Question

If the arm is 6.75 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 4.80×1022kg and its diameter is 3120 km.

Answer 1

The angular speed is 4.22 rpm.

Step-by-step explanation:

Given that,

Length of arm = 6.75 m

Mass of Europa
`M= 4.80*10^(22)\ kg`

Diameter = 3120 km

We need to calculate the angular speed

Using formula of gravitational force

`F=(GmM)/(R^2)`

`(mv^2)/(r)=(GmM)/(R^2)`

`(v^2)/(r)=(GM)/(R^2)`

`\omega^2=(GM)/(r* R^2)`

Put the value into the formula

`\omega=\sqrt{(6.67*10^(-11)*4.80*10^(22))/(6.75*(1560000)^2)}`

`\omega=0.441\ rad/s`

`\omega=4.22\ rpm`

Hence, The angular speed is 4.22 rpm.