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3 years ago

Beyond what point must an object be squeezed for it to become a black hole

Physics

1 answer:

Naddik [55]3 years ago

4 0

Answer:

Condition

$\frac{m}{R} \geq \frac{c^{2} }{2*g}$

Explanation:

Basically black hole is an object from which light rays can not escape it means to go out from gravitational field , that body should thrown with speed greater then light.

Let's do some calculation

Gravitational potential at surface = $-\frac{G*M*m}{R}$

If we give kinetic energy equal to magnitude of Potential energy as on surface it will escape.

$\frac{m*v^{2} }{2}$ = $-\frac{G*M*m}{R}$

⇒ $\frac{m}{R} =\frac{c^{2} }{2*G}$

It will be more better for black hole if above ratio (analogous to density ) is more then above calculated

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3 years ago

The energy difference in these two conformations has been measured to be about 23 kj per mole. how much of this energy differenc

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Answer:

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3 years ago

A newly discovered planet is found to have a density if 2/3pe and a radius of 2RE, where PE and RED are the density and radius o

Liono4ka [1.6K]

Missing question in the text of the exercise. Found on internet:
"What is the acceleration due to gravity on the surface of the planet?"

Solution:
The gravitational acceleration at Earth's surface is given by:
$g= \frac{GM}{r^2}$
where
G is the gravitational constant
M is the Earth's mass
r is the Earth's radius

The Earth's mass can be rewritten also as the product between the Earth's density, $d$ , and its volume (the volume of a sphere of radius r):
$M=dV=d ( \frac{4}{3} \pi r^3)= \frac{4}{3} \pi d r^3$

Now let's call M' the mass of the new planet, r' its radius and d' its density. The acceleration due to gravity on the surface of the new planet is
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so we need to find M' and r'.

The problem says the radius of the new planet is twice the Earth's radius:
$r'=2r$ (2)
and that its density is 2/3 of Earth's density:
$d'= \frac{2}{3} d$
so the mass M' of the new planet is, with respect to the Earth's mass:
$M' = d'V' = \frac{4}{3} \pi d' (r')^3 = \frac{4}{3} \pi ( \frac{2}{3}d) (2r)^3 = ( \frac{4}{3} \pi d r^3 )( \frac{16}{3}) = \frac{16}{3} M$ (3)

And if we substitute (2) and (3) into (1), we find the gravitational acceleration on the surface of the new planet:
$g'= \frac{G( \frac{16}{3}M) }{(2r)^2}= \frac{GM}{r^2} \frac{4}{3} = \frac{4}{3}g$
And since $g=9.81 m/s^2$ , we find
$g'= \frac{4}{3}(9.81 m/s^2)=13.1 m/s^2$

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Explanation:

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3 years ago

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