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4 years ago

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Suppose that Superman wants to stop Earth so it does not rotate. He exerts a force on Earth at Earth's equator tangent to its su

rface for a time interval of 1 year. The mass of Earth is 5.98�1024 kg, the radius is 6.37�106 m. Assume that Earth is a solid sphere with its mass distributed uniformly.Part AWhat magnitude force must he exert to stop Earth's rotation?

1 answer:

MatroZZZ [7]4 years ago

6 0

Answer:

$F = 3.514 * 10^{19}N$

Explanation:

We know that torque exerted by superman is given by:

$T = F_s * R_t = I_t * \alpha_t$ where It is earth's inertia, and αt is earth acceleration.

The inertia of a solid sphere is calculated as:

$I_t = \frac{2}{5}*m_t*R_t^2$

Earth's acceleration is:

$\omega_f = \omega_o + \alpha_t * t$

Where t is the lapse of 1 year. t = 365*24*3600 = 31536000s

$\omega_f = 0$ $\omega_o = \frac{2*\pi}{24h * 3600s/h}$

Solving for the acceleration and replacing the values:

$\alpha_t = \frac{\omega_f - \omega_o}{t}$ Replacing this value on the torque equation:

$F_s = \frac{2}{5} m_t * R_t*\frac{2\pi}{365*(24*3600)^2} =3.514*10^{19}N$

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