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Re: escape velocity question



John, you have nicely resolved the order of magnitude error posed
originally. I hope you won't be offended if I try to fine-tune your
result, since I think you have underestimated the white dwarf radius
slightly. I know you were just going for the quick rough estimate.

Since g = G * m / (R^2) we can find the radius R = sqrt( G * m / g ).

With g = 37000 * g_earth = 3.6E+05 m/s^2 and m = 1 solar mass =
2.0E+30 kg, we get R = 1.9E+07 m or about triple the earth's radius. If
you assume less mass, say 0.8 solar mass, you get a slightly smaller
radius, perhaps 1.7E+07 m but still over 2.5 earth radii. This agrees
pretty well with what little wd radius data I have seen (e.g. Sirius B and
Procyon B); 2-3 times wider than Earth, smaller than Neptune.

With this radius and escape velocity v = sqrt( 2 * g * R ) we get
v = 3.7E+06 m/s for a 1 solar mass star, 3.5E+06 m/s for 0.8 solar mass.
Higher than your estimate by a factor of about 1.6.
Just slightly over 0.01c .

Best wishes,

Larry

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Larry Cartwright
Physics, Physical Science, Internet Teacher
Charlotte High School, 378 State Street, Charlotte MI 48813
<physics@scnc.cps.k12.mi.us> or <science@scnc.cps.k12.mi.us>

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
I am very good at learning from my mistakes.
Undoubtedly, I shall learn a great deal today.
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On Mon, 15 Mar 1999, John Mallinckrodt wrote:

No relativity required here. Escape velocity is given classically by

v_esc = (2 g R)^(1/2)

For an escape velocity equal to the speed of light with g = 37,000 g_earth
we would find R = 120 million km (the size of a red giant) and
M = 42 million solar masses. A white dwarf has (I think) a radius closer
to that of Earth so the escape velocity would be "only"
37000^(1/2) x 11.2 km/s = 2.2 x 10^6 m/s, less than 1/100 the speed of
light.