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

Acceleration,
`a=2.22* 10^(-3)\ m/s^2`

Step-by-step explanation:

It is given that,

Time period of revolution of the moon,
`T=2060* 10^3\ s`

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

The radius of the planet,
`r=3.9* 10^6\ m`

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

`a=R* \omega^2`, R is the radius of orbit

Since,
`\omega=(2\pi)/(T)`

The radius of orbit is,

`R=r+h`

`R=3.9* 10^6\ m+235* 10^6\ m=238900000\ m`

So,
`a=(4\pi^2 R)/(T^2)`

`a=(4\pi^2 * 238900000)/((2060* 10^3)^2)`

`a=2.22* 10^(-3)\ m/s^2`

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