How Small Must Earth's Mass Be Compressed to Form a Black Hole?

[ticels]

a black hole is an object so heavy that neither matter nor even light can secape the influence of its gravitational field. Since no light can scape from it, it appears black. Suppose a mass apporxmiately the size of the Earth's mass 5.56x10^24 kg is packed into a small unifrom sphere of radius r.

*Use speed of light c=3.0x10^8 and Universal Gravitation G
*Escape speed must be the speed of light
*Relative equation
g=sq(G/r2); F=GMm/r^2; escape velocity=sq(2GM/r)

1) based on Newtonian mechanics, determine the limiting radius r0 where this mass (approximately the size of the Earth's mass) becomes a black hole. Answer in units of m.

2)Using Newtonian mechanics, how much would a mass of 4.26μg weigh at the surface of this super-dense sphere? Answer in units of N.

thx for help in advance.

 

[Mentia]

Hey hey. This is a pretty cool problem I think. What you're doing is calculating the Schwartzchild radius. You can get more info on that from Wikipedia: http://en.wikipedia.org/wiki/Schwarzschild_radius

Basically, you set the escape velocity to be the speed of light and then solve for "r".

You plug in c for the velocity, the given mass for M, and G for G and voila you've got the Schwartzchild radius.

For the 2nd problem, now that you have "r", you can solve for F from Newton's law of universal gravitation: F=GMm/r^2

 

[thomate1]

I am happy to assist with your gravitation problem. Based on the information provided, we can use the equations of Universal Gravitation and the speed of light to determine the limiting radius and weight of the super-dense sphere.

1) According to the equation for escape velocity, we can set the escape velocity equal to the speed of light, c, and solve for the radius, r. This gives us the equation:

c = √(2GM/r)

Where G is the Universal Gravitational Constant and M is the mass of the sphere. Plugging in the values given, we get:

3.0x10^8 = √(2 x 6.67x10^-11 x 5.56x10^24 / r)

Solving for r, we get:

r = 2GM/c^2 = 2 x 6.67x10^-11 x 5.56x10^24 / (3.0x10^8)^2 = 8.22x10^3 m

Therefore, the limiting radius, r0, where the mass becomes a black hole is approximately 8.22 kilometers.

2) To determine the weight of a mass of 4.26μg at the surface of the super-dense sphere, we can use the equation for gravitational force:

F = GMm/r^2

Where G is the Universal Gravitational Constant, M is the mass of the sphere, m is the mass of the object, and r is the radius of the sphere. Plugging in the values given, we get:

F = 6.67x10^-11 x 5.56x10^24 x 4.26x10^-6 / (8.22x10^3)^2 = 2.92x10^-2 N

Therefore, the weight of a mass of 4.26μg at the surface of the super-dense sphere is approximately 2.92x10^-2 Newtons.

I hope this helps with your gravitation problem. Please let me know if you have any further questions.