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There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think we should land a vehicle there to search for life. Before launching it, we would want to test such a lander under the gravity conditions at the surface of Europa. One way to do this is to put the lander at the end of a rotating arm in an orbiting earth satellite.
If the arm is 6.00 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 4.8x10^22 kg and its diameter is 3138 km .

Respuesta :

boffeemadrid

Answer:

4.44 rpm

Explanation:

[tex]\omega[/tex] = Angular speed

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Europa = [tex]\frac{3138000}{2}\ m[/tex]

R = Radius of arm = 6 m

The acceleration due to gravity is given by

[tex]g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 4.8\times 10^{22}}{\left(\frac{3138000}{2}\right)^2}\\\Rightarrow g=1.3\ m/s^2[/tex]

Here the centripetal acceleration of the arm and acceleration due to gravity are equal

[tex]a_c=\omega^2R[/tex]

[tex]a_c=g\\\Rightarrow \omega^2R=1.3\\\Rightarrow \omega^2\times 6=1.3\\\Rightarrow \omega=\sqrt{\frac{1.3}{6}}\\\Rightarrow \omega=0.46547\ rad/s[/tex]

Converting to rpm

[tex]1\ rad/s=\frac{60}{2\pi}\ rpm[/tex]

[tex]0.46547\ rad/s=0.46547\times \frac{60}{2\pi}\ rpm=4.44\ rpm[/tex]

The angular speed of the arm is 4.44 rpm

onyebuchinnaji

The angular speed of the lander at the end of the rotating arm is 4.5 rpm.

The given parameters;

  • radius of the arm, r = 6 m
  • mass of the Europa, m = 4.8 x 10²² kg
  • diameter, D = 3138 km = 3,138,000 m
  • Radius of the Europa, R = 1,569,000 m

The acceleration due to gravity of the Europa is calculated as follows;

[tex]F = Mg = \frac{GmM}{R^2} \\\\g = \frac{Gm}{R^2} \\\\g = \frac{(6.67 \times 10^{-11}) \times (4.8\times 10^{22})}{(1,569,000)^2} \\\\g = 1.3 \ m/s^2[/tex]

The centripetal acceleration of the lander =  acceleration due to gravity

[tex]a_c = g\\\\a_c = 1.3 \ m/s^2[/tex]

The angular speed of the lander is calculated as follows;

[tex]a_c = \omega^2 r\\\\\omega^2 = \frac{a_c}{r} \\\\\omega = \sqrt{\frac{a_c}{r}} \\\\\omega = \sqrt{\frac{1.3}{6}} \\\\\omega = 0.466 \ rad/s\\\\\omega = 0.466 \ \frac{rad}{s} \ \times \frac{1 \ rev}{2 \pi \ rad} \times \frac{60 \ s }{1 \min} = 4.5 \ rpm[/tex]

Thus, the angular speed of the lander at the end of the rotating arm is 4.5 rpm.

Learn more here:https://brainly.com/question/14453709

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