Distance (pc)'''||
+|| 10*freefall || 30 || 6.8x10^-7^ ||
+|| 10*freefall || .01 || .00002 ||
+|| 100*freefall || 30 || 7.4x10^-6^ ||
+|| 100*freefall || .01 || .0002 ||
+|| 1000*freefall || 30 || 7.5x10^-5^ ||
+|| 1000*freefall || .01 || .002 ||
+
+
+Now, what if a gas parcel started from rest, a distance r away from the star surface? Now we are solving, 
+
+[[latex($-\frac{G*M}{r}+\frac{G*M}{R} = .99 (\frac{GM}{R})$)]]
+
+[[latex($-\frac{G*M}{r}+\frac{G*M}{R} = .70 (\frac{GM}{R})$)]]
+
+|| '''ke(r)''' || '''Error (%)''' || '''Distance (pc)'''||
+|| 0 || .01 || 2.2x10^-6^ ||
+|| 0 || 30 || 7.5x10^-8^ ||
+
+
+Lastly, what if the parcel was moving, however, it was moving ''slower'' than freefall? Now ke(r) will be a fraction of the freefall energy in the table below. In particular, what if ke(r) was 1/2, 1/5, 1/10 the freefall kinetic energy... at what distance, r, would I see a 0.01% error? What about a 30% error?
+
+[[latex($ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = .99 (\frac{GM}{R})$)]]
+
+[[latex($ke(r)-\frac{G*M}{r}+\frac{G*M}{R} = .70 (\frac{GM}{R})$)]]
+