Question

What should be the radius of a solid sphere of mass 9
times that of the mass of the earth to be a black hole?
Gravitational constant G = 6.67 x 10-11 N-m? / kg?
mass of the earth = 6 x 1024 kg, velocity of light
c = 3 x 108 m/s)​

Answer 1

Answer:

Explanation:

where

G = Newton's universal constant of gravity = 6.67 x 10-11 N-m2/kg3

M = the mass of the 'attracting object' [the planet] [in units of kg]

m = the mass of the object trying to escape [e.g., me or a ball or a rocket or a molecule] [in kg]

R = the distance between the centers of objects M and m [in units of m]

note: provided we do everything in the same units, we don't have to worry about units

while the kinetic energy we know from above:

Ekinetic=0.5 m v2

where

m = mass of the moving object [in kg]

v = the velocity of object m [in m/sec]

If we set these two energies equal to each other, and solve for v, we find the exact velocity needed to escape from the energy well:

0.5 m v2= GMm/R

v= (2GM/R)0.5

and since this velocity is exactly what is needed to 'escape,' it is called the escape velocity:

vescape= (2GM/R)0.5

Note what extremely important parameter is not in the escape velocity equation: the mass of the moving object.  The escape velocity depends only on the mass and size of the object from which something is trying to escape.  The escape velocity from the Earth is the same for a pebble as it would be for the Space Shuttle.

example #1: What is the escape velocity from the Earth?

MEarth=5.97 x 1024 kg

REarth=6378 km = 6.378 x 106 m

vescape= (2 x 6.67 x 10-11 N-m2/kg3 x 5.97 x 1024 kg / 6.378 x 106 m)0.5

vescape=1.12 x 104m/sec=11.2 km/sec

(this is equivalent to about 7 miles/sec or 25,200 miles per hour)