Question

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

Answer 1

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

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

      $a_r = 3.81 *10^{-3} m/s^2$