(a) Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of earth due to earth’s rotation (period 1 day). What is acceleration of object on the surface of the earth (at equator) towards its centre ? How does these accelerations compare with g = 9.8 m/s2?
(b) Earth also moves in circular orbit around sun once every year with on orbital radius of 11 1.5 10 × m . What is the acceleration of earth (or any object on the surface of the earth) towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s2?
(a) Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of earth due to earth’s rotation (period 1 day). What is acceleration of object on the surface of the earth (at equator) towards its centre ? How does these accelerations compare with g = 9.8 m/s2?
(b) Earth also moves in circular orbit around sun once every year with on orbital radius of 11 1.5 10 × m . What is the acceleration of earth (or any object on the surface of the earth) towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s2?
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1 Answer
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This is a Short Answer Type Question as classified in NCERT Exemplar
Explanation – a) radius of earth =6400km= 6.4
Time period = 1 day = 24 = 86400s
Centripetal acceleration a= w2r= R(2 )2=4 2R/T
= = 0.034m/s2
b) time = 1yr=365 days= 365=3.15
centripetal acceleration = Rw2=
2
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Please find the solution below:
after 10 kicks,
v? = 3tî v? = 24cos 60°î + 24sin 60°? = 12î + 12√3?
v? = v? – v? = (12 – 3t)î + 12√3?
It is minimum when 12 - 3t = 0 ⇒ t = 4sec
ω = θ² + 2θ
α = (ωdω)/dθ = (θ² + 2θ) (2θ + 2)
At θ = 1rad.
ω = 3rad/s and α = 12rad/s²
a? = αR = 12 m/s² a? = ω²R = 9 m/s² A? = √ (a? ² + a? ²) = 15 m/s²
a? = v? ²/4r
a_A? = (v? ²/r²) × r = v? ²/r
a_A = 3v? ²/4r
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