what is the orbital velocity of an object 4.431844x10^-03 m from the surface of a "earth mass" blackhole.
2007-07-15 04:28:42 · 3 answers · asked by ftm821 2 in Science & Mathematics ➔ Physics
Using the equation
v^2/r = GM/r^2
G = gravitational constant (6.67 x 10-3 Nm^2kg-2)
v = orbital velocity
r = distance between bodies
M = mass of back hole
Subtituting
G(6.0 X 10^24kg)/[(4.43 x 10-3m)^2] = v^2/r
v^2/r = 2.04 x 10^19
v^2 = 9.03 x 10^16
v = 3.01 x 10^8 m/s
2007-07-15 04:51:39 · answer #1 · answered by Mandél M 3 · 0⤊ 0⤋
The standard Newtonian formula is GM / r = v^2
GM for a earth mass is about 3.986E14 m^3 s^-2
GM / r = 8.994E+16 m^2 s^-2
v = 299,899,965 m / s
This is about the same as the speed of light. Therefore, the Newtonian formula is not applicable, and the General Relativity formula must be used. The definition of "distance from the surface" needs to be clarified -- is it the Schwarzschild r-coordinate from the event horizon? Also, is "velocity" the GR proper velocity, or coordinate velocity?
2007-07-15 11:51:36 · answer #2 · answered by morningfoxnorth 6 · 0⤊ 0⤋
The distance you cite is inside the hole's event horizon, so I don't think the concept of "orbital velocity" has any meaning in there.
2007-07-15 11:55:58 · answer #3 · answered by RickB 7 · 0⤊ 0⤋