Question

If the arm is 4.25 m long and pivots about one end, at what angular speed (in rpm) should it spin so that the acceleration of the lander is the same as the acceleration due to gravity at the surface of Europa? The mass of Europa is 4.8×1022kg and its diameter is 3138 km.

Answer 1

To solve this problem it is necessary to apply the concepts related to the acceleration of gravity due to the force exerted by a start and the calculation of angular velocity as a function of acceleration and radius.

By definition we know that the acceleration exerted by the celestial body is given under the equation

`g = (GM)/(R^2)`

Where,

G = Gravitational Universal Constant

M = Mass

R = Radius

The radius of Europa is

`R = (D)/(2) = (31838*10^3)/(2)`

`R  = 1569*10^3m`

Applying the gravitational equation,

`g = (GM)/(R^2)`

`g = ((6.67*10^(-11))(4.8*10^(22)))/((1569*10^3)^2)`

`g = 1.3m/s^2`

Therefore the angular acceleration can be obtained through the kinematic equation

`a = r\omega^2`

Where,

a = acceleration

r = length of the arm

`\omega =` Angular acceleration

As a = g then,

`g = r\omega^2`

Where,

`\omega = \sqrt{(g)/(r)}`

`\omega = \sqrt{(1.3)/(4.25)}`

`\omega = 0.553rad/s`

Therefore the angular speed of arm is 0.553rad/s