A particle would take a time t to move down a straight tunnel from the surface of earth (supposed to be a homogeneous sphere) to its centre. If gravity were to remain constant this time would be t'. The ratio of t'/t will be
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Explanation:
Given A particle would take a time t to move down a straight tunnel from the surface of earth (supposed to be a homogeneous sphere) to its centre. If gravity were to remain constant this time would be t'. The ratio of The ratio of t'/t will be
- Now there is a particle. So the particle moves through the tunnel and falls under the earth and the force is applied towards the centre of the earth. So the particle stops at the centre and the acceleration due to gravity at the centre is zero. This acceleration due to gravity varies at times and so g is given by
- g’ = g (1 – d/R)
- So d is the depth from the surface of earth. So d varies with time along with g.
- Suppose acceleration due to gravity is constant, so time will be t’
- So we need to find the ratio of t / t’.
- Let a particle be x distance from the centre of earth.
- So we have g’ = g(R – d / R)
- So g’ = g(x) / R ------1
- Now g’ = f(x)
- Now if the particle is slightly shifted upwards, then there will be a downward force and particle is attracted towards centre ,acceleration due to gravity will be towards the earth and so similarly in the case of particle shifted slightly downwards. So simple harmonic motion is being executed.
- So a = - ω^2 x ----------2
- Comparing 1 and 2 we have
- ω^2 = g / R
- So ω = √g / R
- Now 2π / T = √g / R
- So T = 2π√R / g (Formula for Time period for shm through tunnel of earth)
- So t = T / 4 ( since 4 times the particle moves towards centre)
- t = π / 2 √R / g -------------1
- When acceleration due to gravity is constant we have
- s = ut + ½ at^2
- R = 0 + ½ g t’^2
- So t’ = √2R / g -------------2
- So t / t’ = π /2 √R / g / √2 R / g
- t / t’ = π / 2√2
- So t’ / t = 2√ 2 / π
Reference link will be
https://brainly.in/question/7503580
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