Calculate the angular momentum of a star of mass 2 ｘ10³⁰ kg and radius 7 ｘ 10⁵ Km. If it makes one complete rotation about its axis once in 20 days, what is its kinetic energy? (Ans: 14ｘ10⁴² J s,  2.5 ｘ10³⁶ J)



Data Given:


Mass of the star =m = 2 ｘ10³⁰ kg
 
Radius = r = 7 ｘ 10⁵ Km = 7 ｘ 10⁸ m

Time Period = T = 20 days = 20 x 24 x 60 x 60 s = 1728000 s = 1.728 ｘ 10⁶ s

To Find:


Angular Momentum  = →L = ?

Kinetic Energy  = K.E = ?

Solution:


Using formula for Angular momentum L in terms of angular velocity  ധ and Moment of inertia I

→L = Iധ  

Where ധ = 2ㅈT and I = 2MR25, Thus

→L = 4ㅈMR25T

By putting values

→L = 4ｘ3.1416ｘ2ｘ10³⁰kgｘ(7ｘ10⁸m)²5ｘ1.728ｘ10⁶s

→L = 25.1328ｘ10³⁰kgｘ49ｘ10¹⁶m²5ｘ1.728ｘ10⁶s

→L = 1,231.5072ｘ10⁴⁶kgm²8.64ｘ10⁶s

→L = 142.535 ｘ10⁴⁰ kg m² s⁻¹  

or

→L = 1.425 ｘ10⁴² J s  -----------Ans.

Now To Find 
Kinetic Energy  = K.E = ? we know that 

K.E = →L = 12 Iധ²}

Where ധ = 2ㅈT ⇒ ധ² = 4ㅈ²T² and I = 2MR25, Thus

K.E = 4ㅈ²MR²5T²

By putting the corresponding values

K.E = 4(3.1416)²ｘ2ｘ10³⁰kgｘ(7ｘ10⁸m)²5ｘ(1.728ｘ10⁶s)²

K.E = 78.95720448ｘ10³⁰kgｘ49ｘ10¹⁶m²5ｘ2.985984ｘ10¹²s² 

K.E = 3,868.90301952ｘ10⁴⁶kgm²14.92992ｘ10¹²s²

K.E = 259.137 ｘ10³⁴ kg m² s⁻² 

or

K.E = 2.59.137 ｘ10³⁶ J ------------Ans. (2)



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