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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 surface 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?

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lcmendozaf

Answer:

[tex]F = 3.514 * 10^{19}N[/tex]

Explanation:

We know that torque exerted by superman is given by:

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

The inertia of a solid sphere is calculated as:

[tex]I_t = \frac{2}{5}*m_t*R_t^2[/tex]

Earth's acceleration is:

[tex]\omega_f = \omega_o + \alpha_t * t[/tex]

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

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

Solving for the acceleration and replacing the values:

[tex]\alpha_t = \frac{\omega_f - \omega_o}{t}[/tex] Replacing this value on the torque equation:

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

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