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Deffense [45]
3 years ago
15

Why does a black hole have a stronger gravitational pull than the star that collapse to form it?​

Physics
1 answer:
Studentka2010 [4]3 years ago
8 0

Answer:

We consider Black Holes as an object that possesses extreme gravitational pull, but wait aren’t they have the same mass(or less) as that of their parent star. And we know that gravitational pull ‘F’ is directly proportional to the mass of an object, so if the mass is same(or less) then why do black holes have stronger gravity than the stars they evolved from.

The above consideration that F is directly proportional to the mass is partially correct, one should also mention that F is also inversely proportional to the square of the distance between the considered objects.

F = G*(M*m)/(r^2)

Where:

· F is the force acting on you due to star

· M is the mass of Parent star / Black Hole

· m is the mass of an observer, here it is you

· r is the radial distance between the star and you

We know that black hole formed, has much smaller size than that of its parent star and all that mass is compressed to a much smaller scale. If you consider a Star as having a size of an earth then the black hole formed will have a size of small city.

Let us say that you are standing at an r distance away from a star (r>R1), where R1 is the radius of the star, of course (R1>R2), where R2 is the radius of Black Hole.

The Force by which the star in case 1 attracts you will be equal(or less) to the force by which black hole in case 2. So, there is nothing increase in gravitational pull, it is same(or less) as that of the parent star.

Wait a minute, then why people say that black holes have massive gravitational pull.

The gravitational pull increases as we move closer to the black hole, and when we are at its surface, it is enormous as compare to its star surface, because of the difference in the size.

We know that gravitational pull not only depends upon the mass but also depends upon the radial distance between the concerned objects here, it is you and the black hole.

Here, the size of the black hole is much smaller than that of its parent star, i.e (R1>>>R2), and thus we get F1<<<F2, and that is why we say that the black hole has enormous gravitational pull, such that nothing can escape, not even light.

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A proton, starting from rest, accelerates through a potential difference of 1.0 kV and then moves into a magnetic field of 0.040
Minchanka [31]

Answer:

r = 0.11 m

Explanation:

The radius of the proton's resulting orbit can be calculated equaling the force centripetal (Fc) with the Lorentz force (F_{B}), as follows:

F_{c} = F_{B} \rightarrow \frac{m*v^{2}}{r} = qvB (1)

<u>Where:</u>

<em>m: is the proton's mass =  1.67*10⁻²⁷ kg</em>

<em>v: is the proton's velocity</em>

<em>r: is the radius of the proton's orbit</em>

<em>q: is the proton charge = 1.6*10⁻¹⁹ C</em>

<em>B: is the magnetic field = 0.040 T </em>

Solving equation (1) for r, we have:

r = \frac{mv}{qB}   (2)

By conservation of energy, we can find the velocity of the proton:

K = U \rightarrow \frac{1}{2}mv^{2} = q*\Delta V   (3)

<u>Where:</u>

<em>K: is kinetic energy</em>

<em>U: is electrostatic potential energy</em>

<em>ΔV: is the potential difference = 1.0 kV </em>

Solving equation (3) for v, we have:

v = \sqrt{\frac{2q\Dela V}{m}} = \sqrt{\frac{2*1.6 \cdot 10^{-19} C*1.0 \cdot 10^{3} V}{1.67 \cdot 10^{-27} kg}} = 4.38 \cdot 10^{5} m/s  

Now, by introducing v into equation (2), we can find the radius of the proton's resulting orbit:

r = \frac{mv}{qB} = \frac{1.67 \cdot 10^{-27} kg*4.38 \cdot 10^{5} m/s}{1.6 \cdot 10^{-19} C*0.040 T} = 0.11 m

Therefore, the radius of the proton's resulting orbit is 0.11 m.

I hope it helps you!  

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3 years ago
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