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 = 2×10^{30} kg).

Asked by Pragya Singh | 1 year ago | 149

Any matter/ object will remain stuck to the surface if the outward centrifugal force is lesser than the inward gravitational pull.

Gravitational force, f_{G}

\( \dfrac{GM\;m}{R^{2}}\)= [Neglecting negative sign]

Here,

M = Mass of the star = 2.5× 2 × 10^{30} = 5 × 10^{31 }kg

m = Mass of the object

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

Therefore, f_{G}

= \( \dfrac{(6.67\times 10^{-11})\times (5\times 10^{31})\;m}{(1.2\times 10^{4})^{2}}\)

\( = 1.334\times 10^{13}m\;N\)

Now, Centrifugal force, f_{c}= m r ω^{2}

Here, ω = Angular speed = 2πν

ν = Angular frequency = 10 rev s^{–1}

f_{c} = m R (2πν)^{2}

f_{c} = m × (1.2 x 10^{4}) × 4 × (3.14)^{2} × (1.2)^{2}

= (6.82 ×10^{5}m) N

As f_{G} > f_{C}, the object will remain stuck to the surface of black hole.

A rocket is fired ‘vertically’ from the surface of mars with a speed of 2 km s^{–1}. If 20% of its initial energy is lost due to martian atmospheric resistance, how far will the rocket go from the surface of mars before returning to it? Mass of mars = 6.4×10^{23} kg; radius of mars = 3395 km; G = 6.67×10^{-11} N m^{2} kg^{–2}.

A spaceship is stationed on Mars. How much energy must be expended on the spaceship to launch it out of the solar system? Mass of the space ship = 1000 kg; mass of the sun = 2×1030 kg; mass of mars = 6.4×1023 kg; radius of mars = 3395 km; radius of the orbit of mars = 2.28 ×108 km; G = 6.67×10-11 N m2kg–2.

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