Question

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?

Answer 1

`F = 3.514 * 10^(19)N`

Step-by-step 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 = (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 = (2*\pi)/(24h * 3600s/h)`

Solving for the acceleration and replacing the values:

`\alpha_t = (\omega_f - \omega_o)/(t)` Replacing this value on the torque equation:

`F_s = (2)/(5) m_t * R_t*(2\pi)/(365*(24*3600)^2) =3.514*10^(19)N`