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il63 [147K]
3 years ago
9

An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 17601760 × 103 seconds (ab

out 2.0×1012.0×101 days) on average to complete one nearly circular revolution around the unnamed planet. If the distance from the center of the moon to the surface of the planet is 295.0295.0 × 106 m and the planet has a radius of 3.603.60 × 106 m, calculate the moon's radial acceleration acac .
Physics
1 answer:
Yuliya22 [10]3 years ago
4 0

Answer:

The value is a_r  = 3.81 *10^{-3} m/s^2

Explanation:

Generally the moon's radial acceleration is mathematically represented as

a_r  =  r *  w^2

Here w is the angular velocity which is mathematically represented as

w =\frac{2 \pi }{ T}

substituting 1760 * 10^3 \ seconds for T(i.e the period of the moon ) we have

w =\frac{2  * 3.142 }{  1760 * 10^3}

=> w = 3.57 *10^{-6} \  rad/s

From the question r(which is the radius of the orbit ) is evaluated as

r =  R + H

substitute 3.60 * 10^6 m for R and 295.0 * 10^6  \  m H

       r =  295.0 * 10^6   +3.60 * 10^6

=>   r =  2.986 *10^{8} \  m

So

    a_r  =   2.986 *10^{8} *  ( 3.57 *10^{-6} )^2

      a_r  = 3.81 *10^{-3} m/s^2

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4.7\mu m

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3 years ago
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5.24 m/s

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So for the rod to be able to rise upward to the straight up position, the kinetic energy caused by linear speed v0 must be just enough to convert into the potential energy.

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