Question

# When a large star becomes a supernova, its core may be compressed so tightly that it becomes a neutron star, with a radius of about 20 km (about the size of the San Francisco area). If a neutron star rotates once every second, (a) what is the speed of a particle on the star’s equator and (b) what is the magnitude of the particle’s centripetal acceleration? (c) If the neutron star rotates faster, do the answers to (a) and (b) increase, decrease, or remain the same?

Solution

## (a)  we know $$v=2πr/T$$putting values according to question$$=2π(20km)/1.0s$$$$=126km/s$$(b) The magnitude of the acceleration is  $$a=\dfrac { { v }^{ 2 } }{ r }$$putting values according to question$$=\dfrac { { (126km/s) }^{ 2 } }{ 20km } =7.9\times { 10 }^{ 5 }m/{ s }^{ 2 }.$$(c) Clearly,$$v\propto1/T$$  and $$a\propto v^2$$$$a\propto1/T^2$$both $$v$$ and $$a$$ will increase if $$T$$ is reduced. Physics

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