#### Question

A star 2.5 times the mass of the sun and collapsed to a size of 12 km rotates with a speed of 1.2 rev. per second. (Extremely compact stars of this kind are known as neutron stars. Certain stellar objects called pulsars belong to this category). Will an object placed on its equator remain stuck to its surface due to gravity? (mass of the sun = 2 x 10^{30} kg).

#### Solution 1

Yes

A body gets stuck to the surface of a star if the inward gravitational force is greater than the outward centrifugal force caused by the rotation of the star.

Gravitational force, `f_g = (GMm)/R^2`

Where

*M *= Mass of the star = 2.5 × 2 × 10^{30} = 5 × 10^{30 }kg

*m *= Mass of the body

*R =* Radius of the star = 12 km = 1.2 ×10^{4} m

`:.f_g = (6.67xx10^(-11)xx5xx10^(30)xxm)/(1.2xx10^4)^2 = 2.31xx 10^(11) m N`

Centrifugal force, *f*_{c}^{ }= *mr**ω*^{2}*ω* = Angular speed = 2π*ν*

ν = Angular frequency = 1.2 rev s^{–1}

*f*_{c} = *mR* (2π*ν*)^{2}

= *m* × (1.2 ×10^{4}) × 4 × (3.14)^{2} × (1.2)^{2} = 1.7 ×10^{5}*m*^{ }N

Since *f*_{g} > *f*_{c}, the body will remain stuck to the surface of the star

#### Solution 2

Acceleration due to gravity of the star,g= GM/R^{2 }…………(i)

Here M is the mass and R is the radius of the star.

The outward centrifugal force acting on a body of mass m at the equator of the star =mv2/R =mR w2——-(ii)

From equation (i), the acceleration due to the gravity of the star

`=(6.67xx10^(-11) xx2.5xx2xx10^(30))/(12xx10^3)^2 = 2.316 xx 10^(12) "m/s"^2`

∴Inward force due to gravity on a body of mass m

=`m xx 2.316 xx 10^12 N`

From equation ii the outward centrifugal force = `mRomega^2`

`=mxx (12xx10^3) xx ((2pixx1.5)/(-1))^2`

= `mxx 1.06 xx 10^6 N`

Since the inward force due to gravity on a body at the equator of the star is about 2.2 million times more than the outward centrifugal force, the body will remain stuck to the surface of the star.