Question

Black holes have all their mass packed into a single point (known as singularity). How does this affect the gravitational force on an object close to the black hole? Newton’s formula for the force of gravity between two masses (m1 and m2), where r is the distance between the centers of the two objects, is F=Gm1m2r2F=Gm1m2r2. A main-sequence star of 3 solar masses has a radius of approximately 2.5 times the radius of our Sun. According to Newton’s formula, what would happen to the gravitational force felt by an object on the surface of this 3-solar-mass star if the star’s entire mass gradually collapsed to a singularity? Assume the object remains on the surface of the star throughout the collapse.

Answer 1

a) the gravitational force will be so big that the object would never be able to leave

b) the gravitational pull will have a similar effect

Step-by-step explanation:

a)

Newtons approach for the gravitational pull is:

`F = (Gm1m2)/(r^2)`

let m1 be the mass of the black hole, and m2 the mass of the poor object which is brave enough to get close to it. r is the distance from the black hole to the object.

The mass of a black hole is tremendously big and compact, even our sun is light weight compared to a random black hole. The gravitational force decreases with the square of the distance, however, m1 can sometimes be so big that even an object that is far away would be attracted. The object will get attracted untill r is 0 and the gravitational force is equal to infinite ( no return in this case)

b)

Now a collapsing star with 3 solar masses and a radius of 2.5 times the radius of our sun will do the same, as the radius of the star decreases the gravitational pull increases with the square of the distance, until the singularity (r=0) which means that the gravitational force is infinite