Manipal MET2013PhysicsGravitation
The speed with which the earth have to rotate on its axis so that a person on the equator would weight (3 / 5)^ th as much as present. [Radius of earth =6400 ~km ]
Options
- A4.83 10⁻³ rads ⁻¹
- B5.41 10⁻³ rads ⁻¹
- C7.82 10⁻⁴ rads ⁻¹
- D8.88 10⁻¹⁴ rads ⁻¹
Correct answer
C. 7.82 10⁻⁴ rads ⁻¹
Step-by-step solution
The apparent weight of person on the equator (latitude =0 ) is given by ^ =w-m R ^2 Here, ^ =(3 / 5) w =(3 / 5) mg [ =m g] or m R ^2=m g-(3 / 5) m g= ( 2 5 ) m g or ^2= 2 g 5 R = 2 g 5 R Here, g=9.8 ~ms ⁻² and R=6400 ~km =6400 10^3 ~m = ( 2 5 9.8 6400 10^3 ) rads ⁻¹=7.82 10⁻⁴ rads ⁻¹