Q. 233.5( 6 Votes )

A star 2.5 times

Answer :

Now if an object is on a star placed at equator it will move with the surface of planet and will actually cover a circular path with certain speed depending upon angular speed or rate of revolution of star, now due to movement in circular path with some speed the object will experience a outward centrifugal force pushing it out of the planet but gravitational force on the object due to star is pulling it inwards or towards the surface so for object to be stuck on the star the gravitational force must be greater than the centrifugal force

The situation has been explained in the figure

We know gravitational force on a body is given as

Where F is the gravitational force

G is universal gravitational Constant

G = 6.67 × 10-11 Nm2Kg-2

M is mass of the first body which in this case is a star

Mass of star is 2.5 times mass of the sun, and mass of the sun is

Ms = 2 × 1030 Kg

So the mass of the star is

M = 2.5 × Ms = 2.5 × 2 × 1030 Kg = 5 × 1030 Kg

m is the mass of second body which in this case is the object, mass of an object is unknown let it be m

and R is the distance between the two bodies which here will be equal to the radius of star r as the object is on its surface i.e.

R = r = 12 Km = 1.2 × 104 m

So putting values of M, m, R, and G we get the magnitude of the inward gravitational force on object FG as

In inward direction, here m is mass of the object

Now body on the surface of the star is actually covering a circular path with a radius equal to the radius of the sun, the sun is revolving with a frequency of 1.2 revolutions per second so its rate of change of angle w.r.t center or angular frequency will be given as

ω = 2πf

where ω is the angular frequency of a body revolving with frequency f revolutions per second

here f = 1.2 rev/s

so we get the angular frequency of revolution as

ω = 2π × 1.2 rev/s = 7.539 rad/s

now centrifugal force Fc on the object which is undergoing circular motion in a circle of radius r with angular speed or frequency ω is given as

Fc = mrω2

Here radius of circular path or radius of the star is

r = 12 Km = 1.2 × 104 m

The angular speed of the object is

ω = 7.539 rad/s

mass of the object is m, mass is unknown

so putting values we get outward centrifugal force as

Fc = m × 1.2 × 104 m × (7.539 rad/s)2 = m × 6.8 × 105 N

In outward direction so clearly we can see comparing centrifugal force FC and Gravitational Force FG that gravitational force is much greater as

FG = m × 2.31 × 1012 N, in an inward direction

FC = m × 6.8 × 105 N, in outward direction

Value of m i.e. mass of an object is same in both the equations

So we can say


i.e. effectively object is pushed inward towards the surface of the star so it will remain stuck on the surface of the star

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