Respuesta :
Answer:
a) τ = 4.47746 * 10^25 N-m
b) E = 2.06301 * 10^13 J
c) P = 3.25511*10^21 W
Explanation:
Given:
- The radius of earth r = 6.3781×10^6 m
- The angular speed of earth w = 7.27*10^-5 rad/s
- The time taken to reach above speed t = 5 yrs = 1.57784760 * 10^8 s
- The mass of earth m = 5.972 × 10^24 kg
- The inertia of sphere I = 2/5 * m* r^2
Solution:
- The angular acceleration of the earth from rest to w is given by α:
α = w / t
α = (7.27*10^-5) / (1.57784760 * 10^8)
α = 4.60754*10^-13 rad/s^2
- The required torque τ is given by:
τ = I*α
τ = 2/5 * m* r^2 * α
τ = 2/5 *(5.972 × 10^24) * (6.3781×10^6)^2 * (4.60754*10^-13)
τ = 4.47746 * 10^25 N-m
- The power required P to turn the earth to the speed w is:
P = τ*w
P = (4.47746 * 10^25)*(7.27*10^-5)
P = 3.25511*10^21 W
- The energy E required is :
E = P / t
E = (3.25511*10^21) / (1.57784760 * 10^8)
E = 2.06301 * 10^13 J
(a) The torque required is 4.47×10²⁵ N-m
(b) The energy required is 2.06×10¹³ J
(c) And the power required is 3.25×10²¹ W
The radius of earth r = 6.3781×10⁶ m
The angular speed of earth v = 7.27×10⁻⁵ rad/s
The time taken to reach the given speed is t = 5 yrs = 1.577 ×10⁸ s
The mass of earth m = 5.972 × 10²⁴ kg
Torque:
(1) The angular acceleration of the earth:
[tex]\alpha = v / t\\\\\alpha= \frac{7.27\times10^-5} {1.577 \times 10^8}\\\\\alpha= 4.6\times10^{-13} rad/s^2[/tex]
The torque of is given by:
τ = Iα
where I is the inertia of the sphere [tex]I = \frac{2}{5} mr^2[/tex]
[tex]\tau = \frac{2}{5} (5.972 \times 10^24) \times(6.3781\times10^6)^2 \times (4.60754*10^-{13})\\\\\tau = 4.47 \times 10^{25} Nm[/tex]
(2) The power required to turn the earth to the speed v is:
P = τv
P = (4.47 × 10²⁵)(7.27×10⁻⁵)
P = 3.25×10²¹ W
(3) The energy E required is given by:
[tex]E = \frac{P}{t}\\\\E = \frac{3.25\times10^{21}} {1.57784760 \times 10^8}\\\\E = 2.06 \times 10^{13} J[/tex]
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