Question

An undiscovered planet, many lightyears from Earth, has one moon in a periodic orbit. This moon takes 2060 × 103 seconds (about 24 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 235.0 × 106 m and the planet has a radius of 3.90 × 106 m, calculate the moon's radial acceleration ????c .

Answer 1

Answer:

Acceleration, $a=2.22\times 10^{-3}\ m/s^2$

Explanation:

It is given that,

Time period of revolution of the moon, $T=2060\times 10^3\ s$

If the distance from the center of the moon to the surface of the planet is, $h=235\times 10^6\ m$

The radius of the planet, $r=3.9\times 10^6\ m$

Let a is the moon's radial acceleration. Mathematically, it is given by :

$a=R\times \omega^2$, R is the radius of orbit

Since, $\omega=\dfrac{2\pi}{T}$

The radius of orbit is,

$R=r+h$

$R=3.9\times 10^6\ m+235\times 10^6\ m=238900000\ m$

So, $a=\dfrac{4\pi^2 R}{T^2}$

$a=\dfrac{4\pi^2 \times 238900000}{(2060\times 10^3)^2}$

$a=2.22\times 10^{-3}\ m/s^2$

Hence, this is the required solution for the radial acceleration of the moon.