A level h2 physics 2009 P3 Q5

G123 SG

I'll just start with part c) cos I don't get c)iii)

c) the mass of the star in b) may be considered to be a point mass at its centre

i) calculate the gravitational field strength at the surface of the star. Ans: 1.2 x 10^17 ms^-2

ii) determine the centripetal acceleration of a particle moving in a circular path of radius 1.7 x 10^4 m and with a period of rotation of 0.21s. Ans: 1.5 x 10^7 ms^-2

iii) the star rotates about its axis with a period of 0.21s. Use your answers in i) and ii) to suggest whether particles on the surface of the star leave the surface owing to the high speed of rotation of the star.

For c)iii), what my friend told me was that Since the acceleration of free fall (1.2 x 10^12 ms^-2) is much larger than the centripetal acceleration (1.5 x 10^7 ms^-2), the gravitational pull is more than what is required to provide the centripetal force. Therefore, the particles will not leave the star.

Ok guys this is what I don't get: if the star's gravitational pull provides the centripetal force for the particles to stay on the star, how do we know when the particle leaves the star? Aren't both acting in the same direction?(toward the centre of the star)

MUFC SG

Hi, I'm not an expert in this, but this is how I think it can be intepreted. You can draw a free body diagram where the particle is in between the star and the axis:

Star Particle Axis

Then the star exerts a gravitational force on the particle towards the left, while the centripetal force is towards the right. Since the gravitational force is larger, the particle continues "sticking" to the star instead of flying off due to the high rotational speed (and hence high centripetal force).

Lau_boy SG

testing

Lau_boy SG

gravitational force is a force that exist between 2 bodies due to their mass. In this case between the star and the particle. centripetal force is the force needed for a body to sustain in circular motion. When the star rotates, the particle on the surface will be in a circular motion. The C force needed by the particle to maintain this circular motion is provided by the G force, thus if the G force > C force, the particle will remain on the surface.

Imagine a ball tied to a string making a circular motion. The C force needed by the ball for this circular motion is provided by the tension force in the string. If you swing the ball faster then the tension in the string will increase. If the tension is too high and string break, the ball will fly away. This is analogy to the particle on star's surface. If the star rotates faster then the particle's weight on the surface will reduce because the C force needed has increased while the G force remain the same, if the C force > G force then the particle will fly away.

G123 SG

Ah i see. Thanks a lot guys!