Physics, 08.04.2020 00:34 semily74

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.

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Answer from: itsnotxavier

The gravitational force on the object would gradually increase toward infinity.

Explanation:

Given:

Mass of the main-sequence star = 3M ...where $M=2\times 10^3^0$ kg

Radius of the star = 2.5R ...where $R=6.95700\times 10^8$ m

According to the question:

The entire star mass gradually collapse to a singularity.

Using Newtons law we have to find the gravitational force.

⇒ Gravitational force, $g=\frac{GM}{r^2}$ ..equation (i)

From the above equation we can say that mass is in direct proportion whereas radius is inverse proportion of the gravitational force.

Note:

Whenever any star's mass collapsed to singularity, virtually in mathematical term the volume of the star is said to be zero and its density becomes infinite.

Density infinite means mass infinite as $\rho = \frac{m}{V}$ .

Now:

As mass or the density tends to infinity,then gravitational force also tends to infinity.

Mathematically:

Lets take the mass as one unit then and radius as 'x'.

As the radius will shrink to zero.

⇒ $\lim_{x \to 0} \frac{1}{x} = \frac{1}{0} = \infty$

The gravitational force on the object would gradually increase toward infinity.

Answer from: Quest

Iwannasay c is the most accurate

Answer from: Quest

force is based upon mass and acceleration.the statement is not always true.

it can be explained from newton's second law of motion.

as per newton's second law of motion the rate of change of momentum is directly proportional to the applied force and takes place along the direction of force.

mathematically it can be written as -

$f\alpha \frac{dp}{dt}$

⇒ $f= k \frac{dp}{dt}$

here k is the proportionality constant whose value is 1.

$f=\frac{dp}{dt}$

$f= \frac{d[mv]}{dt}$

$f= m\frac{dv}{dt} +v\frac{dm}{dt}$

$f=ma+v\frac{dm}{dt}$ [a is the acceleration=dv/dt]

hence we see that force not only depends on mass and acceleration but also on velocity and rate of change of mass. force f will depend on mass and acceleration only when mass is constant.

hence the statement that force is based upon mass and acceleration is false.

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